Unformatted text preview: 5E16(pp 10161025) 1/18/06 4:13 PM Page 1016 CHAPTER 16
If we approximate a solid by rectangular columns
and let the number of columns increase, the
limit of sums of volumes of columns is the volume
of the solid. Multiple Integrals 5E16(pp 10161025) 1/18/06 4:13 PM Page 1017 In this chapter we extend the idea of a deﬁnite integral to double
and triple integrals of functions of two or three variables. These
ideas are then used to compute volumes, surface areas, masses,
and centroids of more general regions than we were able to consider in Chapters 6 and 9. We also use double integrals to calculate probabilities when two random variables are involved.  16.1 Double Integrals over Rectangles
In much the same way that our attempt to solve the area problem led to the deﬁnition of a
deﬁnite integral, we now seek to ﬁnd the volume of a solid and in the process we arrive at
the deﬁnition of a double integral. Review of the Definite Integral
First let’s recall the basic facts concerning deﬁnite integrals of functions of a single variable. If f ͑x͒ is deﬁned for a ഛ x ഛ b, we start by dividing the interval ͓a, b͔ into n subintervals ͓x iϪ1, x i ͔ of equal width ⌬x ͑b Ϫ a͒͞n and we choose sample points x* in these
i
subintervals. Then we form the Riemann sum
n ͚ f ͑x*͒ ⌬x 1 i i1 and take the limit of such sums as n l ϱ to obtain the deﬁnite integral of f from a to b : y 2 b a n f ͑x͒ dx lim ͚ f ͑x*͒ ⌬x
i n l ϱ i1 In the special case where f ͑x͒ ജ 0, the Riemann sum can be interpreted as the sum of the
areas of the approximating rectangles in Figure 1, and xab f ͑x͒ dx represents the area under
the curve y f ͑x͒ from a to b.
y Îx f(x*)
i 0 FIGURE 1 a
x*
¡ ⁄ ¤
x*
™ ‹ xi1 x*
£ xi
x*
i b xn1 x *
xn Volumes and Double Integrals
In a similar manner we consider a function f of two variables deﬁned on a closed rectangle Խ R ͓a, b͔ ϫ ͓c, d͔ ͕͑x, y͒ ʦ 2 ޒa ഛ x ഛ b, c ഛ y ഛ d͖
1017 5E16(pp 10161025) 1018 ❙❙❙❙ 1/18/06 4:14 PM Page 1018 CHAPTER 16 MULTIPLE INTEGRALS z and we ﬁrst suppose that f ͑x, y͒ ജ 0. The graph of f is a surface with equation z f ͑x, y͒.
Let S be the solid that lies above R and under the graph of f, that is, z=f(x, y) Խ S ͕͑x, y, z͒ ʦ 0 3 ޒഛ z ഛ f ͑x, y͒, ͑x, y͒ ʦ R͖
0 c a (See Figure 2.) Our goal is to ﬁnd the volume of S.
The ﬁrst step is to divide the rectangle R into subrectangles. We do this by dividing the
interval ͓a, b͔ into m subintervals ͓x iϪ1, x i ͔ of equal width ⌬x ͑b Ϫ a͒͞m and dividing
͓c, d͔ into n subintervals ͓yjϪ1, yj ͔ of equal width ⌬y ͑d Ϫ c͒͞n. By drawing lines parallel to the coordinate axes through the endpoints of these subintervals as in Figure 3, we
form the subrectangles d
y b x R FIGURE 2 Խ Rij ͓x iϪ1, x i ͔ ϫ ͓yjϪ1, yj ͔ ͕͑x, y͒ x iϪ1 ഛ x ഛ x i, yjϪ1 ഛ y ഛ yj͖
each with area ⌬A ⌬x ⌬y.
y R ij d (xi, yj) (x * , y* )
ij ij yj Îy yj_1
›
c
(x * *
£™, y £™) FIGURE 3 0 a ⁄ ¤ xi_1 xi Dividing R into subrectangles b x Îx * *
If we choose a sample point ͑x ij , y ij ͒ in each Rij , then we can approximate the part of
S that lies above each Rij by a thin rectangular box (or “column”) with base Rij and height
* *
f ͑x ij , yij ͒ as shown in Figure 4. (Compare with Figure 1.) The volume of this box is the
height of the box times the area of the base rectangle:
* *
f ͑x ij , yij ͒ ⌬A
z z 0 c a 0 f(x * , y* ) ij ij
d y y
b
x x Rij
FIGURE 4 FIGURE 5 5E16(pp 10161025) 1/18/06 4:14 PM Page 1019 SECTION 16.1 DOUBLE INTEGRALS OVER RECTANGLES ❙❙❙❙ 1019 If we follow this procedure for all the rectangles and add the volumes of the corresponding boxes, we get an approximation to the total volume of S:
m VϷ 3 n ͚ ͚ f ͑x *, y *͒ ⌬A
ij ij i1 j1 (See Figure 5.) This double sum means that for each subrectangle we evaluate f at the chosen point and multiply by the area of the subrectangle, and then we add the results.
Our intuition tells us that the approximation given in (3) becomes better as m and n
become larger and so we would expect that
 The meaning of the double limit in Equation
4 is that we can make the double sum as close
as we like to the number V [for any choice of
* *
͑x ij , yij ͒] by taking m and n sufﬁciently large. m n ͚ ͚ f ͑x *, y *͒ ⌬A V lim 4 ij m, n l ϱ i1 j1 ij We use the expression in Equation 4 to deﬁne the volume of the solid S that lies under the
graph of f and above the rectangle R. (It can be shown that this deﬁnition is consistent with
our formula for volume in Section 6.2.)
Limits of the type that appear in Equation 4 occur frequently, not just in ﬁnding volumes but in a variety of other situations as well—as we will see in Section 16.5—even
when f is not a positive function. So we make the following deﬁnition.
5 Definition The double integral of f over the rectangle R is
 Notice the similarity between Deﬁnition 5
and the deﬁnition of a single integral in
Equation 2. m yy f ͑x, y͒ dA
R n ͚ ͚ f ͑x *, y *͒ ⌬A lim ij m, n l ϱ i1 j1 ij if this limit exists.
The precise meaning of the limit in Deﬁnition 5 is that for every number Ͼ 0 there is
an integer N such that Ϳ yy m f ͑x, y͒ dA Ϫ n ͚ ͚ f ͑x *, y *͒ ⌬A
ij ij i1 j1 R Ϳ Ͻ * *
for all integers m and n greater than N and for any choice of sample points ͑x ij , yij ͒ in Rij.
It can be proved that the limit in Deﬁnition 5 exists if f is a continuous function. (It also
exists for some discontinuous functions as long as they are reasonably “well behaved.”)
* *
The sample point ͑x ij , yij ͒ can be chosen to be any point in the subrectangle Rij , but if
we choose it to be the upper righthand corner of Rij [namely ͑x i, yj ͒, see Figure 3], then
the expression for the double integral looks simpler:
m 6 yy f ͑x, y͒ dA
R lim n ͚ ͚ f ͑x , y ͒ ⌬A m, n l ϱ i1 j1 i j By comparing Deﬁnitions 4 and 5, we see that a volume can be written as a double
integral:
If f ͑x, y͒ ജ 0, then the volume V of the solid that lies above the rectangle R and
below the surface z f ͑x, y͒ is
V yy f ͑x, y͒ dA
R 5E16(pp 10161025) 1020 ❙❙❙❙ 1/18/06 4:14 PM Page 1020 CHAPTER 16 MULTIPLE INTEGRALS The sum in Deﬁnition 5,
m n ͚ ͚ f ͑x *, y *͒ ⌬A
ij ij i1 j1 is called a double Riemann sum and is used as an approximation to the value of the
double integral. [Notice how similar it is to the Riemann sum in (1) for a function of a
single variable.] If f happens to be a positive function, then the double Riemann sum
represents the sum of volumes of columns, as in Figure 5, and is an approximation to the
volume under the graph of f .
y (1, 2) R¡™
1 R™™
(2, 1) (1, 1) R¡¡
0 EXAMPLE 1 Estimate the volume of the solid that lies above the square R ͓0, 2͔ ϫ ͓0, 2͔
and below the elliptic paraboloid z 16 Ϫ x 2 Ϫ 2y 2. Divide R into four equal squares
and choose the sample point to be the upper right corner of each square Rij . Sketch the
solid and the approximating rectangular boxes. (2, 2) 2 SOLUTION The squares are shown in Figure 6. The paraboloid is the graph of R™¡
1 f ͑x, y͒ 16 Ϫ x 2 Ϫ 2y 2 and the area of each square is 1. Approximating the volume
by the Riemann sum with m n 2, we have x 2 2 VϷ FIGURE 6 2 ͚ ͚ f ͑x , y ͒ ⌬A
i j i1 j1 f ͑1, 1͒ ⌬A ϩ f ͑1, 2͒ ⌬A ϩ f ͑2, 1͒ ⌬A ϩ f ͑2, 2͒ ⌬A z
16 z=16≈2¥ 13͑1͒ ϩ 7͑1͒ ϩ 10͑1͒ ϩ 4͑1͒ 34
This is the volume of the approximating rectangular boxes shown in Figure 7.
We get better approximations to the volume in Example 1 if we increase the number of
squares. Figure 8 shows how the columns start to look more like the actual solid and the
corresponding approximations become more accurate when we use 16, 64, and 256
squares. In the next section we will be able to show that the exact volume is 48. 2
2 y x FIGURE 7 FIGURE 8 The Riemann sum approximations to
the volume under z=16≈2¥
become more accurate as m and
n increase. (a) m=n=4, VÅ41.5 (b) m=n=8, VÅ44.875 Խ (c) m=n=16, VÅ46.46875 EXAMPLE 2 If R ͕͑x, y͒ Ϫ1 ഛ x ഛ 1, Ϫ2 ഛ y ഛ 2͖, evaluate the integral yy s1 Ϫ x
R 2 dA 5E16(pp 10161025) 1/18/06 4:15 PM Page 1021 SECTION 16.1 DOUBLE INTEGRALS OVER RECTANGLES z S (1, 0, 0) 1021 SOLUTION It would be very difﬁcult to evaluate this integral directly from Deﬁnition 5 but,
because s1 Ϫ x 2 ജ 0, we can compute the integral by interpreting it as a volume. If
z s1 Ϫ x 2, then x 2 ϩ z 2 1 and z ജ 0, so the given double integral represents the
volume of the solid S that lies below the circular cylinder x 2 ϩ z 2 1 and above the
rectangle R. (See Figure 9.) The volume of S is the area of a semicircle with radius 1
times the length of the cylinder. Thus (0, 0, 1) x ❙❙❙❙ (0, 2, 0) y yy s1 Ϫ x FIGURE 9 2 dA 1 ͑1͒2 ϫ 4 2
2 R The Midpoint Rule
The methods that we used for approximating single integrals (the Midpoint Rule, the
Trapezoidal Rule, Simpson’s Rule) all have counterparts for double integrals. Here we
consider only the Midpoint Rule for double integrals. This means that we use a double
* *
Riemann sum to approximate the double integral, where the sample point ͑x ij , yij ͒ in Rij is
chosen to be the center ͑xi , yj͒ of Rij. In other words, xi is the midpoint of ͓x iϪ1, x i ͔ and yj
is the midpoint of ͓yjϪ1, yj ͔.
Midpoint Rule for Double Integrals
m n yy f ͑x, y͒ dA Ϸ ͚ ͚ f ͑x , y ͒ ⌬A
i j i1 j1 R where xi is the midpoint of ͓x iϪ1, x i ͔ and yj is the midpoint of ͓yjϪ1, yj ͔.
EXAMPLE 3 Use the Midpoint Rule with m n 2 to estimate the value of the integral xxR ͑x Ϫ 3y 2 ͒ dA, where R ͕͑x, y͒ Խ 0 ഛ x ഛ 2, 1 ഛ y ഛ 2͖. SOLUTION In using the Midpoint Rule with m n 2, we evaluate f ͑x, y͒ x Ϫ 3y 2 y
(2, 2) 2
3
2 at the centers of the four subrectangles shown in Figure 10. So x1 1 , x2 3 , y1 5 ,
2
2
4
and y2 7 . The area of each subrectangle is ⌬A 1 . Thus
4
2 R¡™ R™™ R¡¡ R™¡ 2 yy ͑x Ϫ 3y 1 2 ͒ dA Ϸ 2 ͚ ͚ f ͑x , y ͒ ⌬A
i j i1 j1 R f ͑x1, y1͒ ⌬A ϩ f ͑x1, y2 ͒ ⌬A ϩ f ͑x2 , y1 ͒ ⌬A ϩ f ͑x2 , y2 ͒ ⌬A
0 1 2 f ( 1 , 5 ) ⌬A ϩ f ( 1 , 7 ) ⌬A ϩ f ( 3 , 5 ) ⌬A ϩ f ( 3 , 7 ) ⌬A
2 4
2 4
2 4
2 4 x (Ϫ 67 ) 1 ϩ (Ϫ 139 ) 1 ϩ (Ϫ 51) 1 ϩ (Ϫ 123 ) 1
16 2
16 2
16 2
16 2 FIGURE 10 Ϫ 95 Ϫ11.875
8
Thus, we have yy ͑x Ϫ 3y 2 ͒ dA Ϸ Ϫ11.875 R NOTE
In the next section we will develop an efﬁcient method for computing double
integrals and then we will see that the exact value of the double integral in Example 3 is
Ϫ12. (Remember that the interpretation of a double integral as a volume is valid only when
the integrand f is a positive function. The integrand in Example 3 is not a positive function, so its integral is not a volume. In Examples 2 and 3 in Section 16.2 we will discuss
how to interpret integrals of functions that are not always positive in terms of volumes.) If
we keep dividing each subrectangle in Figure 10 into four smaller ones with similar shape,
■ 5E16(pp 10161025) 1022 ❙❙❙❙ 1/18/06 4:15 PM Page 1022 CHAPTER 16 MULTIPLE INTEGRALS Number of
subrectangles Midpoint Rule
approximations 1
4
16
64
256
1024 Ϫ11.5000
Ϫ11.8750
Ϫ11.9687
Ϫ11.9922
Ϫ11.9980
Ϫ11.9995 we get the Midpoint Rule approximations displayed in the chart in the margin. Notice how
these approximations approach the exact value of the double integral, Ϫ12. Average Value
Recall from Section 6.5 that the average value of a function f of one variable deﬁned on
an interval ͓a, b͔ is
1
b
fave
ya f ͑x͒ dx
bϪa
In a similar fashion we deﬁne the average value of a function f of two variables deﬁned
on a rectangle R to be
fave 1
A͑R͒ yy f ͑x, y͒ dA
R where A͑R͒ is the area of R.
If f ͑x, y͒ ജ 0, the equation
A͑R͒ ϫ fave yy f ͑x, y͒ dA
R says that the box with base R and height fave has the same volume as the solid that lies
under the graph of f . [If z f ͑x, y͒ describes a mountainous region and you chop off the
tops of the mountains at height fave, then you can use them to ﬁll in the valleys so that the
region becomes completely ﬂat. See Figure 11.] FIGURE 11 EXAMPLE 4 The contour map in Figure 12 shows the snowfall, in inches, that fell on the state of Colorado on December 24, 1982. (The state is in the shape of a rectangle that
measures 388 mi west to east and 276 mi south to north.) Use the contour map to estimate the average snowfall for Colorado as a whole on December 24. 24
22 18
16 14
12 10
8
6
4
0 FIGURE 12 2 20 5E16(pp 10161025) 1/18/06 4:15 PM Page 1023 SECTION 16.1 DOUBLE INTEGRALS OVER RECTANGLES ❙❙❙❙ 1023 SOLUTION Let’s place the origin at the southwest corner of the state. Then 0 ഛ x ഛ 388, 0 ഛ y ഛ 276, and f ͑x, y͒ is the snowfall, in inches, at a location x miles to the east and
y miles to the north of the origin. If R is the rectangle that represents Colorado, then the
average snowfall for the state on December 24 was
fave 1
A͑R͒ yy f ͑x, y͒ dA
R where A͑R͒ 388 ؒ 276. To estimate the value of this double integral let’s use the Midpoint Rule with m n 4. In other words, we divide R into 16 subrectangles of equal
size, as in Figure 13. The area of each subrectangle is
1
⌬A 16 ͑388͒͑276͒ 6693 mi2 y
276 24
22 18
16 20 14 10 12
8
6
4
2 0 0 388 x FIGURE 13 Using the contour map to estimate the value of f at the center of each subrectangle,
we get
4 4 yy f ͑x, y͒ dA Ϸ ͚ ͚ f ͑x , y ͒ ⌬A
i R j i1 j1 Ϸ ⌬A͓0.4 ϩ 1.2 ϩ 1.8 ϩ 3.9 ϩ 0 ϩ 3.9 ϩ 4.0 ϩ 6.5
ϩ 0.1 ϩ 6.1 ϩ 16.5 ϩ 8.8 ϩ 1.8 ϩ 8.0 ϩ 16.2 ϩ 9.4͔
͑6693͒͑88.6͒
͑6693͒͑88.6͒
Ϸ 5.5
͑388͒͑276͒
On December 24, 1982, Colorado received an average of approximately 5 1 inches of
2
snow.
Therefore fave Ϸ 5E16(pp 10161025) 1024 ❙❙❙❙ 1/18/06 4:16 PM Page 1024 CHAPTER 16 MULTIPLE INTEGRALS Properties of Double Integrals
We list here three properties of double integrals that can be proved in the same manner as
in Section 5.2. We assume that all of the integrals exist. Properties 7 and 8 are referred to
as the linearity of the integral.
7 yy ͓ f ͑x, y͒ ϩ t͑x, y͔͒ dA yy f ͑x, y͒ dA ϩ yy t͑x, y͒ dA
R  Double integrals behave this way because
the double sums that deﬁne them behave
this way. 8 R R yy c f ͑x, y͒ dA c yy f ͑x, y͒ dA
R where c is a constant R If f ͑x, y͒ ജ t͑x, y͒ for all ͑x, y͒ in R, then 9 yy f ͑x, y͒ dA ജ yy t͑x, y͒ dA R  16.1 R Exercises 1. (a) Estimate the volume of the solid that lies below the surface z xy and above the rectangle
R ͕͑x, y͒ 0 ഛ x ഛ 6, 0 ഛ y ഛ 4͖. Use a Riemann sum
with m 3, n 2, and take the sample point to be the
upper right corner of each subrectangle.
(b) Use the Midpoint Rule to estimate the volume of the solid
in part (a). Խ 2. If R ͓Ϫ1, 3͔ ϫ ͓0, 2͔, use a Riemann sum with m 4, n 2 to estimate the value of xxR ͑y Ϫ 2x ͒ dA. Take the
sample points to be the upper left corners of the subrectangles.
2 2 3. (a) Use a Riemann sum with m n 2 to estimate the value of xxR sin͑x ϩ y͒ dA, where R ͓0, ͔ ϫ ͓0, ͔. Take the
sample points to be lower left corners.
(b) Use the Midpoint Rule to estimate the integral in part (a). 1 2 3 4 1.0 2 0 Ϫ3 Ϫ6 Ϫ5 1.5 3 1 Ϫ4 Ϫ8 Ϫ6 2.0 4 3 0 Ϫ5 Ϫ8 2.5 5 5 3 Ϫ1 Ϫ4 3.0 7 8 6 3 0 6. A 20ftby30ft swimming pool is ﬁlled with water. The depth is measured at 5ft intervals, starting at one corner of the pool,
and the values are recorded in the table. Estimate the volume of
water in the pool. 4. (a) Estimate the volume of the solid that lies below the surface z x ϩ 2y 2 and above the rectangle R ͓0, 2͔ ϫ ͓0, 4͔.
Use a Riemann sum with m n 2 and choose the
sample points to be lower right corners.
(b) Use the Midpoint Rule to estimate the volume in part (a).
5. A table of values is given for a function f ͑x, y͒ deﬁned on R ͓1, 3͔ ϫ ͓0, 4͔.
(a) Estimate xxR f ͑x, y͒ dA using the Midpoint Rule with
m n 2.
(b) Estimate the double integral with m n 4 by choosing
the sample points to be the points farthest from the origin. y 0 x 0
0
5
10
15
20 5 10 15 20 25 30 2
2
2
2
2 3
3
4
3
2 4
4
6
4
2 6
7
8
5
2 7
8
10
6
3 8
10
12
8
4 8
8
10
7
4 7. Let V be the volume of the solid that lies under the graph of f ͑x, y͒ s52 Ϫ x 2 Ϫ y 2 and above the rectangle given by
2 ഛ x ഛ 4, 2 ഛ y ഛ 6. We use the lines x 3 and y 4 to 5E16(pp 10161025) 1/18/06 4:16 PM Page 1025 SECTION 16.2 ITERATED INTEGRALS divide R into subrectangles. Let L and U be the Riemann sums
computed using lower left corners and upper right corners,
respectively. Without calculating the numbers V, L , and U,
arrange them in increasing order and explain your reasoning. ❙❙❙❙ 1025 54
58
62 8. The ﬁgure shows level curves of a function f in the square R ͓0, 1͔ ϫ ͓0, 1͔. Use them to estimate xxR f ͑x, y͒ dA to the
nearest integer. 66 y
68 1
14
13 12 74 76 70
70 11 68 10 74 9 0 1 x 9. A contour map is shown for a function f on the square R ͓0, 4͔ ϫ ͓0, 4͔.
(a) Use the Midpoint Rule with m n 2 to estimate the
value of xxR f ͑x, y͒ dA.
(b) Estimate the average value of f .
y 11–13  Evaluate the double integral by ﬁrst identifying it as the
volume of a solid. 11.
12.
13. 4 ■ 10 0 0 10 20 30 xxR 3 dA, R ͕͑x, y͒ Խ Ϫ2 ഛ x ഛ 2, 1 ഛ y ഛ 6͖
xxR ͑5 Ϫ x͒ dA, R ͕͑x, y͒ Խ 0 ഛ x ഛ 5, 0 ഛ y ഛ 3͖
xxR ͑4 Ϫ 2y͒ dA, R ͓0, 1͔ ϫ ͓0, 1͔
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 14. The integral xxR s9 Ϫ y dA, where R ͓0, 4͔ ϫ ͓0, 2͔,
2 represents the volume of a solid. Sketch the solid.
15. Use a programmable calculator or computer (or the sum 2 command on a CAS) to estimate 10 Ϫx 2Ϫy 2 20
30 0 yy e
R 2 4 x 10. The contour map shows the temperature, in degrees Fahrenheit, at 3:00 P.M. on May 1, 1996, in Colorado. (The state measures
388 mi east to west and 276 mi north to south.) Use the Midpoint Rule with m n 4 to estimate the average temperature
in Colorado at that time.  16.2 dA where R ͓0, 1͔ ϫ ͓0, 1͔. Use the Midpoint Rule with the
following numbers of squares of equal size: 1, 4, 16, 64, 256,
and 1024.
16. Repeat Exercise 15 for the integral xxR cos͑x 4 ϩ y 4 ͒ dA.
17. If f is a constant function, f ͑x, y͒ k, and R ͓a, b͔ ϫ ͓c, d͔, show that xxR k dA k͑b Ϫ a͒͑d Ϫ c͒. 18. If R ͓0, 1͔ ϫ ͓0, 1͔, show that 0 ഛ xxR sin͑x ϩ y͒ dA ഛ 1. Iterated Integrals
Recall that it is usually difﬁcult to evaluate single integrals directly from the deﬁnition of
an integral, but the Fundamental Theorem of Calculus provides a much easier method. The
evaluation of double integrals from ﬁrst principles is even more difﬁcult, but in this section we see how to express a double integral as an iterated integral, which can then be evaluated by calculating two single integrals. 5E16(pp 10261035) 1026 ❙❙❙❙ 1/18/06 4:54 PM Page 1026 CHAPTER 16 MULTIPLE INTEGRALS Suppose that f is a function of two variables that is continuous on the rectangle
R ͓a, b͔ ϫ ͓c, d͔. We use the notation xcd f ͑x, y͒ dy to mean that x is held ﬁxed and
f ͑x, y͒ is integrated with respect to y from y c to y d. This procedure is called partial integration with respect to y. (Notice its similarity to partial differentiation.) Now
xcd f ͑x, y͒ dy is a number that depends on the value of x, so it deﬁnes a function of x :
d A͑x͒ y f ͑x, y͒ dy
c If we now integrate the function A with respect to x from x a to x b, we get y 1 b a A͑x͒ dx y ͫy b a ͬ d f ͑x, y͒ dy dx c The integral on the right side of Equation 1 is called an iterated integral. Usually the
brackets are omitted. Thus
b y y 2 a d c f ͑x, y͒ dy dx y ͫy b a ͬ d f ͑x, y͒ dy dx c means that we ﬁrst integrate with respect to y from c to d and then with respect to x from
a to b.
Similarly, the iterated integral
d y y 3 c b a f ͑x, y͒ dx dy y ͫy d c ͬ b f ͑x, y͒ dx dy a means that we ﬁrst integrate with respect to x (holding y ﬁxed) from x a to x b and
then we integrate the resulting function of y with respect to y from y c to y d. Notice
that in both Equations 2 and 3 we work from the inside out.
EXAMPLE 1 Evaluate the iterated integrals. (a) 3 yy
0 2 1 x 2y dy dx (b) 2 yy
1 3 0 x 2 y dx dy SOLUTION (a) Regarding x as a constant, we obtain y ͫ ͬ
ͩͪ ͩͪ 2 x 2 y dy x 2 1 x2 y2
2 22
2 y2 y1 Ϫ x2 12
2 3 x2
2 Thus, the function A in the preceding discussion is given by A͑x͒ 3 x 2 in this example.
2
We now integrate this function of x from 0 to 3:
3 yy
0 2 1 x 2 y dy dx y 3 0 ͫy 3 3
2
0 y 2 1 ͬ
ͬ x 2 y dy dx x 2 dx x3
2 3 0 27
2 5E16(pp 10261035) 1/18/06 4:55 PM Page 1027 SECTION 16.2 ITERATED INTEGRALS ❙❙❙❙ 1027 (b) Here we ﬁrst integrate with respect to x :
2 yy
1 3 0 x y dx dy y
2 ͫy 2 1 y 3 0 ͬ x y dx dy
2 y2
9y dy 9
2 2 1 ͬ y 2 1 2 ͫ ͬ
x3
y
3 x3 dy x0 27
2 1 Notice that in Example 1 we obtained the same answer whether we integrated with
respect to y or x ﬁrst. In general, it turns out (see Theorem 4) that the two iterated integrals
in Equations 2 and 3 are always equal; that is, the order of integration does not matter.
(This is similar to Clairaut’s Theorem on the equality of the mixed partial derivatives.)
The following theorem gives a practical method for evaluating a double integral by
expressing it as an iterated integral (in either order).
 Theorem 4 is named after the Italian mathematician Guido Fubini (1879–1943), who proved
a very general version of this theorem in 1907.
But the version for continuous functions was
known to the French mathematician AugustinLouis Cauchy almost a century earlier. 4 Fubini’s Theorem If f is continuous on the rectangle
R ͕͑x, y͒ a ഛ x ഛ b, c ഛ y ഛ d͖, then Խ b yy f ͑x, y͒ dA y y
a d c f ͑x, y͒ dy dx y d y c b a f ͑x, y͒ dx dy R More generally, this is true if we assume that f is bounded on R, f is discontinuous only on a ﬁnite number of smooth curves, and the iterated integrals exist.
z The proof of Fubini’s Theorem is too difﬁcult to include in this book, but we can at least
give an intuitive indication of why it is true for the case where f ͑x, y͒ ജ 0. Recall that
if f is positive, then we can interpret the double integral xxR f ͑x, y͒ dA as the volume V of
the solid S that lies above R and under the surface z f ͑x, y͒. But we have another formula that we used for volume in Chapter 6, namely, C
0
x
x a A(x) b y V y A͑x͒ dx b a where A͑x͒ is the area of a crosssection of S in the plane through x perpendicular to the
xaxis. From Figure 1 you can see that A͑x͒ is the area under the curve C whose equation
is z f ͑x, y͒, where x is held constant and c ഛ y ഛ d. Therefore FIGURE 1 Visual 16.2 illustrates Fubini’s Theorem
by showing an animation of Figures 1
and 2. d A͑x͒ y f ͑x, y͒ dy
c and we have yy f ͑x, y͒ dA V y z b a A͑x͒ dx y b a y d c f ͑x, y͒ dy dx R 0 c A similar argument, using crosssections perpendicular to the yaxis as in Figure 2, shows
that
y d
y d yy f ͑x, y͒ dA y y
c f ͑x, y͒ dx dy R x EXAMPLE 2 Evaluate the double integral
FIGURE 2 b a Խ xxR ͑x Ϫ 3y 2 ͒ dA, where R ͕͑x, y͒ 0 ഛ x ഛ 2, 1 ഛ y ഛ 2͖. (Compare with Example 3 in Section 16.1.) 5E16(pp 10261035) 1028 ❙❙❙❙ 1/18/06 4:55 PM CHAPTER 16 MULTIPLE INTEGRALS  Notice the negative answer in Example 2;
nothing is wrong with that. The function f in
that example is not a positive function, so its
integral doesn’t represent a volume. From
Figure 3 we see that f is always negative on
R, so the value of the integral is the negative of
the volume that lies above the graph of f and
below R. SOLUTION 1 Fubini’s Theorem gives yy ͑x Ϫ 3y 2 ͒ dA y 1
y 1 ͑x Ϫ 3y 2 ͒ dy dx 2 y2 0 2 ͬ x2
͑x Ϫ 7͒ dx
Ϫ 7x
2 2 Ϫ12 0 SOLUTION 2 Again applying Fubini’s Theorem, but this time integrating with respect to x
ﬁrst, we have z=x3¥
0.5 2 y [ xy Ϫ y 3]y1 dx 0 _4 0 y R y z
_8 2 0 R 0 _12 Page 1028 1.5 2 2 yy ͑x Ϫ 3y 0 1
x 2 ͒ dA y 2 1 R FIGURE 3 y 2 1 y 2 0 ͫ ͑x Ϫ 3y 2 ͒ dx dy ͬ x2
Ϫ 3xy 2
2 x2 dy x0 2 2 y ͑2 Ϫ 6y 2 ͒ dy 2y Ϫ 2y 3]1 Ϫ12
1 EXAMPLE 3 Evaluate xxR y sin͑xy͒ dA, where R ͓1, 2͔ ϫ ͓0, ͔. SOLUTION 1 If we ﬁrst integrate with respect to x, we get
yy y sin͑xy͒ dA y y
0 2 1 y sin͑xy͒ dx dy R y 0 x2
[Ϫcos͑xy͒]x1 dy y ͑Ϫcos 2y ϩ cos y͒ dy
0 Ϫ 1 sin 2y ϩ sin y]0 0
2
SOLUTION 2 If we reverse the order of integration, we get
 For a function f that takes on both positive
and negative values, xxR f ͑x, y͒ dA is a difference of volumes: V1 Ϫ V2, where V1 is the volume above R and below the graph of f and V2 is
the volume below R and above the graph. The
fact that the integral in Example 3 is 0 means
that these two volumes V1 and V2 are equal.
(See Figure 4.) 2 yy y sin͑xy͒ dA y y
1 y sin͑xy͒ dy dx 0 R To evaluate the inner integral we use integration by parts with
uy dv sin͑xy͒ dy du dy
1
z 0
_1 and so
z=y sin(xy) 0 FIGURE 4 1 y 1
x
2 3 2 y 0 y sin͑xy͒ dy Ϫ vϪ y cos͑xy͒
x ͬ y y0 ϩ cos͑xy͒
x 1
x y 0 cos͑xy͒ dy Ϫ cos x
1
y
ϩ 2 [sin͑xy͒]y0
x
x Ϫ cos x
sin x
ϩ
x
x2 5E16(pp 10261035) 1/18/06 4:56 PM Page 1029 SECTION 16.2 ITERATED INTEGRALS ❙❙❙❙ 1029 If we now integrate the ﬁrst term by parts with u Ϫ1͞x and dv cos x dx, we get
du dx͞x 2, v sin x, and y ͩ Ϫ 2 yy and so cos x
x Ϫ y Therefore
 In Example 2, Solutions 1 and 2 are equally
straightforward, but in Example 3 the ﬁrst solution is much easier than the second one. Therefore, when we evaluate double integrals it is
wise to choose the order of integration that gives
simpler integrals. ͩ 0 1 ͪ dx Ϫ sin x
sin x
Ϫy
dx
x
x2 cos x
sin x
ϩ
x
x2 ͫ ͪ dx Ϫ sin x
y sin͑xy͒ dy dx Ϫ
x sin x
x ͬ 2 1 sin 2
ϩ sin 0
2 Ϫ EXAMPLE 4 Find the volume of the solid S that is bounded by the elliptic paraboloid x 2 ϩ 2y 2 ϩ z 16, the planes x 2 and y 2, and the three coordinate planes.
SOLUTION We ﬁrst observe that S is the solid that lies under the surface z 16 Ϫ x 2 Ϫ 2y 2 16 and above the square R ͓0, 2͔ ϫ ͓0, 2͔. (See Figure 5.) This solid was considered in
Example 1 in Section 16.1, but we are now in a position to evaluate the double integral
using Fubini’s Theorem. Therefore 12
z 8
4
0 0 0.5 1
y 1.5 V yy ͑16 Ϫ x 2 Ϫ 2y 2 ͒ dA y 0
1 0.5
2 2 1.5 x 2 0 y 2 0 ͑16 Ϫ x 2 Ϫ 2y 2 ͒ dx dy R
2 x2 y [16x Ϫ 1 x 3 Ϫ 2y 2x]x0 dy
3 FIGURE 5 0 y 2 0 ( 88 Ϫ 4y 2 ) dy [ 88 y Ϫ 4 y 3 ]2 48
3
3
3
0 In the special case where f ͑x, y͒ can be factored as the product of a function of x only
and a function of y only, the double integral of f can be written in a particularly simple
form. To be speciﬁc, suppose that f ͑x, y͒ t͑x͒h͑y͒ and R ͓a, b͔ ϫ ͓c, d͔. Then
Fubini’s Theorem gives
d yy f ͑x, y͒ dA y y
c b a t͑x͒h͑y͒ dx dy y d c R ͫy b a ͬ t͑x͒h͑y͒ dx dy In the inner integral y is a constant, so h͑y͒ is a constant and we can write
d ͫ y y
c b a ͬ t͑x͒h͑y͒ dx dy y d c ͫ ͩy b h͑y͒ b a ͪͬ t͑x͒ dx dy d y t͑x͒ dx y h͑y͒ dy
a c since xab t͑x͒ dx is a constant. Therefore, in this case, the double integral of f can be written as the product of two single integrals: yy t͑x͒h͑y͒ dA y b a R d t͑x͒ dx y h͑y͒ dy
c where R ͓a, b͔ ϫ ͓c, d͔ 5E16(pp 10261035) ❙❙❙❙ 1030 1/18/06 4:57 PM Page 1030 CHAPTER 16 MULTIPLE INTEGRALS EXAMPLE 5 If R ͓0, ͞2͔ ϫ ͓0, ͞2͔, then yy sin x cos y dA y ͞2 0 sin x dx y ͞2 0 cos y dy R [ ͞2
0 ͞2
0 [sin y] Ϫcos x 1ؒ11 z
 The function f ͑x, y͒ sin x cos y in
Example 5 is positive on R, so the integral represents the volume of the solid that lies above R
and below the graph of f shown in Figure 6. 0
y
x FIGURE 6  16.2
1–2 Exercises Find x03 f ͑x, y͒ dx and x04 f ͑x, y͒ dy.  y
2. f ͑x, y͒
xϩ2 1. f ͑x, y͒ 2x ϩ 3x y
2 ■ ■ ■ ■ 16. ■ ■ ■ ■ ■ ■ 1 ϩ x2
dA,
1 ϩ y2 yy
R ■ ■ 17. Խ R ͕͑x, y͒ 0 ഛ x ഛ 1, 0 ഛ y ഛ 1͖
R ͓0, ͞6͔ ϫ ͓0, ͞3͔ yy x sin͑x ϩ y͒ dA,
R 3–12 Calculate the iterated integral. 
3 3. yy 5. yy 7. 9.
11.
12. 1 1 0 2 0 ͑1 ϩ 4xy͒ dx dy ͞2 0 2 yy
0 1 0 4 yy
1 2 1 ln 2 ͑2x ϩ y͒ 8 dx dy
x
y
ϩ
y
x y y
0 yy
0 1 0 ■ 13–20 e 0 1 ■ ln 5 18. ͑x 2 ϩ y 2 ͒ dy dx yy y y (x ϩ sy ) dx dy 2 Ϫ1 4 1 0 1 0 19. 2 yy 8. 2 1 xe x
dy dx
y 20. 2 yy 10. dy dx 1 1 0 ■ ͑x ϩ y͒Ϫ2 dx dy 2 3 dx dy ■  1 21. ■ ■ ■ ■ ■ ■ Խ Ϫ 5y 4 ͒ dA, R ͕͑x, y͒ 0 ഛ x ഛ 3, 0 ഛ y ഛ 1͖ yy cos͑x ϩ 2y͒ dA,
R ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ Sketch the solid whose volume is given by the iterated ■ yy 22. yy 0 1 0 1 0 1 0 ͑4 Ϫ x Ϫ 2y͒ dx dy
͑2 Ϫ x 2 Ϫ y 2 ͒ dy dx Calculate the double integral. yy ͑6x y yy ■ R ͓1, 2͔ ϫ ͓0, 1͔ integral. Խ R ͕͑x, y͒ 0 ഛ x ഛ , 0 ഛ y ഛ ͞2͖ R 15. R ͓0, 1͔ ϫ ͓0, 2͔ dA, 2 21–22 ■ x2y x
dA,
x ϩ y2 yy
R R 14. yy xye R ͓0, 1͔ ϫ ͓0, 1͔ R ■ 13. x
dA,
1 ϩ xy yy
R xy
dy dx
sx 2 ϩ y 2 ϩ 1
■  2xϪy 1 6. x sin y dy dx ͩ ͪ 4 4. xy 2
dA,
2
x ϩ1 Խ R ͕͑x, y͒ 0 ഛ x ഛ 1, Ϫ3 ഛ y ഛ 3͖ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 23. Find the volume of the solid that lies under the plane 3x ϩ 2y ϩ z 12 and above the rectangle
R ͕͑x, y͒ 0 ഛ x ഛ 1, Ϫ2 ഛ y ഛ 3͖. Խ 24. Find the volume of the solid that lies under the hyperbolic paraboloid z 4 ϩ x 2 Ϫ y 2 and above the square
R ͓Ϫ1, 1͔ ϫ ͓0, 2͔. ■ 5E16(pp 10261035) 1/18/06 4:57 PM Page 1031 SECTION 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS 25. Find the volume of the solid lying under the elliptic CAS paraboloid x 2͞4 ϩ y 2͞9 ϩ z 1 and above the rectangle
R ͓Ϫ1, 1͔ ϫ ͓Ϫ2, 2͔. 32. Graph the solid that lies between the surfaces
2 Խ Խ z 1 ϩ e x sin y and the planes x Ϯ1, y 0, y ,
and z 0. 33–34  R has vertices ͑Ϫ1, 0͒, ͑Ϫ1, 5͒, ͑1, 5͒, ͑1, 0͒ 34. f ͑x, y͒ e sx ϩ e y,
y z x sx 2 ϩ y and the planes x 0, x 1, y 0, y 1,
and z 0. ■ 28. Find the volume of the solid bounded by the elliptic paraboloid CAS ■ ■ 1 yy
0 29. Find the volume of the solid in the ﬁrst octant bounded by the ■ ■ ■ ■ ■ ■ ■ ■ 1 0 xϪy
dy dx
͑x ϩ y͒3 1 yy and 0 1 0 xϪy
dx dy
͑x ϩ y͒3 Do the answers contradict Fubini’s Theorem? Explain what is
happening. cylinder z 9 Ϫ y 2 and the plane x 2.
30. (a) Find the volume of the solid bounded by the surface ■ R ͓0, 4͔ ϫ ͓0, 1͔ 35. Use your CAS to compute the iterated integrals z 1 ϩ ͑x Ϫ 1͒2 ϩ 4y 2, the planes x 3 and y 2, and the
coordinate planes. 36. (a) In what way are the theorems of Fubini and Clairaut z 6 Ϫ xy and the planes x 2, x Ϫ2, y 0, y 3,
and z 0.
(b) Use a computer to draw the solid. similar?
(b) If f ͑x, y͒ is continuous on ͓a, b͔ ϫ ͓c, d ͔ and
t͑x, y͒ y 31. Use a computer algebra system to ﬁnd the exact value of the x a integral xxR x 5y 3e x y dA, where R ͓0, 1͔ ϫ ͓0, 1͔. Then use the
CAS to draw the solid whose volume is given by the integral.  16.3 Խ Խ Find the average value of f over the given rectangle. 33. f ͑x, y͒ x 2 y, 27. Find the volume of the solid bounded by the surface CAS 1031 z eϪx cos͑x 2 ϩ y 2 ͒ and z 2 Ϫ x 2 Ϫ y 2 for x ഛ 1,
y ഛ 1. Use a computer algebra system to approximate the
volume of this solid correct to four decimal places. 26. Find the volume of the solid enclosed by the surface ; ❙❙❙❙ y y c f ͑s, t͒ dt ds for a Ͻ x Ͻ b, c Ͻ y Ͻ d, show that txy tyx f ͑x, y͒. Double Integrals over General Regions
For single integrals, the region over which we integrate is always an interval. But for
double integrals, we want to be able to integrate a function f not just over rectangles but
also over regions D of more general shape, such as the one illustrated in Figure 1. We suppose that D is a bounded region, which means that D can be enclosed in a rectangular
region R as in Figure 2. Then we deﬁne a new function F with domain R by
F͑x, y͒ 1 ͭ f ͑x, y͒ if ͑x, y͒ is in D
0
if ͑x, y͒ is in R but not in D y y R
D 0 FIGURE 1 D x 0 FIGURE 2 x 5E16(pp 10261035) 1032 ❙❙❙❙ 1/18/06 4:58 PM Page 1032 CHAPTER 16 MULTIPLE INTEGRALS z If the double integral of F exists over R, then we deﬁne the double integral of f over
D by graph of f 0 2
y yy f ͑x, y͒ dA yy F͑x, y͒ dA
D where F is given by Equation 1 R D
x Deﬁnition 2 makes sense because R is a rectangle and so xxR F͑x, y͒ dA has been previously deﬁned in Section 16.1. The procedure that we have used is reasonable because the
values of F͑x, y͒ are 0 when ͑x, y͒ lies outside D and so they contribute nothing to the integral. This means that it doesn’t matter what rectangle R we use as long as it contains D.
In the case where f ͑x, y͒ ജ 0 we can still interpret xxD f ͑x, y͒ dA as the volume of the
solid that lies above D and under the surface z f ͑x, y͒ (the graph of f ). You can see that
this is reasonable by comparing the graphs of f and F in Figures 3 and 4 and remembering that xxR F͑x, y͒ dA is the volume under the graph of F .
Figure 4 also shows that F is likely to have discontinuities at the boundary points of
D. Nonetheless, if f is continuous on D and the boundary curve of D is “well behaved”
(in a sense outside the scope of this book), then it can be shown that xxR F͑x, y͒ dA exists
and therefore xxD f ͑x, y͒ dA exists. In particular, this is the case for the following types
of regions.
A plane region D is said to be of type I if it lies between the graphs of two continuous
functions of x, that is, FIGURE 3
z graph of F 0
y D
x FIGURE 4 Խ D ͕͑x, y͒ a ഛ x ഛ b, t1͑x͒ ഛ y ഛ t 2͑x͖͒
where t1 and t 2 are continuous on ͓a, b͔. Some examples of type I regions are shown in
Figure 5.
y y y=g™(x) y y=g™(x) y=g™(x)
D D D y=g¡(x)
y=g¡(x)
0 a y=g¡(x)
b x 0 a b x 0 a b x FIGURE 5 Some type I regions y In order to evaluate xxD f ͑x, y͒ dA when D is a region of type I, we choose a rectangle
R ͓a, b͔ ϫ ͓c, d͔ that contains D, as in Figure 6, and we let F be the function given by
Equation 1; that is, F agrees with f on D and F is 0 outside D. Then, by Fubini’s Theorem, y=g™(x) d
b yy f ͑x, y͒ dA yy F͑x, y͒ dA y y D a D c y=g¡(x)
0 FIGURE 6 a x b x d c F͑x, y͒ dy dx R Observe that F͑x, y͒ 0 if y Ͻ t1͑x͒ or y Ͼ t 2͑x͒ because ͑x, y͒ then lies outside D.
Therefore y d c F͑x, y͒ dy y t 2͑x͒ t1͑x͒ F͑x, y͒ dy y t 2͑x͒ t1͑x͒ f ͑x, y͒ dy 5E16(pp 10261035) 1/18/06 4:58 PM Page 1033 SECTION 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS ❙❙❙❙ 1033 because F͑x, y͒ f ͑x, y͒ when t1͑x͒ ഛ y ഛ t 2͑x͒. Thus, we have the following formula
that enables us to evaluate the double integral as an iterated integral.
3 If f is continuous on a type I region D such that Խ D ͕͑x, y͒ a ഛ x ഛ b, t1͑x͒ ഛ y ഛ t 2͑x͖͒
b yy f ͑x, y͒ dA y y then D y D x=h™(y) c Խ D ͕͑x, y͒ c ഛ y ഛ d, h1͑y͒ ഛ x ഛ h2͑y͖͒ 4
0 f ͑x, y͒ dy dx The integral on the right side of (3) is an iterated integral that is similar to the ones we
considered in the preceding section, except that in the inner integral we regard x as being
constant not only in f ͑x, y͒ but also in the limits of integration, t1͑x͒ and t 2͑x͒.
We also consider plane regions of type II, which can be expressed as d x=h¡(y) t 2͑x͒ t1͑x͒ a x where h1 and h2 are continuous. Two such regions are illustrated in Figure 7.
Using the same methods that were used in establishing (3), we can show that y
d x=h¡(y) D x=h™(y) d yy f ͑x, y͒ dA y y 5
0
c h1͑ y͒ c D x h 2͑ y͒ f ͑x, y͒ dx dy where D is a type II region given by Equation 4. FIGURE 7 xxD ͑x ϩ 2y͒ dA, where D is the region bounded by the parabolas
y 2x 2 and y 1 ϩ x 2. Some type II regions EXAMPLE 1 Evaluate SOLUTION The parabolas intersect when 2x 2 1 ϩ x 2, that is, x 2 1, so x Ϯ1. We y y=1+≈ (_1, 2) note that the region D, sketched in Figure 8, is a type I region but not a type II region
and we can write (1, 2) Խ D ͕͑x, y͒ Ϫ1 ഛ x ഛ 1, 2x 2 ഛ y ഛ 1 ϩ x 2 ͖
D Since the lower boundary is y 2x 2 and the upper boundary is y 1 ϩ x 2, Equation 3
gives y=2≈ 1 _1 FIGURE 8 1 x yy ͑x ϩ 2y͒ dA y y
D Ϫ1 1ϩx 2 2x 2 ͑x ϩ 2y͒ dy dx 1 y1ϩx
[ xy ϩ y 2]y2x
Ϫ1 y 2 2 dx 1 y ͓x͑1 ϩ x 2 ͒ ϩ ͑1 ϩ x 2 ͒2 Ϫ x͑2x 2 ͒ Ϫ ͑2x 2 ͒2 ͔ dx
Ϫ1
1 y ͑Ϫ3x 4 Ϫ x 3 ϩ 2x 2 ϩ x ϩ 1͒ dx
Ϫ1 ͬ x5
x4
x3
x2
Ϫ3
Ϫ
ϩ2
ϩ
ϩx
5
4
3
2 1 Ϫ1 32
15 5E16(pp 10261035) ❙❙❙❙ 1034 1/18/06 4:59 PM Page 1034 CHAPTER 16 MULTIPLE INTEGRALS y NOTE
When we set up a double integral as in Example 1, it is essential to draw a
diagram. Often it is helpful to draw a vertical arrow as in Figure 8. Then the limits of
integration for the inner integral can be read from the diagram as follows: The arrow
starts at the lower boundary y t1͑x͒, which gives the lower limit in the integral, and
the arrow ends at the upper boundary y t 2͑x͒, which gives the upper limit of integration.
For a type II region the arrow is drawn horizontally from the left boundary to the right
boundary.
■ (2, 4) y=2x
y=≈
D EXAMPLE 2 Find the volume of the solid that lies under the paraboloid z x 2 ϩ y 2 and 0 1 above the region D in the xyplane bounded by the line y 2x and the parabola y x 2. x 2 SOLUTION 1 From Figure 9 we see that D is a type I region and
FIGURE 9 Խ D as a type I region D ͕͑x, y͒ 0 ഛ x ഛ 2, x 2 ഛ y ഛ 2x͖ y
4 Therefore, the volume under z x 2 ϩ y 2 and above D is (2, 4) V yy ͑x 2 ϩ y 2 ͒ dA y x= 1 y
2 2 0 D x=œ„
y yͫ
yͩ
2 x2y ϩ 0 D x 0 2 0 FIGURE 10 y3
3 ͬ y 2x x2 ͑x 2 ϩ y 2 ͒ dy dx y2x dx yͫ
2 x 2͑2x͒ ϩ 0 yx 2 x6
14x 3
Ϫ
Ϫ x4 ϩ
3
3 ͪ ͑2x͒3
͑x 2 ͒3
Ϫ x 2x 2 Ϫ
3
3 x7
x5
7x 4
dx Ϫ
Ϫ
ϩ
21
5
6 ͬ 2 0 ͬ dx 216
35 SOLUTION 2 From Figure 10 we see that D can also be written as a type II region: D as a type II region
 Figure 11 shows the solid whose volume
is calculated in Example 2. It lies above the
xyplane, below the paraboloid z x 2 ϩ y 2,
and between the plane y 2x and the
parabolic cylinder y x 2.
z y=≈ Խ D {͑x, y͒ 0 ഛ y ഛ 4, 1 y ഛ x ഛ sy }
2
Therefore, another expression for V is
V yy ͑x 2 ϩ y 2 ͒ dA y 4 0 D z=≈+¥ y 4 0 ͫ ͬ x3
ϩ y 2x
3 y sy
1
2 y ͑x 2 ϩ y 2 ͒ dx dy xsy dy x1 y
2 y 4 0 ͩ y 3͞2
y3
y3
ϩ y 5͞2 Ϫ
Ϫ
3
24
2 ͪ dy 2
15 y 5͞2 ϩ 2 y 7͞2 Ϫ 13 y 4 40 216
7
96
35 x FIGURE 11 y=2x y EXAMPLE 3 Evaluate xxD xy dA, where D is the region bounded by the line y x Ϫ 1 and
the parabola y 2 2x ϩ 6.
SOLUTION The region D is shown in Figure 12. Again D is both type I and type II, but
the description of D as a type I region is more complicated because the lower boundary
consists of two parts. Therefore, we prefer to express D as a type II region: Խ D {(x, y) Ϫ2 ഛ y ഛ 4, 1 y 2 Ϫ 3 ഛ x ഛ y ϩ 1}
2 5E16(pp 10261035) 1/18/06 4:59 PM Page 1035 SECTION 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS y 1035 y
(5, 4) (5, 4) ¥ x= 3
2 y=œ„„„„„
2x+6
y=x1 x=y+1
x 0 _3 (_1, _2) _2 (_1, _2) (a) D as a type I region (b) D as a type II region Then (5) gives
4 yy xy dA y y
Ϫ2 D x 0 y=_ œ„„„„„
2x+6 FIGURE 12 ❙❙❙❙ yϩ1
y 2Ϫ3 1
2 4 xy dx dy y 4 Ϫ2 ͫ ͬ xyϩ1 x2
y
2 [ dy x1 y 2Ϫ3
2 1 y y ͑y ϩ 1͒ 2 Ϫ ( 1 y 2 Ϫ 3) 2 dy
2
2
Ϫ2 1
2 z y 4 Ϫ2 ͫ (0, 0, 2) ͩ Ϫ ͪ
ͬ y5
ϩ 4y 3 ϩ 2y 2 Ϫ 8y dy
4 1
y6
y3
Ϫ
ϩ y4 ϩ 2
Ϫ 4y 2
2
24
3
x+2y+z=2 x=2y
T 36 Ϫ2 If we had expressed D as a type I region using Figure 12(a), then we would have
obtained
Ϫ1 y yy xy dA y y (0, 1, 0) 0 4 D s2xϩ6 Ϫs2xϩ6 Ϫ3 xy dy dx ϩ y 5 Ϫ1 y s2xϩ6 xϪ1 xy dy dx 1 ”1, 2 , 0’ but this would have involved more work than the other method. x EXAMPLE 4 Find the volume of the tetrahedron bounded by the planes x ϩ 2y ϩ z 2, FIGURE 13 y
1 x 2y, x 0, and z 0. x+2y=2
x
”or y=1 ’
2 1
”1, 2 ’ D
x
y= 2 0 FIGURE 14 1 x SOLUTION In a question such as this, it’s wise to draw two diagrams: one of the threedimensional solid and another of the plane region D over which it lies. Figure 13 shows
the tetrahedron T bounded by the coordinate planes x 0, z 0, the vertical plane
x 2y, and the plane x ϩ 2y ϩ z 2. Since the plane x ϩ 2y ϩ z 2 intersects the
xyplane (whose equation is z 0) in the line x ϩ 2y 2, we see that T lies above the
triangular region D in the xyplane bounded by the lines x 2y, x ϩ 2y 2, and x 0.
(See Figure 14.)
The plane x ϩ 2y ϩ z 2 can be written as z 2 Ϫ x Ϫ 2y, so the required volume
lies under the graph of the function z 2 Ϫ x Ϫ 2y and above Խ D {͑x, y͒ 0 ഛ x ഛ 1, x͞2 ഛ y ഛ 1 Ϫ x͞2} 5E16(pp 10361043) 1036 ❙❙❙❙ 1/18/06 4:20 PM Page 1036 CHAPTER 16 MULTIPLE INTEGRALS Therefore
V yy ͑2 Ϫ x Ϫ 2y͒ dA y 1 0 y 1Ϫx͞2 x͞2 ͑2 Ϫ x Ϫ 2y͒ dy dx D
1 [ 0 y 1 0 y 1 0 y1Ϫx͞2 y 2y Ϫ xy Ϫ y 2 ͫ dx yx͞2 ͩ ͪ ͩ ͪ
ͬ 2ϪxϪx 1Ϫ x
2 y 2 Ϫxϩ 1 x2
x2
ϩ
2
4 ͬ dx 1
3 0 x01 xx1 sin͑y 2 ͒ dy dx. SOLUTION If we try to evaluate the integral as it stands, we are faced with the task of ﬁrst
evaluating x sin͑y 2 ͒ dy. But it’s impossible to do so in ﬁnite terms since x sin͑y 2 ͒ dy is
not an elementary function. (See the end of Section 8.5.) So we must change the order
of integration. This is accomplished by ﬁrst expressing the given iterated integral as a
double integral. Using (3) backward, we have y=1 D
y=x 1 y y
1 0 x 1 x sin͑y 2 ͒ dy dx yy sin͑y 2 ͒ dA
D Խ D ͕͑x, y͒ 0 ഛ x ഛ 1, x ഛ y ഛ 1͖ where
FIGURE 15 We sketch this region D in Figure 15. Then from Figure 16 we see that an alternative
description of D is
D ͕͑x, y͒ 0 ഛ y ഛ 1, 0 ഛ x ഛ y͖ D as a type I region Խ y This enables us to use (5) to express the double integral as an iterated integral in the
reverse order: 1 1 x=0 x
2 x3
͑x Ϫ 2x ϩ 1͒ dx
Ϫ x2 ϩ x
3
2 EXAMPLE 5 Evaluate the iterated integral 0 Ϫ 1Ϫ yy D 0 x=y 1 x sin͑y 2 ͒ dy dx yy sin͑y 2 ͒ dA
D y
0 1 0 x y y 0 1 xy sin͑y 2 ͒ dx dy y [ x sin͑y 2 ͒]x0 dy
0 1 1 y y sin͑y 2 ͒ dy Ϫ 1 cos͑y 2 ͒]0
2
0 FIGURE 16 1 ͑1 Ϫ cos 1͒
2 D as a type II region Properties of Double Integrals
We assume that all of the following integrals exist. The ﬁrst three properties of double integrals over a region D follow immediately from Deﬁnition 2 and Properties 7, 8, and 9 in
Section 16.1.
6 yy ͓ f ͑x, y͒ ϩ t͑x, y͔͒ dA yy f ͑x, y͒ dA ϩ yy t͑x, y͒ dA
D 7 D D yy c f ͑x, y͒ dA c yy f ͑x, y͒ dA
D D 5E16(pp 10361043) 1/18/06 4:20 PM Page 1037 SECTION 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS ❙❙❙❙ 1037 If f ͑x, y͒ ജ t͑x, y͒ for all ͑x, y͒ in D, then y D yy f ͑x, y͒ dA ജ yy t͑x, y͒ dA 8
D¡ D™ D x 0 FIGURE 17 The next property of double integrals is similar to the property of single integrals given
by the equation xab f ͑x͒ dx xac f ͑x͒ dx ϩ xcb f ͑x͒ dx.
If D D1 ʜ D2 , where D1 and D2 don’t overlap except perhaps on their boundaries
(see Figure 17), then 9 y D yy f ͑x, y͒ dA yy f ͑x, y͒ dA ϩ yy f ͑x, y͒ dA
D1 D D2 Property 9 can be used to evaluate double integrals over regions D that are neither type I
nor type II but can be expressed as a union of regions of type I or type II. Figure 18 illustrates this procedure. (See Exercises 49 and 50.)
The next property of integrals says that if we integrate the constant function f ͑x, y͒ 1
over a region D, we get the area of D : D 0 x (a) D is neither type I nor type II. yy 1 dA A͑D͒ 10 D y Figure 19 illustrates why Equation 10 is true: A solid cylinder whose base is D and whose
height is 1 has volume A͑D͒ ؒ 1 A͑D͒, but we know that we can also write its volume
as xxD 1 dA.
Finally, we can combine Properties 7, 8, and 10 to prove the following property. (See
Exercise 53.) D™
D¡ 0 x (b) D=D¡ ʜ D™,
D¡ is type I, D™ is type II. 11 If m ഛ f ͑x, y͒ ഛ M for all ͑x, y͒ in D, then
mA͑D͒ ഛ yy f ͑x, y͒ dA ഛ MA͑D͒
D FIGURE 18 EXAMPLE 6 Use Property 11 to estimate the integral z xxD e sin x cos y dA, where D is the disk with center the origin and radius 2.
z=1 SOLUTION Since Ϫ1 ഛ sin x ഛ 1 and Ϫ1 ഛ cos y ഛ 1, we have Ϫ1 ഛ sin x cos y ഛ 1
and therefore eϪ1 ഛ e sin x cos y ഛ e 1 e 0 D
x FIGURE 19 Cylinder with base D and height 1 y Thus, using m e Ϫ1 1͞e, M e, and A͑D͒ ͑2͒2 in Property 11, we obtain
4
ഛ yy e sin x cos y dA ഛ 4e
e
D 5E16(pp 10361043) ❙❙❙❙ 1038 1–6  1. yy 3. yy x2 1 sx dx dy 4. yy ey y 0 ͞2 y y cos 0 0 e sin dr d ■ ■ 6.
■ ■ 3 2 yy x y x 1 yy v 0 0 ■ 23. Bounded by the planes x 0, y 0, z 0, and xϩyϩz1 s1 Ϫ v 2 du dv
■ ■ ■ 24. Bounded by the planes z x, y x, x ϩ y 2, and z 0
■ 25. Enclosed by the cylinders z x 2, y x 2 and the planes z 0, y 4 26. Bounded by the cylinder y 2 ϩ z 2 4 and the planes x 2y, x 0, z 0 in the ﬁrst octant 27. Bounded by the cylinder x 2 ϩ y 2 1 and the planes y z, 2y
dA,
x2 ϩ 1 yy y 1, y x, z 0 ͑x 2 Ϫ y͒ dy dx D ͕͑x, y͒ 1 ഛ x ഛ 2, 0 ഛ y ഛ 2x͖
D {͑x, y͒ 0 ഛ x ഛ 1, 0 ഛ y ഛ sx } Խ 3 yy e y2 dA, yy e x͞y dA, x 0, z 0 in the ﬁrst octant 28. Bounded by the cylinders x 2 ϩ y 2 r 2 and y 2 ϩ z 2 r 2 Խ ■ Խ D ͕͑x, y͒ 1 ഛ y ഛ 2, y ഛ x ഛ y 3 ͖ yy x sy Ϫ x dA,
2 Խ D is bounded by y 0, y x 2, x 1 yy x cos y dA, 31–32 yy ͑x ϩ y͒ dA,
yy y 3 yy xy dA, ■ ■ ■ ■ 2 dA, D is enclosed by x 0 and x s1 Ϫ y CAS  ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 33–36  Use a computer algebra system to ﬁnd the exact volume
of the solid. 34. Between the paraboloids z 2x 2 ϩ y 2 and z 8 Ϫ x 2 Ϫ 2y 2 D is the triangular region with vertices ͑0, 0͒, and inside the cylinder x 2 ϩ y 2 1 D ■ ■ bounded by the curves y x 3 Ϫ x and y x 2 ϩ x for x ജ 0 35. Enclosed by z 1 Ϫ x 2 Ϫ y 2 and z 0 ͑1, 2͒, and ͑0, 3͒
■ ■ 33. Under the surface z x 3 y 4 ϩ xy 2 and above the region D is bounded by the circle with center the origin and radius 2 yy Find the volume of the solid by subtracting two volumes. 2 D 2xy dA,  planes z 3y, z 2 ϩ y yy ͑2x Ϫ y͒ dA, 19–28 ■ 32. The solid enclosed by the parabolic cylinder y x 2 and the D ■ ■ y x 2 Ϫ 1 and the planes x ϩ y ϩ z 2,
2x ϩ 2y Ϫ z ϩ 10 0 ■ 18. ■ 31. The solid enclosed by the parabolic cylinders y 1 Ϫ x 2, D is bounded by y sx and y x 2 D is the triangular region with vertices (0, 2), (1, 1), and ͑3, 2͒ 17. ■ that is bounded by the planes y x, z 0, and z x and the
cylinder y cos x. (Use a graphing device to estimate the
points of intersection.) D ͕͑x, y͒ 0 ഛ y ഛ 1, 0 ഛ x ഛ y͖ D 16. ■ ; 30. Find the approximate volume of the solid in the ﬁrst octant
2 D 15. ■ xcoordinates of the points of intersection of the curves y x 4
and y 3x Ϫ x 2. If D is the region bounded by these curves,
estimate xxD x dA. D 14. ■ ; 29. Use a graphing calculator or computer to estimate the Խ D ͕͑x, y͒ 0 ഛ y ഛ 1, 0 ഛ x ഛ y͖ D 13. 0 2Ϫx Խ D 12. 1 22. Enclosed by the paraboloid z x 2 ϩ 3y 2 and the planes x 0, D ͕͑x, y͒ 0 ഛ x ഛ 2, Ϫx ഛ y ഛ x͖ dA, D 11. y ͑1, 1͒, ͑4, 1͒, and ͑1, 2͒ xy dx dy 4y
dA,
x ϩ2 yy D 10. ■ 1 2 Evaluate the double integral.  D 9. 2 yy D 8. 21. Under the surface z xy and above the triangle with vertices
2. ■ 7. Exercises ͑x ϩ 2y͒ dy dx 0 0 7–18 Page 1038 Evaluate the iterated integral. 1 ■ 4:21 PM CHAPTER 16 MULTIPLE INTEGRALS  16.3 5. 1/18/06 ■ ■ ■ ■ ■ ■ ■ Find the volume of the given solid. 19. Under the plane x ϩ 2y Ϫ z 0 and above the region bounded by y x and y x 4 20. Under the surface z 2x ϩ y 2 and above the region bounded by x y 2 and x y 3 ■ 36. Enclosed by z x 2 ϩ y 2 and z 2y
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 37– 42 ■  Sketch the region of integration and change the order of
integration. 37. 4 yy
0 sx 0 f ͑x, y͒ dy dx 38. 1 yy
0 4 4x f ͑x, y͒ dy dx ■ 5E16(pp 10361043) 1/18/06 4:22 PM Page 1039 SECTION 16.4 DOUBLE INTEGRALS IN POLAR COORDINATES 39.
41. 3 yy s9Ϫy 2 Ϫs9Ϫy 2 0 2 yy
1 ■ ln x f ͑x, y͒ dx dy 40. 0 f ͑x, y͒ dy dx 0 ■ ■ ■ 42.
■ 3 yy ■ 1 yy
0 s9Ϫy 0 ͞4 arctan x ■ 51–52 f ͑x, y͒ dx dy ■ 51. f ͑x, y͒ dy dx
■ ■ ■ ■ 52.  Evaluate the integral by reversing the order of
integration.
1 yy 45. y y 47. yy 48. 0 3 0 2 e x dx dy 9 1 ͞2 arcsin y 8 yy 2 46. 1 yy
0 1 yy
0 1 sy
1 x2 2 x 2ϩy 2 D is the disk with center the origin and radius 1
2 dA, ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ iterated integrals was obtained as follows:
1 0 2y 0 f ͑x, y͒ dx dy ϩ y 3 1 y 3Ϫy 0 f ͑x, y͒ dx dy Sketch the region D and express the double integral as an
iterated integral with reversed order of integration. ■ ■ ■ ■ ■ ■ ■ ■ ■ 55. Evaluate xxD ͑x 2 tan x ϩ y 3 ϩ 4͒ dA, where Խ 50. dA D ͕͑x, y͒ x 2 ϩ y 2 ഛ 2͖.
[Hint: Exploit the fact that D is symmetric with respect to both
axes.] yy xy dA 56. Use symmetry to evaluate xxD ͑2 Ϫ 3x ϩ 4y͒ dA, where D D y is the region bounded by the square with vertices ͑Ϯ5, 0͒
and ͑0, Ϯ5͒. y=1+≈ 1 (1, 1) D x=1
0 1 x x=_1 D 57. Compute xxD s1 Ϫ x 2 Ϫ y 2 dA, where D is the disk x 2 ϩ y 2 ഛ 1, by ﬁrst identifying the integral as the volume
of a solid. x=¥
0 _1 CAS x y=_1
■ yy e yy f ͑x, y͒ dA y y y ■ D ͓0, 1͔ ϫ ͓0, 1͔ D D _1 ϩ y 3 dA, 54. In evaluating a double integral over a region D, a sum of x 3 sin͑ y 3 ͒ dy dx cos x s1 ϩ cos 2 x dx dy ■ yy x 3 53. Prove Property 11. sx 3 ϩ 1 dx dy 49–50  Express D as a union of regions of type I or type II and
evaluate the integral. 49. yy sx ■ e x dx dy ■ Use Property 11 to estimate the value of the integral. D 4 3
sy 0 44. y cos͑x 2 ͒ dx dy y2 0 ■ 3 3y 1039 D 43–48 43.  ❙❙❙❙ ■  16.4 ■ ■ ■ ■ ■ ■ ■ ■ ■ 58. Graph the solid bounded by the plane x ϩ y ϩ z 1 and the paraboloid z 4 Ϫ x 2 Ϫ y 2 and ﬁnd its exact volume.
(Use your CAS to do the graphing, to ﬁnd the equations of the
boundary curves of the region of integration, and to evaluate
the double integral.) Double Integrals in Polar Coordinates
Suppose that we want to evaluate a double integral xxR f ͑x, y͒ dA, where R is one of the
regions shown in Figure 1. In either case the description of R in terms of rectangular coordinates is rather complicated but R is easily described using polar coordinates.
y y ≈+¥=4 ≈+¥=1
R R
0 x
0 FIGURE 1 (a) R=s(r, ¨)  0¯r¯1, 0¯¨¯2πd ≈+¥=1 (b) R=s(r, ¨)  1¯r¯2, 0¯¨¯πd x 5E16(pp 10361043) 1040 ❙❙❙❙ 1/18/06 4:23 PM CHAPTER 16 MULTIPLE INTEGRALS y P (r, ¨ ) =P (x, y)
r Recall from Figure 2 that the polar coordinates ͑r, ͒ of a point are related to the rectangular coordinates ͑x, y͒ by the equations x x r cos r2 x2 ϩ y2 y ¨
O Page 1040 x y r sin (See Section 11.3.)
The regions in Figure 1 are special cases of a polar rectangle Խ R ͕͑r, ͒ a ഛ r ഛ b, ␣ ഛ ഛ ͖ FIGURE 2 which is shown in Figure 3. In order to compute the double integral xxR f ͑x, y͒ dA, where
R is a polar rectangle, we divide the interval ͓a, b͔ into m subintervals ͓riϪ1, ri ͔ of equal
width ⌬r ͑b Ϫ a͒͞m and we divide the interval ͓␣, ͔ into n subintervals ͓jϪ1, j ͔ of
equal width ⌬ ͑ Ϫ ␣͒͞n. Then the circles r ri and the rays j divide the polar
rectangle R into the small polar rectangles shown in Figure 4.
¨=¨ j
¨=¨ j _1
r=b R ij ¨=∫ (r i*, ¨ j*) R
Î¨
r=a r=ri ¨=å r=ri _1 ∫
å
O O FIGURE 3 Polar rectangle FIGURE 4 Dividing R into polar subrectangles The “center” of the polar subrectangle Խ Rij ͕͑r, ͒ riϪ1 ഛ r ഛ ri , jϪ1 ഛ ഛ j ͖
has polar coordinates j* 1 ͑jϪ1 ϩ j ͒
2 r i* 1 ͑riϪ1 ϩ ri ͒
2 We compute the area of Rij using the fact that the area of a sector of a circle with radius r
1
and central angle is 2 r 2. Subtracting the areas of two such sectors, each of which has
central angle ⌬ j Ϫ jϪ1 , we ﬁnd that the area of Rij is
⌬Ai 1 r2i ⌬ Ϫ 21 r iϪ1
2
2 ⌬ 1 ͑r2i Ϫ2 r iϪ1 ͒ ⌬
2 1 ͑ri ϩ riϪ1 ͒͑ri Ϫ riϪ1 ͒ ⌬ r* ⌬r ⌬
i
2
Although we have deﬁned the double integral xxR f ͑x, y͒ dA in terms of ordinary rectangles, it can be shown that, for continuous functions f , we always obtain the same
answer using polar rectangles. The rectangular coordinates of the center of Rij are
͑r* cos j*, r i* sin j* ͒, so a typical Riemann sum is
i
m 1 n ͚ ͚ f ͑r* cos *, r* sin *͒ ⌬A
i i1 j1 j i j i m n ͚ ͚ f ͑r* cos *, r* sin * ͒ r* ⌬r ⌬
i i1 j1 j i j i 5E16(pp 10361043) 1/18/06 4:23 PM Page 1041 SECTION 16.4 DOUBLE INTEGRALS IN POLAR COORDINATES ❙❙❙❙ 1041 If we write t͑r, ͒ rf ͑r cos , r sin ͒, then the Riemann sum in Equation 1 can be written as
m n ͚ ͚ t͑r*, * ͒ ⌬r ⌬
i j i1 j1 which is a Riemann sum for the double integral
 y␣ y b a t͑r, ͒ dr d Therefore, we have
m yy f ͑x, y͒ dA
R lim n ͚ ͚ f ͑r* cos *, r* sin * ͒ ⌬A
i m, n l ϱ i1 j1
m lim
 ␣ y b a i j n  ͚ ͚ t͑r*, * ͒ ⌬r ⌬ y␣ y
i m, n l ϱ i1 j1 y j j b a i t͑r, ͒ dr d f ͑r cos , r sin ͒ r dr d 2 Change to Polar Coordinates in a Double Integral If f is continuous on a polar rectangle R given by 0 ഛ a ഛ r ഛ b, ␣ ഛ ഛ , where 0 ഛ  Ϫ ␣ ഛ 2, then
 yy f ͑x, y͒ dA y␣ y b a f ͑r cos , r sin ͒ r dr d R  The formula in (2) says that we convert from rectangular to polar coordinates in a
double integral by writing x r cos and y r sin , using the appropriate limits of
integration for r and , and replacing dA by r dr d. Be careful not to forget the additional
factor r on the right side of Formula 2. A classical method for remembering this is shown
in Figure 5, where the “inﬁnitesimal” polar rectangle can be thought of as an ordinary rectangle with dimensions r d and dr and therefore has “area” dA r dr d. dA
d¨
dr
r r d¨ O FIGURE 5 EXAMPLE 1 Evaluate xxR ͑3x ϩ 4y 2 ͒ dA, where R is the region in the upper halfplane bounded by the circles x 2 ϩ y 2 1 and x 2 ϩ y 2 4.
SOLUTION The region R can be described as Խ R ͕͑x, y͒ y ജ 0, 1 ഛ x 2 ϩ y 2 ഛ 4͖
It is the halfring shown in Figure 1(b), and in polar coordinates it is given by 1 ഛ r ഛ 2, 5E16(pp 10361043) 1042 ❙❙❙❙ 1/18/06 4:24 PM Page 1042 CHAPTER 16 MULTIPLE INTEGRALS 0 ഛ ഛ . Therefore, by Formula 2, yy ͑3x ϩ 4y 2 ͒ dA y 0 y 2 1 ͑3r cos ϩ 4r 2 sin 2 ͒ r dr d R y 0 y
y 2 1 0  Here we use the trigonometric identity y 0 ͑3r 2 cos ϩ 4r 3 sin 2 ͒ dr d r2
[r 3 cos ϩ r 4 sin 2]r1 d y0 ͑7 cos ϩ 15 sin 2 ͒ d [7 cos ϩ 15 ͑1 Ϫ cos 2 ͒] d
2 1
sin2 2 ͑1 Ϫ cos 2 ͒ 7 sin ϩ See Section 8.2 for advice on integrating
trigonometric functions. 15
15
Ϫ
sin 2
2
4 ͬ 0 15
2 EXAMPLE 2 Find the volume of the solid bounded by the plane z 0 and the paraboloid z 1 Ϫ x 2 Ϫ y 2. SOLUTION If we put z 0 in the equation of the paraboloid, we get x 2 ϩ y 2 1. This z
(0, 0, 1) means that the plane intersects the paraboloid in the circle x 2 ϩ y 2 1, so the solid lies
under the paraboloid and above the circular disk D given by x 2 ϩ y 2 ഛ 1 [see Figures 6
and 1(a)]. In polar coordinates D is given by 0 ഛ r ഛ 1, 0 ഛ ഛ 2. Since
1 Ϫ x 2 Ϫ y 2 1 Ϫ r 2, the volume is 0 V yy ͑1 Ϫ x 2 Ϫ y 2 ͒ dA y D
y x 2 0 D FIGURE 6 y 2 0 d y 1 0 y ͫ 1 0 ͑1 Ϫ r 2 ͒ r dr d r2
r4
͑r Ϫ r ͒ dr 2
Ϫ
2
4
3 ͬ 1 0
2 If we had used rectangular coordinates instead of polar coordinates, then we would have
obtained
V yy ͑1 Ϫ x 2 Ϫ y 2 ͒ dA y 1 Ϫ1 D y s1Ϫx 2 Ϫs1Ϫx 2 ͑1 Ϫ x 2 Ϫ y 2 ͒ dy dx which is not easy to evaluate because it involves ﬁnding the following integrals: y s1 Ϫ x
r=h™(¨) ¨=∫
D ¨=å ∫ 3 2 2 dx y ͑1 Ϫ x ͒ 2 3͞2 If f is continuous on a polar region of the form Խ D ͕͑r, ͒ ␣ ഛ ഛ , h1͑ ͒ ഛ r ഛ h2͑ ͖͒ r=h¡(¨) FIGURE 7
D=s(r, ¨)  å¯¨¯∫, h¡(¨)¯r¯h™(¨)d y x s1 Ϫ x dx dx What we have done so far can be extended to the more complicated type of region
shown in Figure 7. It’s similar to the type II rectangular regions considered in Section 16.3.
In fact, by combining Formula 2 in this section with Formula 16.3.5, we obtain the following formula. å
O 2 then  yy f ͑x, y͒ dA y␣ y
D h 2͑ ͒ h1͑ ͒ f ͑r cos , r sin ͒ r dr d 5E16(pp 10361043) 1/18/06 4:25 PM Page 1043 SECTION 16.4 DOUBLE INTEGRALS IN POLAR COORDINATES ❙❙❙❙ 1043 In particular, taking f ͑x, y͒ 1, h1͑ ͒ 0, and h2͑ ͒ h͑ ͒ in this formula, we see
that the area of the region D bounded by ␣, , and r h͑ ͒ is
A͑D͒ yy 1 dA y y
␣ yͫ ͬ
D ␣ h͑ ͒ 0 h͑ ͒ r2
2   r dr d
 1
2
␣ d y 0 ͓h͑ ͔͒ 2 d and this agrees with Formula 11.4.3.
EXAMPLE 3 Use a double integral to ﬁnd the area enclosed by one loop of the fourleaved
rose r cos 2. π
¨= 4 SOLUTION From the sketch of the curve in Figure 8 we see that a loop is given by the region Խ D {͑r, ͒ Ϫ͞4 ഛ ഛ ͞4, 0 ഛ r ഛ cos 2}
So the area is
π ¨=_ 4 A͑D͒ yy dA y
D FIGURE 8 y ͞4 Ϫ͞4 ͞4 Ϫ͞4 1y
4 y cos 2 0 r dr d ͞4
[ 1 r 2]cos 2 d 1 yϪ͞4 cos 2 2 d
2
2
0 ͞4 ͞4 Ϫ͞4 ͑1 ϩ cos 4 ͒ d 1 [ ϩ 1 sin 4]Ϫ͞4
4
4
8 EXAMPLE 4 Find the volume of the solid that lies under the paraboloid z x 2 ϩ y 2, y above the xyplane, and inside the cylinder x 2 ϩ y 2 2x. (x1)@+¥=1
(or r=2 cos ¨) SOLUTION The solid lies above the disk D whose boundary circle has equation x 2 ϩ y 2 2x or, after completing the square,
͑x Ϫ 1͒2 ϩ y 2 1 D
0 1 x 2 (see Figures 9 and 10). In polar coordinates we have x 2 ϩ y 2 r 2 and x r cos , so
the boundary circle becomes r 2 2r cos , or r 2 cos . Thus, the disk D is given by Խ D {͑r, ͒ Ϫ͞2 ഛ ഛ ͞2, 0 ഛ r ഛ 2 cos }
and, by Formula 3, we have FIGURE 9
z V yy ͑x 2 ϩ y 2 ͒ dA y
D y ͞2 Ϫ͞2 8y ͫͬ ͞2 0 2y x
y FIGURE 10 ͞2 0 r4
4 ͞2 Ϫ͞2 y 2 cos 0 2 cos d 4 y ͞2 Ϫ͞2 0 cos 4 d 8 y ͞2 0 ͩ r 2 r dr d cos 4 d 1 ϩ cos 2
2 ͪ 2 d ͓1 ϩ 2 cos 2 ϩ 1 ͑1 ϩ cos 4 ͔͒ d
2
͞2 ͩ ͪͩ ͪ 2[ 3 ϩ sin 2 ϩ 1 sin 4]0 2
2
8 3
2
2 3
2 5E16(pp 10441053) 1044 ❙❙❙❙ 4:50 PM Page 1044 CHAPTER 16 MULTIPLE INTEGRALS  16.4
1– 6 1/18/06 Exercises
12. A region R is shown. Decide whether to use polar coordinates or rectangular coordinates and write xxR f ͑x, y͒ dA as an iterated integral, where f is an arbitrary continuous function on R.
 xxR s4 Ϫ x 2 Ϫ y 2 dA, Խ where R ͕͑x, y͒ x 2 ϩ y 2 ഛ 4, x ജ 0͖
Ϫx 2Ϫy 2 13.
14. xxR ye x dA, where R is the region in the ﬁrst quadrant enclosed
by the circle x 2 ϩ y 2 25 15. xxR arctan͑ y͞x͒ dA,
where R ͕͑x, y͒ 1 ഛ x 2 ϩ y 2 ഛ 4, 0 ഛ y ഛ x͖
xxD x dA, y 2. 2 2 R R
0 xxD e
dA, where D is the region bounded by the
semicircle x s4 Ϫ y 2 and the yaxis 16. y 1. 0 2 x 2 x ■ ■ 17–20
y 3. 4. y
3 2 Խ where D is the region in the ﬁrst quadrant that lies
between the circles x 2 ϩ y 2 4 and x 2 ϩ y 2 2x
■  ■ ■ ■ ■ ■ ■ ■ ■ ■ Use a double integral to ﬁnd the area of the region. 17. One loop of the rose r cos 3
18. The region enclosed by the curve r 4 ϩ 3 cos R R 1
0 0 x 2 19. The region within both of the circles r cos and r sin 1 3 x 20. The region inside the circle r 4 sin and outside the circle r 2 ■ ■ 21–27 ■  ■ ■ ■ ■ ■ ■ ■ ■ ■ Use polar coordinates to ﬁnd the volume of the given solid.
y R y 6. 5
2 21. Under the paraboloid z x 2 ϩ y 2 and above the 2 5. 22. Inside the sphere x 2 ϩ y 2 ϩ z 2 16 and outside the disk x 2 ϩ y 2 ഛ 9 cylinder x 2 ϩ y 2 4 R
2 5 0 0 x 2 x 23. A sphere of radius a
24. Bounded by the paraboloid z 10 Ϫ 3x 2 Ϫ 3y 2 and the plane z 4 25. Above the cone z sx 2 ϩ y 2 and below the sphere x 2 ϩ y 2 ϩ z2 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 26. Bounded by the paraboloids z 3x 2 ϩ 3y 2 and
7– 8 z 4 Ϫ x2 Ϫ y2 Sketch the region whose area is given by the integral and
evaluate the integral.
7.  2 y y 7 4 ■ ■ 9–16 r dr d
■ 8.
■ ■ ■ ͞2 y y
0 ■ 4 cos 0 27. Inside both the cylinder x 2 ϩ y 2 4 and the ellipsoid 4x 2 ϩ 4y 2 ϩ z 2 64 r dr d
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 28. (a) A cylindrical drill with radius r 1 is used to bore a hole through the center of a sphere of radius r 2 . Find the volume
of the ringshaped solid that remains.
(b) Express the volume in part (a) in terms of the height h of
the ring. Notice that the volume depends only on h, not on
r 1 or r 2 . Evaluate the given integral by changing to polar
coordinates.
9. ■  xxD xy dA,
where D is the disk with center the origin and radius 3 10. xxR ͑x ϩ y͒ dA, where R is the region that lies to the left of
the yaxis between the circles x 2 ϩ y 2 1 and x 2 ϩ y 2 4 29–32  Evaluate the iterated integral by converting to polar
coordinates. 11. xxR cos͑x 2 ϩ y 2 ͒ dA, 29. where R is the region that lies above the
xaxis within the circle x 2 ϩ y 2 9 1 yy
0 s1Ϫx 2 0 ex 2 ϩy 2 dy dx 30. a y y
Ϫa sa 2Ϫy 2 0 ͑x 2 ϩ y 2 ͒3͞2 dx dy 5E16(pp 10441053) 1/18/06 4:51 PM Page 1045 SECTION 16.5 APPLICATIONS OF DOUBLE INTEGRALS 31.
■ 2 yy s4Ϫy 2 Ϫs4Ϫy 0 ■ x 2 y 2 dx dy 32. 2 ■ ■ ■ ■ 2 yy
0 ■ s2xϪx 2 0
■ ■ ■ ■ ϱ is constant along eastwest lines and increases linearly from
2 ft at the south end to 7 ft at the north end. Find the volume of
water in the pool. Ϫϱ Ϫ͑x 2ϩy 2 ͒ yy e Ϫr 1͞s2 x s1Ϫx 2 xy dy dx ϩ y s2 1 y x ޒ 0 y s2 s4Ϫx 2 xy dy dx 0 dA Ϫ͑x 2ϩy 2 ͒ yy e dA Sa where Sa is the square with vertices ͑Ϯa, Ϯa͒. Use this to
show that y ϱ Ϫϱ 2 eϪx dx y ϱ 2 eϪy dy Ϫϱ (c) Deduce that
ϱ 2 Ϫϱ 2 ϩy 2 ͒ alϱ y xy dy dx ϩ y 2 dA lim 2 35. Use polar coordinates to combine the sum
1 eϪ͑x (b) An equivalent deﬁnition of the improper integral in part (a)
is 34. An agricultural sprinkler distributes water in a circular pattern of radius 100 ft. It supplies water to a depth of e feet per hour
at a distance of r feet from the sprinkler.
(a) What is the total amount of water supplied per hour to
the region inside the circle of radius R centered at the
sprinkler?
(b) Determine an expression for the average amount of water
per hour per square foot supplied to the region inside the
circle of radius R. ϱ Ϫϱ y y 33. A swimming pool is circular with a 40ft diameter. The depth y y 1045 where Da is the disk with radius a and center the origin.
Show that sx 2 ϩ y 2 dy dx
■ ❙❙❙❙ eϪx dx s (d) By making the change of variable t s2 x, show that into one double integral. Then evaluate the double integral. y ϱ Ϫϱ 36. (a) We deﬁne the improper integral (over the entire plane ͒ ޒ eϪx 2 ͞2 dx s2 2 I yy eϪ͑x 2 ϩy 2 ͒ dA y Ϫϱ 2ޒ lim alϱ  16.5 ϱ y ϱ Ϫϱ eϪ͑x 2 ϩy 2 ͒ dy dx (This is a fundamental result for probability and statistics.)
37. Use the result of Exercise 36 part (c) to evaluate the following integrals.
Ϫ͑x 2ϩy 2 ͒ yy e dA (a) Da y ϱ 0 2 x 2eϪx dx (b) y ϱ 0 sx eϪx dx Applications of Double Integrals
We have already seen one application of double integrals: computing volumes. Another
geometric application is ﬁnding areas of surfaces and this will be done in the next section.
In this section we explore physical applications such as computing mass, electric charge,
center of mass, and moment of inertia. We will see that these physical ideas are also important when applied to probability density functions of two random variables. Density and Mass
In Section 9.3 we were able to use single integrals to compute moments and the center of
mass of a thin plate or lamina with constant density. But now, equipped with the double
integral, we can consider a lamina with variable density. Suppose the lamina occupies a
region D of the xyplane and its density (in units of mass per unit area) at a point ͑x, y͒ in
D is given by ͑x, y͒, where is a continuous function on D. This means that y
(x, y) D ͑x, y͒ lim
0 FIGURE 1 ⌬m
⌬A x where ⌬m and ⌬A are the mass and area of a small rectangle that contains ͑x, y͒ and the
limit is taken as the dimensions of the rectangle approach 0. (See Figure 1.) 5E16(pp 10441053) 1046 ❙❙❙❙ 1/18/06 4:51 PM Page 1046 CHAPTER 16 MULTIPLE INTEGRALS y * *
(xij , yij ) Rij To ﬁnd the total mass m of the lamina we divide a rectangle R containing D into subrectangles Rij of the same size (as in Figure 2) and consider ͑x, y͒ to be 0 outside D. If
* *
we choose a point ͑x ij , yij ͒ in Rij , then the mass of the part of the lamina that occupies Rij
* *
is approximately ͑x ij , yij ͒ ⌬A, where ⌬A is the area of Rij . If we add all such masses, we
get an approximation to the total mass:
k 0 x FIGURE 2 l ͚ ͚ ͑x*, y* ͒ ⌬A mϷ ij ij i1 j1 If we now increase the number of subrectangles, we obtain the total mass m of the lamina
as the limiting value of the approximations:
k 1 m lim l ͚ ͚ ͑x*, y* ͒ ⌬A yy ͑x, y͒ dA
ij k, l l ϱ i1 j1 ij D Physicists also consider other types of density that can be treated in the same manner.
For example, if an electric charge is distributed over a region D and the charge density (in
units of charge per unit area) is given by ͑x, y͒ at a point ͑x, y͒ in D, then the total charge
Q is given by
Q yy ͑x, y͒ dA 2 D y y=1 1 EXAMPLE 1 Charge is distributed over the triangular region D in Figure 3 so that the
charge density at ͑x, y͒ is ͑x, y͒ xy, measured in coulombs per square meter (C͞m2 ).
Find the total charge. (1, 1) D SOLUTION From Equation 2 and Figure 3 we have Q yy ͑x, y͒ dA y
0 ͫ ͬ D y=1x
x y y2
x
2 1 0 FIGURE 3
Thus, the total charge is 1
2 y 5
24 1 0 1 0 y 1 1Ϫx xy dy dx y1 dx y y1Ϫx 0 1
͑2x Ϫ x ͒ dx
2
2 3 1 x 2
͓1 Ϫ ͑1 Ϫ x͒2 ͔ dx
2 ͫ 2x 3
x4
Ϫ
3
4 ͬ 1 0 5
24 C. Moments and Centers of Mass
In Section 9.3 we found the center of mass of a lamina with constant density; here we consider a lamina with variable density. Suppose the lamina occupies a region D and has density function ͑x, y͒. Recall from Chapter 9 that we deﬁned the moment of a particle about
an axis as the product of its mass and its directed distance from the axis. We divide D into
small rectangles as in Figure 2. Then the mass of Rij is approximately ͑x *, y* ͒ ⌬A, so we
ij
ij
can approximate the moment of Rij with respect to the xaxis by
͓ ͑x *, y* ͒ ⌬A͔ y*
ij
ij
ij
If we now add these quantities and take the limit as the number of subrectangles becomes 5E16(pp 10441053) 1/18/06 4:51 PM Page 1047 SECTION 16.5 APPLICATIONS OF DOUBLE INTEGRALS ❙❙❙❙ 1047 large, we obtain the moment of the entire lamina about the xaxis:
m n ͚ ͚ y* ͑x*, y* ͒ ⌬A yy y ͑x, y͒ dA Mx lim 3 m, n l ϱ i1 j1 ij ij ij D Similarly, the moment about the yaxis is
m (x, y) n ͚ ͚ x* ͑x*, y* ͒ ⌬A yy x ͑x, y͒ dA My lim 4 m, n l ϱ i1 j1 D ij ij ij D As before, we deﬁne the center of mass ͑x, y͒ so that mx My and my Mx . The physical signiﬁcance is that the lamina behaves as if its entire mass is concentrated at its center
of mass. Thus, the lamina balances horizontally when supported at its center of mass (see
Figure 4). FIGURE 4 5 The coordinates ͑x, y͒ of the center of mass of a lamina occupying the region
D and having density function (x, y) are
x My
1
m
m yy x ͑x, y͒ dA Mx
1
m
m y D yy y ͑x, y͒ dA
D where the mass m is given by
m yy ͑x, y͒ dA
D EXAMPLE 2 Find the mass and center of mass of a triangular lamina with vertices ͑0, 0͒, ͑1, 0͒, and ͑0, 2͒ if the density function is x, y͒ 1 ϩ 3x ϩ y.
y
(0, 2) SOLUTION The triangle is shown in Figure 5. (Note that the equation of the upper boundary is y 2 Ϫ 2x.) The mass of the lamina is
y=22x m yy ͑x, y͒ dA y
3 11 D ” , ’
8 16 D y 1 0 0 (1, 0) x 1 0 ͫ y 2Ϫ2x 0 y2
y ϩ 3xy ϩ
2 1 ͑1 ϩ 3x ϩ y͒ dy dx ͬ
ͫ ͬ
y2Ϫ2x dx y0 4 y ͑1 Ϫ x 2 ͒ dx 4 x Ϫ
0 FIGURE 5 1 x3
3 0 8
3 Then the formulas in (5) give
x 1
m 3
8
3
2 1 yy x ͑x, y͒ dA y y
3
8 D y 1 0 y 1 0 ͫ xy ϩ 3x 2 y ϩ x ͑x Ϫ x ͒ dx
3 ͫ 0 y2
2 2Ϫ2x 0 ͑x ϩ 3x 2 ϩ xy͒ dy dx ͬ y2Ϫ2x y0 x2
x4
Ϫ
2
4 dx ͬ 1 0 3
8 5E16(pp 10441053) 1048 ❙❙❙❙ 1/18/06 4:52 PM Page 1048 CHAPTER 16 MULTIPLE INTEGRALS y 1
m 3
8
1
4 1 yy y ͑x, y͒ dA y y
3
8 D y 1 0 ͫ ͫ 0 2Ϫ2x ͑y ϩ 3xy ϩ y 2 ͒ dy dx 0 y2
y2
y3
ϩ 3x
ϩ
2
2
3 ͬ y2Ϫ2x
1 dx 1 y ͑7 Ϫ 9x Ϫ 3x 2 ϩ 5x 3 ͒ dx
4 y0 x2
x4
7x Ϫ 9
Ϫ x3 ϩ 5
2
4 The center of mass is at the point ( , 3 11
8 16 0 ͬ 1 0 11
16 ). EXAMPLE 3 The density at any point on a semicircular lamina is proportional to the distance from the center of the circle. Find the center of mass of the lamina.
SOLUTION Let’s place the lamina as the upper half of the circle x 2 ϩ y 2 a 2 (see Fig y
a D _a ure 6). Then the distance from a point ͑x, y͒ to the center of the circle (the origin) is
sx 2 ϩ y 2. Therefore, the density function is ≈+¥[email protected]
3a ”0, ’
2π
0 ͑x, y͒ Ksx 2 ϩ y 2
a x FIGURE 6 where K is some constant. Both the density function and the shape of the lamina suggest
that we convert to polar coordinates. Then sx 2 ϩ y 2 r and the region D is given by
0 ഛ r ഛ a, 0 ഛ ഛ . Thus, the mass of the lamina is
m yy ͑x, y͒ dA yy Ksx 2 ϩ y 2 dA
D y 0 D y a 0 a ͑Kr͒ r dr d K y d y r 2 dr r3
K
3 ͬ 0 a 0 0 Ka 3
3 Both the lamina and the density function are symmetric with respect to the yaxis, so the
center of mass must lie on the yaxis, that is, x 0. The ycoordinate is given by
y
 Compare the location of the center of mass
in Example 3 with Example 4 in Section 9.3
where we found that the center of mass of a
lamina with the same shape but uniform density
is located at the point ͑0, 4a͑͞3͒͒. 1
m 3 yy y ͑x, y͒ dA Ka y y
3 D 3
a3 y 0 sin d y a 3 0 r dr 0 a 0 r sin ͑⌲r͒ r dr d ͫͬ 4
3
[Ϫcos ] r4
0
a 3 a 0 3 2a 4
3a
3
a 4
2
Therefore, the center of mass is located at the point ͑0, 3a͑͞2͒͒. Moment of Inertia
The moment of inertia (also called the second moment) of a particle of mass m about an
axis is deﬁned to be mr 2, where r is the distance from the particle to the axis. We extend
this concept to a lamina with density function ͑x, y͒ and occupying a region D by proceeding as we did for ordinary moments. We divide D into small rectangles, approximate
the moment of inertia of each subrectangle about the xaxis, and take the limit of the sum 5E16(pp 10441053) 1/18/06 4:52 PM Page 1049 SECTION 16.5 APPLICATIONS OF DOUBLE INTEGRALS ❙❙❙❙ 1049 as the number of subrectangles becomes large. The result is the moment of inertia of the
lamina about the xaxis:
m 6 n ͚ ͚ ͑y * ͒ ͑x *, y * ͒ ⌬A yy y ͑x, y͒ dA I x lim m, n l ϱ i1 j1 ij 2 ij 2 ij D Similarly, the moment of inertia about the yaxis is
m 7 n ͚ ͚ ͑x * ͒ ͑x *, y * ͒ ⌬A yy x ͑x, y͒ dA I y lim m, n l ϱ i1 j1 ij 2 ij 2 ij D It is also of interest to consider the moment of inertia about the origin, also called the
polar moment of inertia:
m 8 I 0 lim n ͚ ͚ ͓͑x* ͒ 2 ij m, n l ϱ i1 j1 ϩ ͑y* ͒2͔ ͑x *, y* ͒ ⌬A yy ͑x 2 ϩ y 2 ͒ ͑x, y͒ dA
ij
ij
ij
D Note that I 0 I x ϩ I y .
EXAMPLE 4 Find the moments of inertia I x , I y , and I 0 of a homogeneous disk D with density ͑x, y͒ , center the origin, and radius a.
SOLUTION The boundary of D is the circle x 2 ϩ y 2 a 2 and in polar coordinates D is described by 0 ഛ ഛ 2, 0 ഛ r ഛ a. Let’s compute I 0 ﬁrst:
I 0 yy ͑x 2 ϩ y 2 ͒ dA y 2 0 y 0 d y a 0 a 0 r 2 r dr d ͫͬ D
2 y r4
r dr 2
4
3 a 0 a 4
2 Instead of computing I x and I y directly, we use the facts that I x ϩ I y I 0 and I x I y
(from the symmetry of the problem). Thus
Ix Iy I0
a 4
2
4 In Example 4 notice that the mass of the disk is
m density ϫ area ͑ a 2 ͒
so the moment of inertia of the disk about the origin (like a wheel about its axle) can be
written as
I 0 1 ma 2
2
Thus, if we increase the mass or the radius of the disk, we thereby increase the moment of 5E16(pp 10441053) 1050 ❙❙❙❙ 1/18/06 4:52 PM Page 1050 CHAPTER 16 MULTIPLE INTEGRALS inertia. In general, the moment of inertia plays much the same role in rotational motion
that mass plays in linear motion. The moment of inertia of a wheel is what makes it difﬁcult to start or stop the rotation of the wheel, just as the mass of a car is what makes it difﬁcult to start or stop the motion of the car.
The radius of gyration of a lamina about an axis is the number R such that
mR 2 I 9 where m is the mass of the lamina and I is the moment of inertia about the given axis.
Equation 9 says that if the mass of the lamina were concentrated at a distance R from the
axis, then the moment of inertia of this “point mass” would be the same as the moment of
inertia of the lamina.
In particular, the radius of gyration y with respect to the xaxis and the radius of gyration x with respect to the yaxis are given by the equations
10 my 2 I x mx 2 I y Thus ͑x, y͒ is the point at which the mass of the lamina can be concentrated without changing the moments of inertia with respect to the coordinate axes. (Note the analogy with the
center of mass.)
EXAMPLE 5 Find the radius of gyration about the xaxis of the disk in Example 4. a 2, so from Equations 10 we have SOLUTION As noted, the mass of the disk is m y2 1
Ix
a 4
a2
4
m
a 2
4 Therefore, the radius of gyration about the xaxis is
y a
2 which is half the radius of the disk. Probability
In Section 9.5 we considered the probability density function f of a continuous random
ϱ
variable X. This means that f ͑x͒ ജ 0 for all x, xϪϱ f ͑x͒ dx 1, and the probability that X
lies between a and b is found by integrating f from a to b:
b P͑a ഛ X ഛ b͒ y f ͑x͒ dx
a Now we consider a pair of continuous random variables X and Y, such as the lifetimes
of two components of a machine or the height and weight of an adult female chosen at random. The joint density function of X and Y is a function f of two variables such that the
probability that ͑X, Y͒ lies in a region D is
P͑͑X, Y͒ ʦ D͒ yy f ͑x, y͒ dA
D 5E16(pp 10441053) 1/18/06 4:53 PM Page 1051 SECTION 16.5 APPLICATIONS OF DOUBLE INTEGRALS ❙❙❙❙ 1051 In particular, if the region is a rectangle, the probability that X lies between a and b and Y
lies between c and d is
P͑a ഛ X ഛ b, c ഛ Y ഛ d͒ y b a y d c f ͑x, y͒ dy dx (See Figure 7.)
z z=f(x, y) c a FIGURE 7
d The probability that X lies between
a and b and Y lies between c and d
is the volume that lies above the
rectangle D=[a, b]x[c, d] and
below the graph of the joint
density function. b
y D x Because probabilities aren’t negative and are measured on a scale from 0 to 1, the joint
density function has the following properties:
f ͑x, y͒ ജ 0 yy f ͑x, y͒ dA 1
2ޒ As in Exercise 36 in Section 16.4, the double integral over 2ޒis an improper integral
deﬁned as the limit of double integrals over expanding circles or squares and we can write
ϱ Ϫϱ Ϫϱ yy f ͑x, y͒ dA y y
2ޒ f ͑x, y͒ dx dy 1 EXAMPLE 6 If the joint density function for X and Y is given by f ͑x, y͒ ͭ C͑x ϩ 2y͒ if 0 ഛ x ഛ 10, 0 ഛ y ഛ 10
0
otherwise ﬁnd the value of the constant C. Then ﬁnd P͑X ഛ 7, Y ജ 2͒.
SOLUTION We ﬁnd the value of C by ensuring that the double integral of f is equal to 1.
Because f ͑x, y͒ 0 outside the rectangle ͓0, 10͔ ϫ ͓0, 10͔, we have
ϱ ϱ Ϫϱ Ϫϱ y y f ͑x, y͒ dy dx y 10 0 y 10 0 C͑x ϩ 2y͒ dy dx C y 10 10 0 C y ͑10x ϩ 100͒ dx 1500C
0 1
Therefore, 1500C 1 and so C 1500 . y10
[ xy ϩ y 2]y0 dx 5E16(pp 10441053) 1052 ❙❙❙❙ 1/18/06 4:53 PM Page 1052 CHAPTER 16 MULTIPLE INTEGRALS Now we can compute the probability that X is at most 7 and Y is at least 2:
P͑X ഛ 7, Y ജ 2͒ y 7 Ϫϱ y ϱ 2
7 f ͑x, y͒ dy dx y 0 [ 1
1500 y xy ϩ y 2
0 7 y10
y2 y 10 1
1500 2 ͑x ϩ 2y͒ dy dx 7 1
dx 1500 y ͑8x ϩ 96͒ dx
0 868
1500 Ϸ 0.5787 Suppose X is a random variable with probability density function f1͑x͒ and Y is a random variable with density function f2͑y͒. Then X and Y are called independent random
variables if their joint density function is the product of their individual density functions:
f ͑x, y͒ f1͑x͒ f2͑y͒
In Section 9.5 we modeled waiting times by using exponential density functions
f ͑t͒ ͭ if t Ͻ 0
if t ജ 0 0
Ϫ1eϪt͞ where is the mean waiting time. In the next example we consider a situation with two
independent waiting times.
EXAMPLE 7 The manager of a movie theater determines that the average time moviegoers
wait in line to buy a ticket for this week’s ﬁlm is 10 minutes and the average time they
wait to buy popcorn is 5 minutes. Assuming that the waiting times are independent, ﬁnd
the probability that a moviegoer waits a total of less than 20 minutes before taking his or
her seat.
SOLUTION Assuming that both the waiting time X for the ticket purchase and the waiting
time Y in the refreshment line are modeled by exponential probability density functions,
we can write the individual density functions as f1͑x͒ ͭ 0
1 Ϫx͞10
10 e if x Ͻ 0
if x ജ 0 f2͑y͒ ͭ if y Ͻ 0
if y ജ 0 0
1 Ϫy͞5
5 e Since X and Y are independent, the joint density function is the product:
f ͑x, y͒ f1͑x͒ f2͑y͒ ͭ 1 Ϫx͞10 Ϫy͞5
50 e if x ജ 0, y ജ 0
otherwise e 0 We are asked for the probability that X ϩ Y Ͻ 20: y
20 P͑X ϩ Y Ͻ 20͒ P͑͑X, Y͒ ʦ D͒
x+y=20 where D is the triangular region shown in Figure 8. Thus D P͑X ϩ Y Ͻ 20͒ yy f ͑x, y͒ dA y
0 FIGURE 8 20 0 20 x y 20Ϫx 1 Ϫx͞10 Ϫy͞5
50 e 0 D 1
50 20 y [e
0 Ϫx͞10 ͑Ϫ5͒eϪy͞5 y20Ϫx
y0 dx e dy dx 5E16(pp 10441053) 1/18/06 4:53 PM Page 1053 SECTION 16.5 APPLICATIONS OF DOUBLE INTEGRALS ❙❙❙❙ 1053 20 1
10 y eϪx͞10͑1 Ϫ e ͑xϪ20͒͞5 ͒ dx
0 20 1
10 y ͑eϪx͞10 Ϫ eϪ4e x͞10 ͒ dx
0 1 ϩ eϪ4 Ϫ 2eϪ2 Ϸ 0.7476
This means that about 75% of the moviegoers wait less than 20 minutes before taking
their seats. Expected Values
Recall from Section 9.5 that if X is a random variable with probability density function f,
then its mean is
ϱ y x f ͑x͒ dx
Ϫϱ Now if X and Y are random variables with joint density function f, we deﬁne the Xmean
and Ymean, also called the expected values of X and Y, to be 1 yy x f ͑x, y͒ dA 11 2 yy yf ͑x, y͒ dA 2ޒ 2ޒ Notice how closely the expressions for 1 and 2 in (11) resemble the moments Mx and My
of a lamina with density function in Equations 3 and 4. In fact, we can think of probability as being like continuously distributed mass. We calculate probability the way we calculate mass—by integrating a density function. And because the total “probability mass”
is 1, the expressions for x and y in (5) show that we can think of the expected values of X
and Y, 1 and 2 , as the coordinates of the “center of mass” of the probability distribution.
In the next example we deal with normal distributions. As in Section 9.5, a single random variable is normally distributed if its probability density function is of the form
f ͑x͒ 1
2
2
eϪ͑xϪ͒ ͑͞2 ͒
s2 where is the mean and is the standard deviation.
EXAMPLE 8 A factory produces (cylindrically shaped) roller bearings that are sold as
having diameter 4.0 cm and length 6.0 cm. In fact, the diameters X are normally distributed with mean 4.0 cm and standard deviation 0.01 cm while the lengths Y are normally
distributed with mean 6.0 cm and standard deviation 0.01 cm. Assuming that X and Y
are independent, write the joint density function and graph it. Find the probability that
a bearing randomly chosen from the production line has either length or diameter that
differs from the mean by more than 0.02 cm.
SOLUTION We are given that X and Y are normally distributed with 1 4.0, 2 6.0,
and 1 2 0.01. So the individual density functions for X and Y are f1͑x͒ 1
2
eϪ͑xϪ4͒ ͞0.0002
0.01s2 f2͑y͒ 1
2
eϪ͑ yϪ6͒ ͞0.0002
0.01s2 5E16(pp 10541063) 1054 ❙❙❙❙ 1/18/06 4:28 PM Page 1054 CHAPTER 16 MULTIPLE INTEGRALS Since X and Y are independent, the joint density function is the product:
1500 f ͑x, y͒ f1͑x͒ f2͑y͒ 1000
500
0
5.95 3.95
y 4 6
6.05 x 1
2
2
eϪ͑xϪ4͒ ͞0.0002eϪ͑yϪ6͒ ͞0.0002
0.0002 5000 Ϫ5000͓͑xϪ4͒2ϩ͑ yϪ6͒2͔
e
4.05 A graph of this function is shown in Figure 9.
Let’s ﬁrst calculate the probability that both X and Y differ from their means by less
than 0.02 cm. Using a calculator or computer to estimate the integral, we have FIGURE 9 Graph of the bivariate normal joint
density function in Example 8 P͑3.98 Ͻ X Ͻ 4.02, 5.98 Ͻ Y Ͻ 6.02͒ y 4.02 3.98 y 6.02 5.98 5000
f ͑x, y͒ dy dx 4.02 y y
3.98 6.02 5.98 2 2 eϪ5000͓͑xϪ4͒ ϩ͑ yϪ6͒ ͔ dy dx Ϸ 0.91
Then the probability that either X or Y differs from its mean by more than 0.02 cm is
approximately
1 Ϫ 0.91 0.09  16.5 Exercises 1. Electric charge is distributed over the rectangle 1 ഛ x ഛ 3, 0 ഛ y ഛ 2 so that the charge density at ͑x, y͒ is
͑x, y͒ 2x y ϩ y 2 (measured in coulombs per square meter).
Find the total charge on the rectangle. 2. Electric charge is distributed over the disk x 2 ϩ y 2 ഛ 4 so that the charge density at ͑x, y͒ is ͑x, y͒ x ϩ y ϩ x 2 ϩ y 2
(measured in coulombs per square meter). Find the total charge
on the disk. 3 –10  Find the mass and center of mass of the lamina that
occupies the region D and has the given density function . Խ
D ͕͑x, y͒ Խ 0 ഛ x ഛ a, 0 ഛ y ഛ b͖; ͑x, y͒ cxy ͑x, y͒ x ϩ y 13. Find the center of mass of a lamina in the shape of an isosceles right triangle with equal sides of length a if the density at any
point is proportional to the square of the distance from the vertex opposite the hypotenuse. Exercise 7. ͑x, y͒ x 16. Find the moments of inertia I x , I y , I 0 for the lamina of 7. D is bounded by y e x, y 0, x 0, and x 1; ͑x, y͒ y Exercise 9. 9. D is bounded by the parabola x y 2 and the line y x Ϫ 2; 18. Consider a square fan blade with sides of length 2 and the ͑x, y͒ 3 Խ 10. D ͕͑x, y͒ 0 ഛ y ഛ cos x, 0 ഛ x ഛ ͞2͖;
■ ■ ■ ■ ■ ■ ͑x, y͒ x
■ Exercise 12.
17. Find the moments of inertia I x , I y , I 0 for the lamina of 8. D is bounded by y sx, y 0, and x 1; ͑x, y͒ x ■ density at any point is proportional to the square of its distance
from the origin. 15. Find the moments of inertia I x , I y , I 0 for the lamina of 6. D is the triangular region with vertices ͑0, 0͒, ͑1, 1͒, ͑4, 0͒; ■ 12. Find the center of mass of the lamina in Exercise 11 if the but outside the circle x 2 ϩ y 2 1. Find the center of mass if
the density at any point is inversely proportional to its distance
from the origin. 5. D is the triangular region with vertices ͑0, 0͒, ͑2, 1͒, ͑0, 3͒; ■ quadrant. Find its center of mass if the density at any point is
proportional to its distance from the xaxis. 14. A lamina occupies the region inside the circle x 2 ϩ y 2 2y 3. D ͕͑x, y͒ 0 ഛ x ഛ 2, Ϫ1 ഛ y ഛ 1͖; ͑x, y͒ xy 2
4. 11. A lamina occupies the part of the disk x 2 ϩ y 2 ഛ 1 in the ﬁrst ■ ■ lower left corner placed at the origin. If the density of the blade
is ͑x, y͒ 1 ϩ 0.1x, is it more difﬁcult to rotate the blade
about the xaxis or the yaxis? 5E16(pp 10541063) 1/18/06 4:29 PM Page 1055 SECTION 16.6 SURFACE AREA CAS Խ ͑x, y͒ xy 20. D is enclosed by the cardioid r 1 ϩ cos ; ͑x, y͒ sx 2 ϩ y 2 ■ ■ ■ CAS ■ ■ ■ ■ ■ ■ ■ ■ ■ 21. A lamina with constant density ͑x, y͒ occupies a square with vertices ͑0, 0͒, ͑a, 0͒, ͑a, a͒, and ͑0, a͒. Find the moments
of inertia I x and I y and the radii of gyration x and y. 22. A lamina with constant density ͑x, y͒ occupies the region under the curve y sin x from x 0 to x . Find the
moments of inertia I x and I y and the radii of gyration x and y. 23. The joint density function for a pair of random variables X and Y is
f ͑x, y͒ ͭ Cx͑1 ϩ y͒ if 0 ഛ x ഛ 1, 0 ഛ y ഛ 2
0
otherwise (a) Find the value of the constant C.
(b) Find P͑X ഛ 1, Y ഛ 1͒.
(c) Find P͑X ϩ Y ഛ 1͒.
24. (a) Verify that f ͑x, y͒ ͭ 4xy if 0 ഛ x ഛ 1, 0 ഛ y ഛ 1
0
otherwise is a joint density function.
(b) If X and Y are random variables whose joint density function is the function f in part (a), ﬁnd
(i) P(X ജ 1 )
(ii) P(X ജ 1 , Y ഛ 1 )
2
2
2
(c) Find the expected values of X and Y.
25. Suppose X and Y are random variables with joint density function
f ͑x, y͒ ͭ 0.1eϪ͑0.5xϩ0.2y͒ if x ജ 0, y ജ 0
0
otherwise (a) Verify that f is indeed a joint density function.
(b) Find the following probabilities.
(i) P͑Y ജ 1͒
(ii) P͑X ഛ 2, Y ഛ 4͒
(c) Find the expected values of X and Y.
26. (a) A lamp has two bulbs of a type with an average lifetime of 1000 hours. Assuming that we can model the probability of
failure of these bulbs by an exponential density function  16.6 1055 with mean 1000, ﬁnd the probability that both of the
lamp’s bulbs fail within 1000 hours.
(b) Another lamp has just one bulb of the same type as in
part (a). If one bulb burns out and is replaced by a bulb
of the same type, ﬁnd the probability that the two bulbs fail
within a total of 1000 hours. 19–20  Use a computer algebra system to ﬁnd the mass, center of
mass, and moments of inertia of the lamina that occupies the region
D and has the given density function. 19. D ͕͑x, y͒ 0 ഛ y ഛ sin x, 0 ഛ x ഛ ͖ ; ❙❙❙❙ 27. Suppose that X and Y are independent random variables, where X is normally distributed with mean 45 and standard deviation
0.5 and Y is normally distributed with mean 20 and standard
deviation 0.1.
(a) Find P͑40 ഛ X ഛ 50, 20 ഛ Y ഛ 25͒.
(b) Find P͑4͑X Ϫ 45͒2 ϩ 100͑Y Ϫ 20͒2 ഛ 2͒.
28. Xavier and Yolanda both have classes that end at noon and they agree to meet every day after class. They arrive at the coffee
shop independently. Xavier’s arrival time is X and Yolanda’s
arrival time is Y , where X and Y are measured in minutes after
noon. The individual density functions are
f1͑x͒ ͭ eϪx if x ജ 0
0
if x Ͻ 0 f2͑ y͒ ͭ 1
50 0 y if 0 ഛ y ഛ 10
otherwise (Xavier arrives sometime after noon and is more likely to arrive
promptly than late. Yolanda always arrives by 12:10 P.M. and is
more likely to arrive late than promptly.) After Yolanda arrives,
she’ll wait for up to half an hour for Xavier, but he won’t wait
for her. Find the probability that they meet.
29. When studying the spread of an epidemic, we assume that the probability that an infected individual will spread the disease to
an uninfected individual is a function of the distance between
them. Consider a circular city of radius 10 mi in which the
population is uniformly distributed. For an uninfected individual at a ﬁxed point A͑x 0 , y0 ͒, assume that the probability function is given by
1
f ͑P͒ 20 ͓20 Ϫ d͑P, A͔͒ where d͑P, A͒ denotes the distance between P and A.
(a) Suppose the exposure of a person to the disease is the sum
of the probabilities of catching the disease from all members of the population. Assume that the infected people are
uniformly distributed throughout the city, with k infected
individuals per square mile. Find a double integral that
represents the exposure of a person residing at A.
(b) Evaluate the integral for the case in which A is the center of
the city and for the case in which A is located on the edge
of the city. Where would you prefer to live? Surface Area  In Section 17.6 we will deal with areas of
more general surfaces, called parametric
surfaces, and so it is possible to omit this section if that later section will be covered. In this section we apply double integrals to the problem of computing the area of a surface.
In Section 9.2 we found the area of a very special type of surface—a surface of revolution—by the methods of singlevariable calculus. Here we compute the area of a surface
with equation z f ͑x, y͒, the graph of a function of two variables. 5E16(pp 10541063) 1056 ❙❙❙❙ 1/18/06 4:30 PM CHAPTER 16 MULTIPLE INTEGRALS z ÎTij Pij
ÎS ij
S Îy 0 y (x i , yj ) Îx R ij D x Page 1056 Let S be a surface with equation z f ͑x, y͒, where f has continuous partial derivatives.
For simplicity in deriving the surface area formula, we assume that f ͑x, y͒ ജ 0 and the
domain D of f is a rectangle. We divide D into small rectangles Rij with area ⌬A ⌬x ⌬y.
If ͑x i, yj ͒ is the corner of Rij closest to the origin, let Pij ͑x i , yj, f ͑x i , yj͒͒ be the point on S
directly above it (see Figure 1). The tangent plane to S at Pij is an approximation to S near
Pij. So the area ⌬Tij of the part of this tangent plane (a parallelogram) that lies directly
above Rij is an approximation to the area ⌬Sij of the part of S that lies directly above Rij.
Thus, the sum ⌬Tij is an approximation to the total area of S, and this approximation
appears to improve as the number of rectangles increases. Therefore, we deﬁne the surface area of S to be ÎA
m m, n l ϱ i1 j1 FIGURE 1
z 0 Îx ij To ﬁnd a formula that is more convenient than Equation 1 for computational purposes,
we let a and b be the vectors that start at Pij and lie along the sides of the parallelogram
with area ⌬Tij. (See Figure 2.) Then ⌬Tij a ϫ b . Recall from Section 15.3 that fx ͑x i , yj ͒
and fy ͑x i , yj ͒ are the slopes of the tangent lines through Pij in the directions of a and b.
Therefore
a ⌬x i ϩ fx ͑x i , yj ͒ ⌬x k Pij
a n ͚ ͚ ⌬T A͑S͒ lim 1 b Խ ÎTij Îy Խ b ⌬y j ϩ fy ͑x i , yj ͒ ⌬y k y and
x Խ i
j
k
a ϫ b ⌬x 0 fx ͑xi , yj ͒ ⌬x
0 ⌬y fy ͑xi , yj ͒ ⌬y FIGURE 2 Խ Ϫfx ͑x i , yj ͒ ⌬x ⌬y i Ϫ fy ͑x i , yj ͒ ⌬x ⌬y j ϩ ⌬x ⌬y k
͓Ϫfx ͑x i , yj ͒i Ϫ fy ͑x i , yj ͒j ϩ k͔ ⌬A
Thus Խ Խ ⌬Tij a ϫ b s͓ fx ͑x i , yj ͔͒ 2 ϩ ͓ fy ͑x i , yj ͔͒ 2 ϩ 1 ⌬A From Deﬁnition 1 we then have
m A͑S͒ lim n ͚ ͚ ⌬T m, n l ϱ i1 j1
m lim ij n ͚ ͚ s͓ f ͑x , y ͔͒ m, n l ϱ i1 j1 x i j 2 ϩ ͓ fy ͑xi , yj ͔͒ 2 ϩ 1 ⌬A and by the deﬁnition of a double integral we get the following formula.
2 The area of the surface with equation z f ͑x, y͒, ͑x, y͒ ʦ D, where fx and fy
are continuous, is A͑S͒ yy s͓ fx ͑x, y͔͒ 2 ϩ ͓ fy ͑x, y͔͒ 2 ϩ 1 dA
D 5E16(pp 10541063) 1/18/06 4:31 PM Page 1057 SECTION 16.6 SURFACE AREA ❙❙❙❙ 1057 We will verify in Section 17.6 that this formula is consistent with our previous formula
for the area of a surface of revolution. If we use the alternative notation for partial derivatives, we can rewrite Formula 2 as follows: A͑S͒ 3 yy
D ͱ ͩ ͪ ͩ ͪ
Ѩz
Ѩx 1ϩ Ѩz
Ѩy 2 ϩ 2 dA Notice the similarity between the surface area formula in Equation 3 and the arc length
formula from Section 9.1:
L y b a ͱ ͩ ͪ 2 dy
dx 1ϩ dx EXAMPLE 1 Find the surface area of the part of the surface z x 2 ϩ 2y that lies above y the triangular region T in the xyplane with vertices ͑0, 0͒, ͑1, 0͒, and ͑1, 1͒. (1, 1) SOLUTION The region T is shown in Figure 3 and is described by
y=x Խ T ͕͑x, y͒ 0 ഛ x ഛ 1, 0 ഛ y ഛ x͖ T
(0, 0) Using Formula 2 with f ͑x, y͒ x 2 ϩ 2y, we get x (1, 0) FIGURE 3 A yy s͑2x͒2 ϩ ͑2͒2 ϩ 1 dA y 1 0 y x 0 s4x 2 ϩ 5 dy dx T z
1 1
y xs4x 2 ϩ 5 dx 1 ؒ 2 ͑4x 2 ϩ 5͒3͞2]1 12 (27 Ϫ 5s5 )
8
3
0
0 Figure 4 shows the portion of the surface whose area we have just computed.
EXAMPLE 2 Find the area of the part of the paraboloid z x 2 ϩ y 2 that lies under the y T x plane z 9. SOLUTION The plane intersects the paraboloid in the circle x 2 ϩ y 2 9, z 9. Therefore, FIGURE 4 the given surface lies above the disk D with center the origin and radius 3 (see Figure 5).
Using Formula 3, we have
z A 9 yy
D ͱ ͩ ͪ ͩ ͪ
1ϩ Ѩz
Ѩx 2 ϩ Ѩz
Ѩy 2 dA yy s1 ϩ ͑2x͒2 ϩ ͑2y͒2 dA
D yy s1 ϩ 4͑x 2 ϩ y 2 ͒ dA
D Converting to polar coordinates, we obtain
Ay D
x 3 FIGURE 5 y 2 0 y 3 0 s1 ϩ 4r 2 r dr d y 2 ( 1 ) 2͑1 ϩ 4r 2 ͒3͞2]3
0
8 3 2 0 d y 3 1
8 0 s1 ϩ 4r 2 ͑8r͒ dr
(37s37 Ϫ 1)
6 5E16(pp 10541063) ❙❙❙❙ 1058  4:32 PM Page 1058 CHAPTER 16 MULTIPLE INTEGRALS  16.6
1–12 1/18/06 Exercises
16. (a) Use the Midpoint Rule for double integrals with Find the area of the surface. 1. The part of the plane z 2 ϩ 3x ϩ 4y that lies above the rectangle ͓0, 5͔ ϫ ͓1, 4͔ CAS 2. The part of the plane 2x ϩ 5y ϩ z 10 that lies inside the cylinder x 2 ϩ y 2 9 3. The part of the plane 3x ϩ 2y ϩ z 6 that lies in the CAS ﬁrst octant
CAS triangle with vertices ͑0, 0͒, ͑0, 1͒, and ͑2, 1͒ rectangle with vertices ͑0, 0͒, ͑4, 0͒, ͑0, 2͒, and ͑4, 2͒ CAS the xyplane
2 between the cylinders x 2 ϩ y 2 1 and x 2 ϩ y 2 4 8. The surface z ͑x 3͞2 ϩy 3͞2 CAS Խ Խ Խ Խ 21. Show that the area of the part of the plane z ax ϩ by ϩ c x ϩy 1
2 that projects onto a region D in the xyplane with area A͑D͒ is
sa 2 ϩ b 2 ϩ 1 A͑D͒. 10. The part of the sphere x ϩ y ϩ z 4 that lies above the
2 2 plane z 1 22. If you attempt to use Formula 2 to ﬁnd the area of the top half 11. The part of the sphere x 2 ϩ y 2 ϩ z 2 a 2 that lies within the cylinder x 2 ϩ y 2 ax and above the xyplane 12. The part of the sphere x 2 ϩ y 2 ϩ z 2 4z that lies inside the paraboloid z x 2 ϩ y 2
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 13–14  Find the area of the surface correct to four decimal places
by expressing the area in terms of a single integral and using your
calculator to estimate the integral. 13. The part of the surface z eϪx 2 Ϫy 2 2 ■ ■ 2 ■ ■ ■ ■ ■ ■ ■ ■ [Hint: Project the surface onto y 2 ϩ z 2 1 intersects the cylinder x 2 ϩ z 2 1. Find the
area of this surface. cylinder x ϩ y 1 ■ 23. Find the area of the ﬁnite part of the paraboloid y x 2 ϩ z 2 24. The ﬁgure shows the surface created when the cylinder 14. The part of the surface z cos͑x 2 ϩ y 2 ͒ that lies inside the
2 of the sphere x 2 ϩ y 2 ϩ z 2 a 2, you have a slight problem
because the double integral is improper. In fact, the integrand
has an inﬁnite discontinuity at every point of the boundary
circle x 2 ϩ y 2 a 2. However, the integral can be computed
as the limit of the integral over the disk x 2 ϩ y 2 ഛ t 2 as
t l a Ϫ. Use this method to show that the area of a sphere of
radius a is 4 a 2.
cut off by the plane y 25.
the xzplane.] that lies above the disk x ϩy ഛ4
2 20. Find, to four decimal places, the area of the part of the surface z ͑1 ϩ x 2 ͒͑͞1 ϩ y 2 ͒ that lies above the square
x ϩ y ഛ 1. Illustrate by graphing this part of the surface. ͒, 0 ഛ x ഛ 1 , 0 ഛ y ഛ 1 9. The part of the surface z xy that lies within the cylinder
2 19. Find, to four decimal places, the area of the part of the surface z 1 ϩ x 2 y 2 that lies above the disk x 2 ϩ y 2 ഛ 1. 7. The part of the hyperbolic paraboloid z y Ϫ x that lies
2
3 Ϫ2 ഛ x ഛ 1 Ϫ1 ഛ y ഛ 1 Illustrate by graphing the surface. 6. The part of the paraboloid z 4 Ϫ x 2 Ϫ y 2 that lies above
2 18. Find the exact area of the surface z 1 ϩ x ϩ y ϩ x2 5. The part of the cylinder y 2 ϩ z 2 9 that lies above the ■ 17. Find the exact area of the surface z 1 ϩ 2x ϩ 3y ϩ 4y 2, 1 ഛ x ഛ 4, 0 ഛ y ഛ 1. 4. The part of the surface z 1 ϩ 3x ϩ 2y 2 that lies above the 2 m n 2 to estimate the area of the surface
z xy ϩ x 2 ϩ y 2, 0 ഛ x ഛ 2, 0 ഛ y ഛ 2.
(b) Use a computer algebra system to approximate the surface
area in part (a) to four decimal places. Compare with the
answer to part (a). ■ z 15. (a) Use the Midpoint Rule for double integrals (see Sec CAS tion 16.1) with four squares to estimate the surface area
of the portion of the paraboloid z x 2 ϩ y 2 that lies above
the square ͓0, 1͔ ϫ ͓0, 1͔.
(b) Use a computer algebra system to approximate the surface
area in part (a) to four decimal places. Compare with the
answer to part (a). x y 5E16(pp 10541063) 1/18/06 4:32 PM Page 1059 SECTION 16.7 TRIPLE INTEGRALS  16.7 ❙❙❙❙ 1059 Triple Integrals
Just as we deﬁned single integrals for functions of one variable and double integrals for
functions of two variables, so we can deﬁne triple integrals for functions of three variables.
Let’s ﬁrst deal with the simplest case where f is deﬁned on a rectangular box: Խ B ͕͑x, y, z͒ a ഛ x ഛ b, c ഛ y ഛ d, r ഛ z ഛ s͖ 1
z The ﬁrst step is to divide B into subboxes. We do this by dividing the interval ͓a, b͔ into
l subintervals ͓x iϪ1, x i ͔ of equal width ⌬x, dividing ͓c, d͔ into m subintervals of width ⌬y,
and dividing ͓r, s͔ into n subintervals of width ⌬z. The planes through the endpoints of
these subintervals parallel to the coordinate planes divide the box B into lmn subboxes B x Bi jk ͓x iϪ1, x i ͔ ϫ ͓yjϪ1, yj ͔ ϫ ͓zkϪ1, zk ͔ y which are shown in Figure 1. Each subbox has volume ⌬V ⌬x ⌬y ⌬z.
Then we form the triple Riemann sum
Bijk
l Îy m n ͚ ͚ ͚ f ͑x * , y * , z * ͒ ⌬V 2 Îz ijk ijk ijk i1 j1 k1 where the sample point ͑x i* , y i* , z i* ͒ is in Bi jk . By analogy with the deﬁnition of a double
jk
jk
jk
integral (16.1.5), we deﬁne the triple integral as the limit of the triple Riemann sums in (2). Îx z 3 Definition The triple integral of f over the box B is
l x y m n ͚ ͚ ͚ f ͑x * , y * , z * ͒ ⌬V yyy f ͑x, y, z͒ dV lim i jk l, m, n l ϱ i1 j1 k1 B i jk i jk if this limit exists.
FIGURE 1 Again, the triple integral always exists if f is continuous. We can choose the sample
point to be any point in the subbox, but if we choose it to be the point ͑x i , yj , zk ͒ we get a
simplerlooking expression for the triple integral:
l yyy f ͑x, y, z͒ dV
B m n ͚ ͚ ͚ f ͑x , y , z ͒ ⌬V lim l, m, n l ϱ i1 j1 k1 i j k Just as for double integrals, the practical method for evaluating triple integrals is to
express them as iterated integrals as follows.
4 Fubini’s Theorem for Triple Integrals If f is continuous on the rectangular box B ͓a, b͔ ϫ ͓c, d͔ ϫ ͓r, s͔, then
s d yyy f ͑x, y, z͒ dV y y y
r B c b a f ͑x, y, z͒ dx dy dz 5E16(pp 10541063) 1060 ❙❙❙❙ 1/18/06 4:33 PM Page 1060 CHAPTER 16 MULTIPLE INTEGRALS The iterated integral on the right side of Fubini’s Theorem means that we integrate ﬁrst
with respect to x (keeping y and z ﬁxed), then we integrate with respect to y (keeping z
ﬁxed), and ﬁnally we integrate with respect to z. There are ﬁve other possible orders in
which we can integrate, all of which give the same value. For instance, if we integrate with
respect to y, then z, and then x, we have
b s yyy f ͑x, y, z͒ dV y y y
a r d c f ͑x, y, z͒ dy dz dx B EXAMPLE 1 Evaluate the triple integral xxxB xyz 2 dV , where B is the rectangular box given by Խ B ͕͑x, y, z͒ 0 ഛ x ഛ 1, Ϫ1 ഛ y ഛ 2, 0 ഛ z ഛ 3͖
SOLUTION We could use any of the six possible orders of integration. If we choose to
integrate with respect to x, then y, and then z, we obtain yyy xyz 2 dV y 3 0 B y 3 0 y 3 0 2 y y
Ϫ1 y 2 Ϫ1 1 0 xyz dx dy dz
2 3 yy
0 yz 2
dy dz
2 3z 2
z3
dz
4
4 ͬ y 3 0 3 0 2 Ϫ1 ͫ ͬ
y 2z 2
4 ͫ ͬ
x 2 yz 2
2 x1 dy dz x0 y2 dz yϪ1 27
4 Now we deﬁne the triple integral over a general bounded region E in threedimensional space (a solid) by much the same procedure that we used for double integrals
(16.3.2). We enclose E in a box B of the type given by Equation 1. Then we deﬁne a
function F so that it agrees with f on E but is 0 for points in B that are outside E. By
deﬁnition, yyy f ͑x, y, z͒ dV yyy F͑x, y, z͒ dV
E B This integral exists if f is continuous and the boundary of E is “reasonably smooth.” The
triple integral has essentially the same properties as the double integral (Properties 6–9 in
Section 16.3).
We restrict our attention to continuous functions f and to certain simple types of regions.
A solid region E is said to be of type 1 if it lies between the graphs of two continuous functions of x and y, that is,
5 z E
z=u¡ (x, y)
0
x Խ E ͕͑x, y, z͒ ͑x, y͒ ʦ D, u 1͑x, y͒ ഛ z ഛ u 2͑x, y͖͒ z=u™ (x, y) D FIGURE 2 A type 1 solid region where D is the projection of E onto the xyplane as shown in Figure 2. Notice that the
upper boundary of the solid E is the surface with equation z u 2͑x, y͒, while the lower
boundary is the surface z u1͑x, y͒.
By the same sort of argument that led to (16.3.3), it can be shown that if E is a type 1
region given by Equation 5, then y 6 ͫ yyy f ͑x, y, z͒ dV yy y
E D u 2͑x, y͒ u1͑x, y͒ ͬ f ͑x, y, z͒ dz dA 5E16(pp 10541063) 1/18/06 4:34 PM Page 1061 SECTION 16.7 TRIPLE INTEGRALS z z=u™(x, y) Խ E ͕͑x, y, z͒ a ഛ x ഛ b, t1͑x͒ ഛ y ഛ t2͑x͒, u1͑x, y͒ ഛ z ഛ u 2͑x, y͖͒
z=u¡(x, y) x b 1061 The meaning of the inner integral on the right side of Equation 6 is that x and y are held
ﬁxed, and therefore u1͑x, y͒ and u 2͑x, y͒ are regarded as constants, while f ͑x, y, z͒ is integrated with respect to z.
In particular, if the projection D of E onto the xyplane is a type I plane region (as in
Figure 3), then E a ❙❙❙❙ and Equation 6 becomes 0 y=g¡(x) D y=g™(x) y
t2͑x͒ b yyy f ͑x, y, z͒ dV y y y 7 a E FIGURE 3 t1͑x͒ u 2͑x, y͒ u1͑x, y͒ f ͑x, y, z͒ dz dy dx A type 1 solid region If, on the other hand, D is a type II plane region (as in Figure 4), then
z Խ E ͕͑x, y, z͒ c ഛ y ഛ d, h1͑y͒ ഛ x ഛ h2͑y͒, u1͑x, y͒ ഛ z ഛ u 2͑x, y͖͒ z=u™(x, y)
E z=u¡(x, y) and Equation 6 becomes x=h¡(y)
0 c d yyy f ͑x, y, z͒ dV y y 8 d c E y x h2͑ y͒ h1͑ y͒ y u 2͑x, y͒ u1͑x, y͒ f ͑x, y, z͒ dz dx dy D
x=h™(y) EXAMPLE 2 Evaluate xxxE z dV , where E is the solid tetrahedron bounded by the four
planes x 0, y 0, z 0, and x ϩ y ϩ z 1. FIGURE 4 Another type 1 solid region z
(0, 0, 1) z=1xy
E
(0, 1, 0) 0 y (1, 0, 0)
x SOLUTION When we set up a triple integral it’s wise to draw two diagrams: one of
the solid region E (see Figure 5) and one of its projection D on the xyplane (see
Figure 6). The lower boundary of the tetrahedron is the plane z 0 and the upper
boundary is the plane x ϩ y ϩ z 1 (or z 1 Ϫ x Ϫ y), so we use u1͑x, y͒ 0 and
u 2͑x, y͒ 1 Ϫ x Ϫ y in Formula 7. Notice that the planes x ϩ y ϩ z 1 and z 0
intersect in the line x ϩ y 1 (or y 1 Ϫ x) in the xyplane. So the projection of E is
the triangular region shown in Figure 6, and we have
9 Խ E ͕͑x, y, z͒ 0 ഛ x ഛ 1, 0 ഛ y ഛ 1 Ϫ x, 0 ഛ z ഛ 1 Ϫ x Ϫ y͖ z=0 This description of E as a type 1 region enables us to evaluate the integral as follows: FIGURE 5 yyy y z dV y 1 1Ϫx y y 0 0 1ϪxϪy 0 z dz dy dx yy
0 E 1 1y
2 y=1x 1 0 D 1
2 y 1 0 0 y=0 1 y 1Ϫx 0 ͫ 0 ͬ
ͫ ͑1 Ϫ x Ϫ y͒3
Ϫ
3 1 y ͑1 Ϫ x͒3 dx
6
FIGURE 6 1Ϫx 0 ͫͬ
z2
2 z1ϪxϪy z0 ͑1 Ϫ x Ϫ y͒2 dy dx x
1 1 1
6 y1Ϫx dx y0 Ϫ ͑1 Ϫ x͒4
4 ͬ 1 0 1
24 dy dx 5E16(pp 10541063) 1062 ❙❙❙❙ 1/18/06 4:35 PM Page 1062 CHAPTER 16 MULTIPLE INTEGRALS z A solid region E is of type 2 if it is of the form y E x Խ E ͕͑x, y, z͒ ͑y, z͒ ʦ D, u1͑y, z͒ ഛ x ഛ u 2͑y, z͖͒ D 0 where, this time, D is the projection of E onto the yzplane (see Figure 7). The back surface is x u1͑y, z͒, the front surface is x u 2͑y, z͒, and we have x=u¡(y, z) E x=u™(y, z) D u 2͑ y, z͒ u1͑ y, z͒ ͬ f ͑x, y, z͒ dx dA Finally, a type 3 region is of the form FIGURE 7 Խ E ͕͑x, y, z͒ ͑x, z͒ ʦ D, u1͑x, z͒ ഛ y ഛ u 2͑x, z͖͒ A type 2 region where D is the projection of E onto the xzplane, y u1͑x, z͒ is the left surface, and
y u 2͑x, z͒ is the right surface (see Figure 8). For this type of region we have z y=u™(x, z) E D u 2͑x, z͒ u1͑x, z͒ ͬ f ͑x, y, z͒ dy dA In each of Equations 10 and 11 there may be two possible expressions for the integral
depending on whether D is a type I or type II plane region (and corresponding to Equations 7 and 8). 0 y=u¡(x, z) ͫ yyy f ͑x, y, z͒ dV yy y 11 E D ͫ yyy f ͑x, y, z͒ dV yy y 10 y x EXAMPLE 3 Evaluate FIGURE 8 A type 3 region xxxE sx 2 ϩ z 2 dV , where E is the region bounded by the paraboloid y x ϩ z and the plane y 4.
2 2 SOLUTION The solid E is shown in Figure 9. If we regard it as a type 1 region, then we
need to consider its projection D1 onto the xyplane, which is the parabolic region in
Figure 10. (The trace of y x 2 ϩ z 2 in the plane z 0 is the parabola y x 2.)
y z y=≈[email protected] y=4
D¡ E Visual 16.7 illustrates how solid regions
(including the one in Figure 9) project
onto coordinate planes. y=≈ 0
4 y
0 x FIGURE 9 FIGURE 10 Region of integration x Projection on xyplane From y x 2 ϩ z 2 we obtain z Ϯsy Ϫ x 2, so the lower boundary surface of E is
z Ϫsy Ϫ x 2 and the upper surface is z sy Ϫ x 2. Therefore, the description of E as
a type 1 region is Խ E {͑x, y, z͒ Ϫ2 ഛ x ഛ 2, x 2 ഛ y ഛ 4, Ϫsy Ϫ x 2 ഛ z ഛ sy Ϫ x 2 }
and so we obtain yyy sx
E 2 ϩ z 2 dV y 2 Ϫ2 4 syϪx 2 x ϪsyϪx 2 y y
2 sx 2 ϩ z 2 dz dy dx 5E16(pp 10541063) 1/18/06 4:36 PM Page 1063 SECTION 16.7 TRIPLE INTEGRALS z ≈[email protected]=4
D£
_2 0 2 x ❙❙❙❙ 1063 Although this expression is correct, it is extremely difﬁcult to evaluate. So let’s
instead consider E as a type 3 region. As such, its projection D3 onto the xzplane is the
disk x 2 ϩ z 2 ഛ 4 shown in Figure 11.
Then the left boundary of E is the paraboloid y x 2 ϩ z 2 and the right boundary is
the plane y 4, so taking u1͑x, z͒ x 2 ϩ z 2 and u 2͑x, z͒ 4 in Equation 11, we have yyy sx 2 ϩ z 2 dV yy E D3 FIGURE 11 ͫy 4 x 2ϩz 2 ͬ sx 2 ϩ z 2 dy dA yy ͑4 Ϫ x 2 Ϫ z 2 ͒sx 2 ϩ z 2 dA Projection on xzplane D3 Although this integral could be written as
2 s4Ϫx 2 Ϫ2 Ϫs4Ϫx 2 y y
 The most difﬁcult step in evaluating a triple
integral is setting up an expression for the region
of integration (such as Equation 9 in Example 2).
Remember that the limits of integration in the
inner integral contain at most two variables, the
limits of integration in the middle integral contain at most one variable, and the limits of integration in the outer integral must be constants. ͑4 Ϫ x 2 Ϫ z 2 ͒sx 2 ϩ z 2 dz dx it’s easier to convert to polar coordinates in the xzplane: x r cos , z r sin . This
gives yyy sx 2 ϩ z 2 dV yy ͑4 Ϫ x 2 Ϫ z 2 ͒sx 2 ϩ z 2 dA E D3 y 2 0 2 y ͫ 2 0 ͑4 Ϫ r 2 ͒r r dr d y 4r 3
r5
Ϫ
3
5 2 0 ͬ 2 0 2 d y ͑4r 2 Ϫ r 4 ͒ dr
0 128
15 Applications of Triple Integrals
Recall that if f ͑x͒ ജ 0, then the single integral xab f ͑x͒ dx represents the area under the
curve y f ͑x͒ from a to b, and if f ͑x, y͒ ജ 0, then the double integral xxD f ͑x, y͒ dA represents the volume under the surface z f ͑x, y͒ and above D. The corresponding interpretation of a triple integral xxxE f ͑x, y, z͒ dV , where f ͑x, y, z͒ ജ 0, is not very useful
because it would be the “hypervolume” of a fourdimensional object and, of course, that
is very difﬁcult to visualize. (Remember that E is just the domain of the function f ; the
graph of f lies in fourdimensional space.) Nonetheless, the triple integral xxxE f ͑x, y, z͒ dV
can be interpreted in different ways in different physical situations, depending on the physical interpretations of x, y, z and f ͑x, y, z͒.
Let’s begin with the special case where f ͑x, y, z͒ 1 for all points in E. Then the triple
integral does represent the volume of E:
V͑E͒ yyy dV 12 E For example, you can see this in the case of a type 1 region by putting f ͑x, y, z͒ 1 in
Formula 6: ͫ yyy 1 dV yy y
E D u 2͑x, y͒ u1͑x, y͒ ͬ dz dA yy ͓u 2͑x, y͒ Ϫ u1͑x, y͔͒ dA
D 5E16(pp 10641073) 1064 ❙❙❙❙ 1/18/06 4:42 PM Page 1064 CHAPTER 16 MULTIPLE INTEGRALS and from Section 16.3 we know this represents the volume that lies between the surfaces
z u1͑x, y͒ and z u 2͑x, y͒.
z EXAMPLE 4 Use a triple integral to ﬁnd the volume of the tetrahedron T bounded by the
planes x ϩ 2y ϩ z 2, x 2y, x 0, and z 0. (0, 0, 2) SOLUTION The tetrahedron T and its projection D on the xyplane are shown in Figures 12
and 13. The lower boundary of T is the plane z 0 and the upper boundary is the plane
x ϩ 2y ϩ z 2, that is, z 2 Ϫ x Ϫ 2y. Therefore, we have x+2y+z=2 x=2y
T V͑T͒ yyy dV y y y 1Ϫx͞2 x͞2 y 2ϪxϪ2y 0 dz dy dx T (0, 1, 0) 0 1 0 y 1 ”1, 2 , 0’ 1 0 y 1Ϫx͞2 x͞2 ͑2 Ϫ x Ϫ 2y͒ dy dx 1
3 x by the same calculation as in Example 4 in Section 16.3.
(Notice that it is not necessary to use triple integrals to compute volumes. They simply give an alternative method for setting up the calculation.) FIGURE 12
y
1 x+2y=2
x
”or y=1 ’
2 All the applications of double integrals in Section 16.5 can be immediately extended to
triple integrals. For example, if the density function of a solid object that occupies the
region E is ͑x, y, z͒, in units of mass per unit volume, at any given point ͑x, y, z͒, then its
mass is 1
”1, 2 ’ D
x y= 2
0 FIGURE 13 1 x m yyy ͑x, y, z͒ dV 13 E and its moments about the three coordinate planes are
14 Myz yyy x ͑x, y, z͒ dV Mxz yyy y ͑x, y, z͒ dV E E Mx y yyy z ͑x, y, z͒ dV
E The center of mass is located at the point ͑x, y, z͒, where
x 15 Myz
m y Mxz
m z Mxy
m If the density is constant, the center of mass of the solid is called the centroid of E. The
moments of inertia about the three coordinate axes are
16 Ix yyy ͑y 2 ϩ z 2 ͒ ͑x, y, z͒ dV
E Iy yyy ͑x 2 ϩ z 2 ͒ ͑x, y, z͒ dV
E Iz yyy ͑x 2 ϩ y 2 ͒ ͑x, y, z͒ dV
E As in Section 16.5, the total electric charge on a solid object occupying a region E and 5E16(pp 10641073) 1/18/06 4:42 PM Page 1065 SECTION 16.7 TRIPLE INTEGRALS ❙❙❙❙ 1065 having charge density ͑x, y, z͒ is
Q yyy ͑x, y, z͒ dV
E If we have three continuous random variables X, Y, and Z, their joint density function
is a function of three variables such that the probability that ͑X, Y, Z͒ lies in E is
P͑͑X, Y, Z͒ ʦ E͒ yyy f ͑x, y, z͒ dV
E In particular,
P͑a ഛ X ഛ b, c ഛ Y ഛ d, r ഛ Z ഛ s͒ y b a d yy
c s r f ͑x, y, z͒ dz dy dx The joint density function satisﬁes
ϱ ϱ ϱ Ϫϱ f ͑x, y, z͒ ജ 0 Ϫϱ Ϫϱ y y y f ͑x, y, z͒ dz dy dx 1 EXAMPLE 5 Find the center of mass of a solid of constant density that is bounded by the
parabolic cylinder x y 2 and the planes x z, z 0, and x 1.
SOLUTION The solid E and its projection onto the xyplane are shown in Figure 14. The
lower and upper surfaces of E are the planes z 0 and z x, so we describe E as a
type 1 region: Խ E ͕͑x, y, z͒ Ϫ1 ഛ y ഛ 1, y 2 ഛ x ഛ 1, 0 ഛ z ഛ x͖
y z x=¥ z=x
E D 0 x 0
x y 1 FIGURE 14 Then, if the density is ͑x, y, z͒ , the mass is
m yyy dV y
E y 1 Ϫ1
2 1 Ϫ1 y y 1 Ϫ1 1 y2 1 y y
y2 x 0 x dx dy dz dx dy y 1 Ϫ1 ͫͬ
x2
2 1 x1 dy xy 2 ͑1 Ϫ y 4 ͒ dy y ͑1 Ϫ y 4 ͒ dy ͫ ͬ yϪ y5
5 0 1 0 4
5 x=1 5E16(pp 10641073) ❙❙❙❙ 1066 1/18/06 4:42 PM Page 1066 CHAPTER 16 MULTIPLE INTEGRALS Because of the symmetry of E and about the xzplane, we can immediately say that
Mxz 0 and, therefore, y 0. The other moments are
Myz yyy x dV y Ϫ1 E y 1 y Ϫ1 2
3 1 y y 1 0
3 1 Ϫ1 2 y y
1 y ͑x, y, z͒ 5. 0 3 1 yyy
0 ■ 0 ■ 7–16 7. xϩz 0  4. 6xz dy dx dz s1Ϫz 0 2 x 0 ͩ 4
7 0 z dz dx dy zx z2
2 1 y7
yϪ
7 dx dy z0
2 1 y y
Ϫ1 1 y 2 x 2 dx dy 2
7 Myz Mxz Mxy
,
,
m
m
m ͪ 5
( 5 , 0, 14 )
7 2 ze y dx dz dy ■ ■ ■ 6.
■ ■ where E is bounded by the planes x 0, y 0,
z 0, and 2x ϩ 2y ϩ z 4 xxxE xy dV ,
xxxE xz dV , 13. xxxE x 2e y dV , 14. Evaluate the iterated integral.
z 1 ͫͬ y dy xy 2 xxxE y dV , xxxE ͑x ϩ 2y͒ dV , where E is bounded by the parabolic cylinder using three different orders of integration. 0 1 y y ͫͬ
ͫ ͬ 12. Խ E ͕͑x, y, z͒ Ϫ1 ഛ x ഛ 1, 0 ഛ y ഛ 2, 0 ഛ z ഛ 1͖ yyy Ϫ1 x3
3 11. 2. Evaluate the integral xxxE ͑xz Ϫ y 3͒ dV , where 3. y x1 1 10. respect to z , then x, and then y.  x dz dx dy Exercises 1. Evaluate the integral in Example 1, integrating ﬁrst with 3–6 1 ͑1 Ϫ y 6 ͒ dy Therefore, the center of mass is  16.7 0 2
͑1 Ϫ y ͒ dy
3 1 0 x 6 Ϫ1 y y 2 x dx dy 2 E 1 y y 2 Mxy yyy z dV y 1 1 2x yy y
0 1 x z y 0 0 yyy
0 ■ y 0 2xyz dz dy dx ■ where E is the solid tetrahedron with vertices
͑0, 0, 0͒, ͑0, 1, 0͒, ͑1, 1, 0͒, and ͑0, 1, 1͒
where E is bounded by the parabolic cylinder
z 1 Ϫ y 2 and the planes z 0, x 1, and x Ϫ1
y x 2 and the planes x z, x y, and z 0 2 zeϪy dx dy dz
■ where E is the solid tetrahedron with vertices
͑0, 0, 0͒, ͑1, 0, 0͒, ͑0, 2, 0͒, and ͑0, 0, 3͒ 15.
■ Evaluate the triple integral. xxxE 2x dV , where Խ xxxE x dV , where E is bounded by the paraboloid x 4y 2 ϩ 4z 2
and the plane x 4 ■ 16. xxxE z dV , where E is bounded by the cylinder y 2 ϩ z 2 9 and
the planes x 0, y 3x, and z 0 in the ﬁrst octant ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ E {͑x, y, z͒ 0 ഛ y ഛ 2, 0 ഛ x ഛ s4 Ϫ y 2, 0 ഛ z ഛ y} 9. xxxE yz cos͑x 5 ͒ dV , where 17–20 E ͕͑x, y, z͒ 0 ഛ x ഛ 1, 0 ഛ y ഛ x, x ഛ z ഛ 2x͖ 8. 17. The tetrahedron enclosed by the coordinate planes and the where E lies under the plane z 1 ϩ x ϩ y
and above the region in the xyplane bounded by the curves
y sx, y 0, and x 1 18. The solid bounded by the cylinder y x 2 and the planes Խ xxxE 6xy dV ,  Use a triple integral to ﬁnd the volume of the given solid. plane 2x ϩ y ϩ z 4
z 0, z 4, and y 9 5E16(pp 10641073) 1/18/06 4:42 PM Page 1067 SECTION 16.7 TRIPLE INTEGRALS 19. The solid enclosed by the cylinder x 2 ϩ y 2 9 and the planes ❙❙❙❙ 1067 31. The ﬁgure shows the region of integration for the integral y ϩ z 5 and z 1 1 20. The solid enclosed by the paraboloid x y 2 ϩ z 2 and the 0 plane x 16 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 1 yy y
■ sx cut from the cylinder y 2 ϩ z 2 1 by the planes y x and
x 1 as a triple integral.
(b) Use either the Table of Integrals (on the back Reference
Pages) or a computer algebra system to ﬁnd the exact value
of the triple integral in part (a). z
1 z=1y 22. (a) In the Midpoint Rule for triple integrals we use a triple Riemann sum to approximate a triple integral over a box
B, where f ͑x, y, z͒ is evaluated at the center ͑ xi , yj , zk ͒
of the box Bijk. Use the Midpoint Rule to estimate
2
2
2
xxxB eϪx Ϫy Ϫz dV , where B is the cube deﬁned by
0 ഛ x ഛ 1, 0 ഛ y ഛ 1, 0 ഛ z ഛ 1. Divide B into eight
cubes of equal size.
(b) Use a computer algebra system to approximate the integral
in part (a) correct to two decimal places. Compare with the
answer to part (a). CAS 0
1
x 32. The ﬁgure shows the region of integration for the integral
1 yy
0 1Ϫx 2 0 1 0 Խ B ͕͑x, y, z͒ 0 ഛ x ഛ 4, 0 ഛ y ഛ 2, 0 ഛ z ഛ 1͖
■ ■ ■ ■ ■ ■ ■ 1
■ ■ x 25–26  Sketch the solid whose volume is given by the iterated 26. 1 1Ϫx yy y
0 0 2 2Ϫy yy y
0 ■ 0 ■ 2Ϫ2z 0
4Ϫy 2 0 dy dz dx 1 34. yy y ■ ■ ■ ■ ■ ■ ■ ■ ■ 0 y x2 1 0 0 ■ 1 ■ f ͑x, y, z͒ dz dx dy 0 y 0
■ f ͑x, y, z͒ dz dy dx
■ ■ ■ ■ ■ ■ ■ ■  y 0, 27. x 2 ϩ z 2 4, 35–38  Find the mass and center of mass of the solid E with the
given density function . y6 28. z 0, x 0, y 2, 29. z 0, z y, 35. E is the solid of Exercise 9; x 1Ϫy z y Ϫ 2x ■ ■ ■ ■ ͑x, y, z͒ 2 36. E is bounded by the parabolic cylinder z 1 Ϫ y 2 and the planes x ϩ z 1, x 0, and z 0; ͑x, y, z͒ 4 2 30. 9x 2 ϩ 4y 2 ϩ z 2 1
■ y yyy dx dz dy Express the integral xxxE f ͑x, y, z͒ dV as an iterated integral in six different ways, where E is the solid bounded by the
given surfaces.
27–30 1 33. ■ y y=1x 33–34  Write ﬁve other iterated integrals that are equal to the
given iterated integral. integral.
25. f ͑x, y, z͒ dy dz dx z=1≈ xxxB sin͑xy 2z 3͒ dV , where
■ 1Ϫx 0 z Խ ■ y Rewrite this integral as an equivalent iterated integral in the
ﬁve other orders. 1
dV , where
23. xxxB
ln͑1 ϩ x ϩ y ϩ z͒
B ͕͑x, y, z͒ 0 ഛ x ഛ 4, 0 ഛ y ഛ 8, 0 ഛ z ഛ 4͖ ■ y y=œ„
x 23–24  Use the Midpoint Rule for triple integrals (Exercise 22)
to estimate the value of the integral. Divide B into eight subboxes
of equal size. 24. f ͑x, y, z͒ dz dy dx Rewrite this integral as an equivalent iterated integral in the
ﬁve other orders. 21. (a) Express the volume of the wedge in the ﬁrst octant that is CAS 1Ϫy 0 37. E is the cube given by 0 ഛ x ഛ a, 0 ഛ y ഛ a, 0 ഛ z ഛ a ;
■ ■ ■ ■ ■ ■ ■ ͑x, y, z͒ x 2 ϩ y 2 ϩ z 2 ■ 5E16(pp 10641073) 1068 ❙❙❙❙ 1/18/06 4:42 PM Page 1068 CHAPTER 16 MULTIPLE INTEGRALS 38. E is the tetrahedron bounded by the planes x 0, y 0, 45. The joint density function for random variables X , Y , and Z is z 0, x ϩ y ϩ z 1; ͑x, y, z͒ y ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ f ͑x, y, z͒ Cxyz if 0 ഛ x ഛ 2, 0 ഛ y ഛ 2, 0 ഛ z ഛ 2, and
f ͑x, y, z͒ 0 otherwise.
(a) Find the value of the constant C.
(b) Find P͑X ഛ 1, Y ഛ 1, Z ഛ 1͒.
(c) Find P͑X ϩ Y ϩ Z ഛ 1͒. ■ 39–40  Set up, but do not evaluate, integral expressions for
(a) the mass, (b) the center of mass, and (c) the moment of inertia
about the z axis. 46. Suppose X , Y , and Z are random variables with joint density ͑x, y, z͒ sx 2 ϩ y 2 39. The solid of Exercise 19; function f ͑x, y, z͒ CeϪ͑0.5xϩ0.2yϩ0.1z͒ if x ജ 0, y ജ 0, z ജ 0,
and f ͑x, y, z͒ 0 otherwise.
(a) Find the value of the constant C.
(b) Find P͑X ഛ 1, Y ഛ 1͒.
(c) Find P͑X ഛ 1, Y ഛ 1, Z ഛ 1͒. 40. The hemisphere x ϩ y ϩ z ഛ 1, z ജ 0;
2 2 2 ͑x, y, z͒ sx 2 ϩ y 2 ϩ z 2 ■ CAS ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 41. Let E be the solid in the ﬁrst octant bounded by the cylinder x 2 ϩ y 2 1 and the planes y z, x 0, and z 0 with the
density function ͑x, y, z͒ 1 ϩ x ϩ y ϩ z. Use a computer
algebra system to ﬁnd the exact values of the following quantities for E.
(a) The mass
(b) The center of mass
(c) The moment of inertia about the zaxis
CAS 42. If E is the solid of Exercise 16 with density function ͑x, y, z͒ x 2 ϩ y 2, ﬁnd the following quantities, correct
to three decimal places.
(a) The mass
(b) The center of mass
(c) The moment of inertia about the zaxis 47–48  The average value of a function f ͑x, y, z͒ over a solid
region E is deﬁned to be fave yyy f ͑x, y, z͒ dV
E where V͑E ͒ is the volume of E. For instance, if is a density function, then ave is the average density of E.
47. Find the average value of the function f ͑x, y, z͒ xyz over the cube with side length L that lies in the ﬁrst octant with one vertex at the origin and edges parallel to the coordinate axes.
48. Find the average value of the function f ͑x, y, z͒ x 2 z ϩ y 2 z over the region enclosed by the paraboloid z 1 Ϫ x 2 Ϫ y 2
and the plane z 0. 43. Find the moments of inertia for a cube of constant density k and side length L if one vertex is located at the origin and three
edges lie along the coordinate axes. 1
V͑E͒ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 49. Find the region E for which the triple integral 44. Find the moments of inertia for a rectangular brick with dimen yyy ͑1 Ϫ x sions a, b, and c, mass M , and constant density if the center of
the brick is situated at the origin and the edges are parallel to
the coordinate axes. 2 Ϫ 2y 2 Ϫ 3z 2 ͒ dV E is a maximum. DISCOVERY PROJECT
Volumes of Hyperspheres
In this project we ﬁnd formulas for the volume enclosed by a hypersphere in ndimensional
space.
1. Use a double integral and trigonometric substitution, together with Formula 64 in the Table of Integrals, to ﬁnd the area of a circle with radius r.
2. Use a triple integral and trigonometric substitution to ﬁnd the volume of a sphere with radius r.
3. Use a quadruple integral to ﬁnd the hypervolume enclosed by the hypersphere
x 2 ϩ y 2 ϩ z 2 ϩ w 2 r 2 in ( .4 ޒUse only trigonometric substitution and the reduction formulas for x sinnx dx or x cosnx dx.)
4. Use an ntuple integral to ﬁnd the volume enclosed by a hypersphere of radius r in ndimensional space ޒn. [Hint: The formulas are different for n even and n odd.] ■ 5E16(pp 10641073) 1/18/06 4:42 PM Page 1069 SECTION 16.8 TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES  16.8 ❙❙❙❙ 1069 Triple Integrals in Cylindrical and Spherical Coordinates
We saw in Section 16.4 that some double integrals are easier to evaluate using polar coordinates. In this section we see that some triple integrals are easier to evaluate using cylindrical or spherical coordinates. Cylindrical Coordinates
Recall from Section 13.7 that the cylindrical coordinates of a point P are ͑r, , z͒, where r,
, and z are shown in Figure 1. Suppose that E is a type 1 region whose projection D on
the xyplane is conveniently described in polar coordinates (see Figure 2). In particular,
suppose that f is continuous and z
P(r, ¨, z) z 0 ¨ Խ E ͕͑x, y, z͒ ͑x, y͒ ʦ D, u1͑x, y͒ ഛ z ഛ u 2͑x, y͖͒ y r where D is given in polar coordinates by x Խ D ͕͑r, ͒ ␣ ഛ ഛ , h1͑ ͒ ഛ r ഛ h 2͑ ͖͒ FIGURE 1 z z=u™(x, y) z=u¡(x, y) 0 r=h¡(¨ ) y D ¨=a x FIGURE 2 ¨=b
r=h™(¨ ) We know from Equation 16.7.6 that ͫ yyy f ͑x, y, z͒ dV yy y 1 E D u 2͑x, y͒ u1͑x, y͒ ͬ f ͑x, y, z͒ dz dA But we also know how to evaluate double integrals in polar coordinates. In fact, combining Equation 1 with Equation 16.4.3, we obtain
z 2  yyy f ͑x, y, z͒ dV y␣ y
E h1͑ ͒ y u 2͑r cos , r sin ͒ u1͑r cos , r sin ͒ f ͑r cos , r sin , z͒ r dz dr d dz d¨
r
r d¨ h2͑ ͒ dr FIGURE 3 Volume element in cylindrical
coordinates: dV=r dz dr d¨ Formula 2 is the formula for triple integration in cylindrical coordinates. It says that
we convert a triple integral from rectangular to cylindrical coordinates by writing
x r cos , y r sin , leaving z as it is, using the appropriate limits of integration for
z, r, and , and replacing dV by r dz dr d. (Figure 3 shows how to remember this.) It is
worthwhile to use this formula when E is a solid region easily described in cylindrical
coordinates, and especially when the function f ͑x, y, z͒ involves the expression x 2 ϩ y 2. 5E16(pp 10641073) 1070 ❙❙❙❙ 1/18/06 4:42 PM Page 1070 CHAPTER 16 MULTIPLE INTEGRALS EXAMPLE 1 A solid E lies within the cylinder x 2 ϩ y 2 1, below the plane z 4, and z z=4 above the paraboloid z 1 Ϫ x 2 Ϫ y 2. (See Figure 4.) The density at any point is proportional to its distance from the axis of the cylinder. Find the mass of E. (0, 0, 4) SOLUTION In cylindrical coordinates the cylinder is r 1 and the paraboloid is z 1 Ϫ r 2, so we can write Խ E ͕͑r, , z͒ 0 ഛ ഛ 2, 0 ഛ r ഛ 1, 1 Ϫ r 2 ഛ z ഛ 4͖
[email protected] Since the density at ͑x, y, z͒ is proportional to the distance from the zaxis, the density
function is
f ͑x, y, z͒ Ksx 2 ϩ y 2 Kr y where K is the proportionality constant. Therefore, from Formula 16.7.13, the mass of
E is (0, 0, 1) 0 (1, 0, 0)
x m yyy Ksx 2 ϩ y 2 dV y FIGURE 4 2 0 1 y y
0 4 1Ϫr 2 ͑Kr͒ r dz dr d E y 2 0 y 1 0 Kr 2 ͓4 Ϫ ͑1 Ϫ r 2 ͔͒ dr d K y ͫ 2K r 3 ϩ EXAMPLE 2 Evaluate s4Ϫx 2 Ϫ2 Ϫs4Ϫx 2 0 y 2 sx 2ϩy 2 0 12K
5
͑x 2 ϩ y 2 ͒ dz dy dx. Խ z=2 E {͑x, y, z͒ Ϫ2 ഛ x ഛ 2, Ϫs4 Ϫ x 2 ഛ y ഛ s4 Ϫ x 2, sx 2 ϩ y 2 ഛ z ഛ 2} 2 z=œ„„„„„
≈+¥ x 1 1 d y ͑3r 2 ϩ r 4 ͒ dr SOLUTION This iterated integral is a triple integral over the solid region z 2 2 y y ͬ r5
5 2 0 2 and the projection of E onto the xyplane is the disk x 2 ϩ y 2 ഛ 4. The lower surface of
E is the cone z sx 2 ϩ y 2 and its upper surface is the plane z 2. (See Figure 5.) This
region has a much simpler description in cylindrical coordinates: Խ E ͕͑r, , z͒ 0 ഛ ഛ 2, 0 ഛ r ഛ 2, r ഛ z ഛ 2͖
y Therefore, we have FIGURE 5 2 s4Ϫx 2 Ϫ2 Ϫs4Ϫx 2 y y y 2 sx 2ϩy 2 ͑x 2 ϩ y 2 ͒ dz dy dx yyy ͑x 2 ϩ y 2 ͒ dV
E y 2 2 y y 0 y 0 2 r 2 r dz dr d 2 d y r 3͑2 Ϫ r͒ dr 0 2 2 r 0 [ 1
2 r4 Ϫ 1r5
5 2
0 16
5 Spherical Coordinates
In Section 13.7 we deﬁned the spherical coordinates ͑ , , ͒ of a point (see Figure 6) and
we demonstrated the following relationships between rectangular coordinates and spherical coordinates:
3 x sin cos y sin sin z cos 5E16(pp 10641073) 1/18/06 4:42 PM Page 1071 SECTION 16.8 TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES z ❙❙❙❙ 1071 In this coordinate system the counterpart of a rectangular box is a spherical wedge Խ P(∏, ¨, ˙) E ͕͑ , , ͒ a ഛ ഛ b, ␣ ഛ ഛ , c ഛ ഛ d͖ ∏ ˙ where a ജ 0,  Ϫ ␣ ഛ 2, and d Ϫ c ഛ . Although we deﬁned triple integrals by dividing solids into small boxes, it can be shown that dividing a solid into small spherical
wedges always gives the same result. So we divide E into smaller spherical wedges Eijk by
means of equally spaced spheres i , halfplanes j , and halfcones k .
Figure 7 shows that Eijk is approximately a rectangular box with dimensions ⌬, i ⌬ (arc
of a circle with radius i , angle ⌬), and i sin k ⌬ (arc of a circle with radius i sin k,
angle ⌬ ). So an approximation to the volume of Eijk is given by 0
y ¨
x FIGURE 6 Spherical coordinates of P ⌬Vijk Ϸ ͑⌬͒͑ i ⌬͒͑ i sin k ⌬ ͒ 2i sin k ⌬ ⌬ ⌬ z ∏ i sin ˙ k Î¨ In fact, it can be shown, with the aid of the Mean Value Theorem (Exercise 39), that the
volume of Eijk is given exactly by Î∏ ෂ ෂ
⌬Vijk i2 sin k ⌬ ⌬ ⌬ ˙k ෂ
ෂ
ෂ
* * *
where ͑ i , j , k ͒ is some point in Eijk . Let ͑x ijk, y ijk , z ijk ͒ be the rectangular coordinates of
this point. Then Î˙ x l ∏ i Î˙ 0 Î¨ ri=∏ i sin ˙ k y yyy f ͑x, y, z͒ dV
E
l ri Î¨=∏ i sin ˙ k Î¨
FIGURE 7 lim m n ͚ ͚ ͚ f ͑x * , y * , z * ͒ ⌬V lim l, m, n l ϱ i1 j1 k1 m n ͚ ͚ ͚ f ͑ ෂ l, m, n l ϱ i1 j1 k1 ෂ i ijk ijk ෂ ijk ijk ෂ ෂ ෂ ෂ ෂ
ෂ
ෂ
sin k cos j, i sin k sin j , i cos k ͒ i2 sin k ⌬ i ⌬ j ⌬ k But this sum is a Riemann sum for the function
F͑ , , ͒ 2 sin f ͑ sin cos , sin sin , cos ͒
Consequently, we have arrived at the following formula for triple integration in spherical
coordinates. 4 yyy f ͑x, y, z͒ dV
E y d c  y␣ y b a f ͑ sin cos , sin sin , cos ͒ 2 sin d d d where E is a spherical wedge given by z ∏ sin ˙ d¨ ˙ Խ E ͕͑ , , ͒ a ഛ ഛ b, ␣ ഛ ഛ , c ഛ ഛ d͖ d∏ Formula 4 says that we convert a triple integral from rectangular coordinates to spherical coordinates by writing ∏ x sin cos ∏ d˙ 0 d¨
x FIGURE 8 Volume element in spherical
coordinates: dV=∏@ sin ˙ d∏ d¨ d˙ y y sin sin z cos using the appropriate limits of integration, and replacing dV by 2 sin d d d. This is
illustrated in Figure 8.
This formula can be extended to include more general spherical regions such as Խ E ͕͑ , , ͒ ␣ ഛ ഛ , c ഛ ഛ d, t1͑, ͒ ഛ ഛ t 2͑, ͖͒ 5E16(pp 10641073) 1072 ❙❙❙❙ 1/18/06 4:42 PM Page 1072 CHAPTER 16 MULTIPLE INTEGRALS In this case the formula is the same as in (4) except that the limits of integration for are
t1͑, ͒ and t 2͑, ͒.
Usually, spherical coordinates are used in triple integrals when surfaces such as cones
and spheres form the boundary of the region of integration.
EXAMPLE 3 Evaluate 2 2 2 3͞2 xxxB e ͑x ϩy ϩz ͒ dV, where B is the unit ball: Խ B ͕͑x, y, z͒ x 2 ϩ y 2 ϩ z 2 ഛ 1͖
SOLUTION Since the boundary of B is a sphere, we use spherical coordinates: Խ B ͕͑ , , ͒ 0 ഛ ഛ 1, 0 ഛ ഛ 2, 0 ഛ ഛ ͖
In addition, spherical coordinates are appropriate because
x 2 ϩ y 2 ϩ z2 2
Thus, (4) gives yyy e ͑x 2ϩy 2ϩz 2 ͒3͞2 dV y 0 2 y y
0 1 e͑ 0 2 3͞2 ͒ 2 sin d d d B y sin d
0 y 2 0
0 [ Ϫcos ͑2͒
NOTE d y 1 0
[e ] 1 3 1
3
0 2e d
3 4 ͑e Ϫ 1͒
3 It would have been extremely awkward to evaluate the integral in Example 3 without spherical coordinates. In rectangular coordinates the iterated integral would have been
■ 1 y y s1Ϫx 2 Ϫ1 Ϫs1Ϫx y
2 s1Ϫx 2Ϫy 2 Ϫs1Ϫx 2Ϫy 2 2 2 e ͑x ϩy ϩz 2 3͞2 ͒ dz dy dx EXAMPLE 4 Use spherical coordinates to ﬁnd the volume of the solid that lies above the
cone z sx 2 ϩ y 2 and below the sphere x 2 ϩ y 2 ϩ z 2 z. (See Figure 9.)
z
(0, 0, 1) ≈+¥[email protected]=z π
4 FIGURE 9 ≈+¥
z=œ„„„„„ y
x ( 1 ) SOLUTION Notice that the sphere passes through the origin and has center 0, 0, 2 . We write the equation of the sphere in spherical coordinates as 2 cos or cos 5E16(pp 10641073) 1/18/06 4:42 PM Page 1073 SECTION 16.8 TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES ❙❙❙❙ 1073 The equation of the cone can be written as  Figure 10 gives another look (this time drawn
by Maple) at the solid of Example 4. cos s 2 sin 2 cos 2 ϩ 2 sin 2 sin 2 sin
This gives sin cos , or ͞4. Therefore, the description of the solid E in
spherical coordinates is Խ E ͕͑ , , ͒ 0 ഛ ഛ 2, 0 ഛ ഛ ͞4, 0 ഛ ഛ cos ͖
Figure 11 shows how E is swept out if we integrate ﬁrst with respect to , then , and
then . The volume of E is
FIGURE 10 V͑E͒ yyy dV y 2 0 ͞4 y y 0 0 E y 2 0 2
3 Visual 16.8 shows an animation of
Figure 11. d y y 0 ͞4 0 z 2 4 2. y yy 3. y y y 4.
■ yy y y y y ͞2 0 r 2 0 ͞6 9Ϫr 2 0 ͞2 0 0 2 ■ 3 0 ͞2 0 2 1 ■ 2
sin cos d
3
3 ͫ cos 4
Ϫ
4 ͬ ͞4
8 0 z x y y ¨ varies from 0 to 2π. 5–6  Set up the triple integral of an arbitrary continuous function
f ͑x, y, z͒ in cylindrical or spherical coordinates over the solid
shown. r dz d dr 1. 0 d 0 Exercises  Sketch the solid whose volume is given by the integral and
evaluate the integral.
4 cos ˙ varies from 0 to π/4 while
¨ is constant. 1–4 0 ͫͬ
3
3 x y ∏ varies from 0 to cos ˙ while
˙ and ¨ are constant.  16.8 sin 2 sin d d d z x FIGURE 11 ͞4 cos z 5. r dz dr d 3 2 sin d d d 2 ■ ■ y x 2 sin d d d
■ z 6. 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ y 2 x
■ ■ ■ ■ ■ 5E16(pp 10741083) 1/18/06 4:43 PM 1074 ❙❙❙❙ CHAPTER 16 MULTIPLE INTEGRALS 7–16  Page 1074 Use cylindrical coordinates. 22. Evaluate xxxE xyz dV , where E lies between the spheres 2 and 4 and above the cone ͞3. 7. Evaluate xxxE sx 2 ϩ y 2 dV , where E is the region that lies 23. Find the volume of the solid that lies above the cone ͞3 inside the cylinder x 2 ϩ y 2 16 and between the planes
z Ϫ5 and z 4. and below the sphere 4 cos . 24. Find the volume of the solid that lies within the sphere 8. Evaluate xxxE ͑x 3 ϩ xy 2 ͒ dV , where E is the solid in the ﬁrst x 2 ϩ y 2 ϩ z 2 4, above the xyplane, and below the cone
z sx 2 ϩ y 2. octant that lies beneath the paraboloid z 1 Ϫ x 2 Ϫ y 2. 25. Find the centroid of the solid in Exercise 21. 9. Evaluate xxxE e z dV , where E is enclosed by the paraboloid z 1 ϩ x 2 ϩ y 2, the cylinder x 2 ϩ y 2 5, and the xyplane. 26. Let H be a solid hemisphere of radius a whose density at any point is proportional to its distance from the center of the base.
(a) Find the mass of H .
(b) Find the center of mass of H .
(c) Find the moment of inertia of H about its axis. 10. Evaluate xxxE x dV , where E is enclosed by the planes z 0 and z x ϩ y ϩ 3 and by the cylinders x ϩ y 4 and
x 2 ϩ y 2 9.
2 2 11. Evaluate xxxE x 2 dV , where E is the solid that lies within 27. (a) Find the centroid of a solid homogeneous hemisphere of the cylinder x 2 ϩ y 2 1, above the plane z 0, and below the
cone z 2 4x 2 ϩ 4y 2. radius a.
(b) Find the moment of inertia of the solid in part (a) about a
diameter of its base. 12. Find the volume of the solid that lies within both the cylinder x 2 ϩ y 2 1 and the sphere x 2 ϩ y 2 ϩ z 2 4. 28. Find the mass and center of mass of a solid hemisphere of radius a if the density at any point is proportional to its
distance from the base. 13. (a) Find the volume of the region E bounded by the parabo loids z x 2 ϩ y 2 and z 36 Ϫ 3x 2 Ϫ 3y 2.
(b) Find the centroid of E (the center of mass in the case where
the density is constant). ■ CAS ■ ■ ■ ■ ■ ■ ■  ■ ■ ■ 32. (a) Find the volume enclosed by the torus sin .
■ (b) Use a computer to draw the torus. ; ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ Use spherical coordinates. 17. Evaluate xxxB ͑x 2 ϩ y 2 ϩ z 2 ͒ dV , where B is the unit ball x 2 ϩ y 2 ϩ z 2 ഛ 1. 18. Evaluate xxxH ͑x 2 ϩ y 2 ͒ dV , where H is the hemispherical region that lies above the xyplane and below the sphere
x 2 ϩ y 2 ϩ z 2 1. 33–34  Evaluate the integral by changing to cylindrical
coordinates.
1 s1Ϫx 2 Ϫ1 Ϫs1Ϫx 2 33. y y 34. yy 1 0 ■ 19. Evaluate xxxE z dV , where E lies between the spheres x 2 ϩ y 2 ϩ z 2 1 and x 2 ϩ y 2 ϩ z 2 4 in the ﬁrst octant. 20. Evaluate xxxE e sx ϩy ϩz dV , where E is enclosed by the sphere
2 2 2 2 21. Evaluate xxxE x 2 dV , where E is bounded by the x zplane and the hemispheres y s9 Ϫ x 2 Ϫ z 2 and y s16 Ϫ x 2 Ϫ z 2. s1Ϫy 2 y 2Ϫx 2Ϫy 2 x 2ϩy 2 y sx 2ϩy 2 x 2ϩy 2 0
■ ■ ͑x 2 ϩ y 2 ͒3͞2 dz dy dx xyz dz dx dy ■ ■ ■ ■ ■ ■ ■ 35–36  Evaluate the integral by changing to spherical
coordinates. 2 x ϩ y ϩ z 9 in the ﬁrst octant.
2 ■ 31. Evaluate xxxE z dV , where E lies above the paraboloid ■ 17–28 ■ z x 2 ϩ y 2 and below the plane z 2y. Use either the
Table of Integrals (on the back Reference Pages) or a
computer algebra system to evaluate the integral. density at any point is proportional to its distance from the
z axis.
■ ■ radius a by two planes that intersect along a diameter at an
angle of ͞6. 2 16. Find the mass of a ball B given by x 2 ϩ y 2 ϩ z 2 ഛ a 2 if the ■ ■ 30. Find the volume of the smaller wedge cut from a sphere of paraboloid z 4x ϩ 4y and the plane z a ͑a Ͼ 0͒ if S has
constant density K . ■ ■ cone z sx 2 ϩ y 2 and below the sphere x 2 ϩ y 2 ϩ z 2 1. 15. Find the mass and center of mass of the solid S bounded by the ■ ■ 29. Find the volume and centroid of the solid E that lies above the cuts out of the sphere of radius a centered at the origin.
(b) Illustrate the solid of part (a) by graphing the sphere and
the cylinder on the same screen.
2 ■ 29–32  Use cylindrical or spherical coordinates, whichever seems
more appropriate. 14. (a) Find the volume of the solid that the cylinder r a cos ; ■ 35.
36.
■ 3 s9Ϫx 2 Ϫ3 Ϫs9Ϫx 2 y y
3 yy
0 s9Ϫy 2 s9Ϫx 2Ϫy 2 0
s18Ϫx 2Ϫy 2 sx 2ϩy 2 0
■ y y ■ ■ z sx 2 ϩ y 2 ϩ z 2 dz dy dx ͑x 2 ϩ y 2 ϩ z 2 ͒ dz dx dy
■ ■ ■ ■ ■ ■ 5E16(pp 10741083) 1/18/06 4:44 PM Page 1075 APPLIED PROJECT ROLLER DERBY CAS 37. In the Laboratory Project on page 880 we investigated the family of surfaces 1 ϩ 1 sin m sin n that have been
5
used as models for tumors. The “bumpy sphere” with m 6
and n 5 is shown. Use a computer algebra system to ﬁnd the
volume it encloses. ❙❙❙❙ 1075 (b) Deduce that the volume of the spherical wedge given by
1 ഛ ഛ 2 , 1 ഛ ഛ 2 , 1 ഛ ഛ 2 is
⌬V 3 Ϫ 3
2
1
͑cos 1 Ϫ cos 2 ͒͑ 2 Ϫ 1 ͒
3 (c) Use the Mean Value Theorem to show that the volume in
part (b) can be written as
ෂ ෂ
⌬V 2 sin ⌬ ⌬ ⌬
ෂ ෂ
where lies between 1 and 2 , lies between 1 and 2 ,
⌬ 2 Ϫ 1 , ⌬ 2 Ϫ 1 , and ⌬ 2 Ϫ 1 . 40. When studying the formation of mountain ranges, geologists 38. Show that
ϱ ϱ ϱ Ϫϱ Ϫϱ Ϫϱ y y y sx 2 ϩ y 2 ϩ z 2 eϪ͑x 2 ϩy2ϩz2 ͒ dx dy dz 2 (The improper triple integral is deﬁned as the limit of a triple
integral over a solid sphere as the radius of the sphere increases
indeﬁnitely.) estimate the amount of work required to lift a mountain from
sea level. Consider a mountain that is essentially in the shape
of a right circular cone. Suppose that the weight density of
the material in the vicinity of a point P is t͑P͒ and the height
is h͑P͒.
(a) Find a deﬁnite integral that represents the total work done
in forming the mountain.
(b) Assume that Mount Fuji in Japan is in the shape of a right
circular cone with radius 62,000 ft, height 12,400 ft, and
density a constant 200 lb͞ft3 . How much work was done in
forming Mount Fuji if the land was initially at sea level? 39. (a) Use cylindrical coordinates to show that the volume of the solid bounded above by the sphere r 2 ϩ z 2 a 2 and below
by the cone z r cot 0 (or 0 ), where
0 Ͻ 0 Ͻ ͞2, is
V P 2a 3
͑1 Ϫ cos 0 ͒
3 APPLIED PROJECT
Roller Derby h
å Suppose that a solid ball (a marble), a hollow ball (a squash ball), a solid cylinder (a steel bar),
and a hollow cylinder (a lead pipe) roll down a slope. Which of these objects reaches the bottom
ﬁrst? (Make a guess before proceeding.)
To answer this question we consider a ball or cylinder with mass m, radius r , and moment of
inertia I (about the axis of rotation). If the vertical drop is h, then the potential energy at the top
is mth. Suppose the object reaches the bottom with velocity v and angular velocity , so v r.
The kinetic energy at the bottom consists of two parts: 1 mv 2 from translation (moving down the
2
slope) and 1 I 2 from rotation. If we assume that energy loss from rolling friction is negligible,
2
then conservation of energy gives
mth 1 mv2 ϩ 1 I 2
2
2
1. Show that
v2 2th
1 ϩ I* where I* I
mr 2 5E16(pp 10741083) 1076 ❙❙❙❙ 1/18/06 4:44 PM Page 1076 CHAPTER 16 MULTIPLE INTEGRALS 2. If y͑t͒ is the vertical distance traveled at time t, then the same reasoning as used in Problem
1 shows that v 2 2ty͑͞1 ϩ I*͒ at any time t. Use this result to show that y satisﬁes the differential equation
dy
dt ͱ 2t
͑sin ␣͒ sy
1 ϩ I* where ␣ is the angle of inclination of the plane.
3. By solving the differential equation in Problem 2, show that the total travel time is T ͱ 2h͑1 ϩ I*͒
t sin 2␣ This shows that the object with the smallest value of I* wins the race.
4. Show that I* 2 for a solid cylinder and I* 1 for a hollow cylinder.
1 5. Calculate I* for a partly hollow ball with inner radius a and outer radius r . Express your answer in terms of b a͞r. What happens as a l 0 and as a l r ?
6. Show that I* 5 for a solid ball and I* 3 for a hollow ball. Thus, the objects ﬁnish in the
2 2 following order: solid ball, solid cylinder, hollow ball, hollow cylinder. DISCOVERY PROJECT
The Intersection of Three Cylinders
The ﬁgure shows the solid enclosed by three circular cylinders with the same diameter that intersect at right angles. In this project we compute its volume and determine how its shape changes
if the cylinders have different diameters. 1. Sketch carefully the solid enclosed by the three cylinders x 2 ϩ y 2 1, x 2 ϩ z 2 1, and y 2 ϩ z 2 1. Indicate the positions of the coordinate axes and label the faces with the equations of the corresponding cylinders. 2. Find the volume of the solid in Problem 1.
CAS 3. Use a computer algebra system to draw the edges of the solid.
4. What happens to the solid in Problem 1 if the radius of the ﬁrst cylinder is different from 1? Illustrate with a handdrawn sketch or a computer graph.
5. If the ﬁrst cylinder is x 2 ϩ y 2 a 2, where a Ͻ 1, set up, but do not evaluate, a double integral for the volume of the solid. What if a Ͼ 1? 5E16(pp 10741083) 1/18/06 4:44 PM Page 1077 SECTION 16.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS  16.9 ❙❙❙❙ 1077 Change of Variables in Multiple Integrals
In onedimensional calculus we often use a change of variable (a substitution) to simplify
an integral. By reversing the roles of x and u, we can write the Substitution Rule (5.5.5) as y 1 b d f ͑x͒ dx y f ͑t͑u͒͒tЈ͑u͒ du a c where x t͑u͒ and a t͑c͒, b t͑d͒. Another way of writing Formula 1 is as follows: y 2 b a d f ͑x͒ dx y f ͑x͑u͒͒
c dx
du
du A change of variables can also be useful in double integrals. We have already seen one
example of this: conversion to polar coordinates. The new variables r and are related to
the old variables x and y by the equations
x r cos y r sin and the change of variables formula (16.4.2) can be written as yy f ͑x, y͒ dA yy f ͑r cos , r sin ͒ r dr d
R S where S is the region in the r plane that corresponds to the region R in the xyplane.
More generally, we consider a change of variables that is given by a transformation T
from the uvplane to the xyplane:
T͑u, v͒ ͑x, y͒
where x and y are related to u and v by the equations
x t͑u, v͒ y h͑u, v͒ x x͑u, v͒ 3 y y͑u, v͒ or, as we sometimes write,
We usually assume that T is a C 1 transformation, which means that t and h have continuous ﬁrstorder partial derivatives.
A transformation T is really just a function whose domain and range are both subsets
of .2 ޒIf T͑u1, v1͒ ͑x 1, y1͒, then the point ͑x 1, y1͒ is called the image of the point ͑u1, v1͒.
If no two points have the same image, T is called onetoone. Figure 1 shows the effect of
a transformation T on a region S in the uvplane. T transforms S into a region R in the
xyplane called the image of S, consisting of the images of all points in S.
√ y T S R (u¡, √¡) 0 FIGURE 1 T –! u (x¡, y¡) 0 x 5E16(pp 10741083) 1078 ❙❙❙❙ 1/18/06 4:44 PM Page 1078 CHAPTER 16 MULTIPLE INTEGRALS If T is a onetoone transformation, then it has an inverse transformation T Ϫ1 from the
xyplane to the uvplane and it may be possible to solve Equations 3 for u and v in terms
of x and y :
u G͑x, y͒
v H͑x, y͒
EXAMPLE 1 A transformation is deﬁned by the equations x u 2 Ϫ v2 y 2uv Խ Find the image of the square S ͕͑u, v͒ 0 ഛ u ഛ 1, 0 ഛ v ഛ 1͖.
√ S£ (0, 1) S¢ (1, 1) S™ S 0 SOLUTION The transformation maps the boundary of S into the boundary of the image. So
we begin by ﬁnding the images of the sides of S. The ﬁrst side, S1 , is given by v 0
͑0 ഛ u ഛ 1͒. (See Figure 2.) From the given equations we have x u 2, y 0, and so
0 ഛ x ഛ 1. Thus, S1 is mapped into the line segment from ͑0, 0͒ to ͑1, 0͒ in the xyplane.
The second side, S 2, is u 1 ͑0 ഛ v ഛ 1͒ and, putting u 1 in the given equations, we
get
x 1 Ϫ v2
y 2v
Eliminating v, we obtain S¡ (1, 0) u T x1Ϫ 4
y ¥
x=1 4 x 5 R
(_1, 0) 0 (1, 0) 0ഛxഛ1 which is part of a parabola. Similarly, S 3 is given by v 1 ͑0 ഛ u ഛ 1͒, whose image is
the parabolic arc (0, 2)
¥
x= 1
4 y2
4 x FIGURE 2 y2
Ϫ1
4 Ϫ1 ഛ x ഛ 0 Finally, S4 is given by u 0 ͑0 ഛ v ഛ 1͒ whose image is x Ϫv 2, y 0, that is,
Ϫ1 ഛ x ഛ 0. (Notice that as we move around the square in the counterclockwise direction, we also move around the parabolic region in the counterclockwise direction.) The
image of S is the region R (shown in Figure 2) bounded by the xaxis and the parabolas
given by Equations 4 and 5.
Now let’s see how a change of variables affects a double integral. We start with a small
rectangle S in the uvplane whose lower left corner is the point ͑u0 , v0 ͒ and whose dimensions are ⌬u and ⌬v. (See Figure 3.)
y √ u=u¸
r (u ¸, √)
Î√ S (u¸, √ ¸) Îu T
(x¸, y¸) √=√¸
0 R r (u, √¸)
u 0 x FIGURE 3 The image of S is a region R in the xyplane, one of whose boundary points is
͑x 0 , y0 ͒ T͑u0 , v0 ͒. The vector
r͑u, v͒ t͑u, v͒ i ϩ h͑u, v͒ j 5E16(pp 10741083) 1/18/06 4:44 PM Page 1079 SECTION 16.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS ❙❙❙❙ 1079 is the position vector of the image of the point ͑u, v͒. The equation of the lower side of S
is v v0 , whose image curve is given by the vector function r͑u, v0͒. The tangent vector
at ͑x 0 , y0 ͒ to this image curve is
ru tu͑u0 , v0 ͒ i ϩ hu͑u0 , v0 ͒ j Ѩx
Ѩy
iϩ
j
Ѩu
Ѩu Similarly, the tangent vector at ͑x 0 , y0 ͒ to the image curve of the left side of S (namely,
u u0 ) is
rv tv͑u0 , v0 ͒ i ϩ hv͑u0 , v0 ͒ j r (u¸, √¸+Î√)
b
R r (u¸, √¸) We can approximate the image region R T͑S͒ by a parallelogram determined by the
secant vectors
a r͑u0 ϩ ⌬u, v0 ͒ Ϫ r͑u0 , v0 ͒ a
r (u¸+Î u, √¸) ru lim ⌬u l 0 r (u¸, √¸) b r͑u0 , v0 ϩ ⌬v͒ Ϫ r͑u0 , v0 ͒ shown in Figure 4. But FIGURE 4 Î√ r√ Ѩx
Ѩy
iϩ
j
Ѩv
Ѩv r͑u0 ϩ ⌬u, v0 ͒ Ϫ r͑u0 , v0 ͒
⌬u and so
Î u ru FIGURE 5 r͑u0 ϩ ⌬u, v0 ͒ Ϫ r͑u0 , v0 ͒ Ϸ ⌬u ru Similarly r͑u0 , v0 ϩ ⌬v͒ Ϫ r͑u0 , v0 ͒ Ϸ ⌬v rv This means that we can approximate R by a parallelogram determined by the vectors
⌬u ru and ⌬v rv . (See Figure 5.) Therefore, we can approximate the area of R by the area
of this parallelogram, which, from Section 13.4, is Խ ͑⌬u r ͒ ϫ ͑⌬v r ͒ Խ Խ r 6 u v u Խ ϫ rv ⌬u ⌬v Computing the cross product, we obtain Խ ԽԽ Խ Խ Խ
i Ѩx
ru ϫ rv Ѩu
Ѩx
Ѩv j Ѩy
Ѩu
Ѩy
Ѩv k Ѩx
Ѩu
Ѩx
0
Ѩv 0 Ѩy
Ѩu
k
Ѩy
Ѩv Ѩx
Ѩu
Ѩy
Ѩu Ѩx
Ѩv
k
Ѩy
Ѩv The determinant that arises in this calculation is called the Jacobian of the transformation
and is given a special notation.  The Jacobian is named after the German
mathematician Carl Gustav Jacob Jacobi
(1804–1851). Although the French mathematician
Cauchy ﬁrst used these special determinants
involving partial derivatives, Jacobi developed
them into a method for evaluating multiple
integrals. 7 Definition The Jacobian of the transformation T given by x t͑u, v͒ and y h͑u, v͒ is Խ Խ Ѩx
Ѩ͑x, y͒
Ѩu
Ѩ͑u, v͒
Ѩy
Ѩu Ѩx
Ѩv
Ѩx Ѩy
Ѩx Ѩy
Ϫ
Ѩy
Ѩu Ѩv
Ѩv Ѩu
Ѩv 5E16(pp 10741083) 1080 ❙❙❙❙ 1/18/06 4:44 PM Page 1080 CHAPTER 16 MULTIPLE INTEGRALS With this notation we can use Equation 6 to give an approximation to the area ⌬A
of R:
⌬A Ϸ 8 Ϳ Ϳ Ѩ͑x, y͒
⌬u ⌬v
Ѩ͑u, v͒ where the Jacobian is evaluated at ͑u0 , v0 ͒.
Next we divide a region S in the uvplane into rectangles Sij and call their images in the
xyplane Rij . (See Figure 6.)
√ y Sij
R ij
S Î√ R Îu T (u i , √ j ) (x i , y j) 0 u 0 x FIGURE 6 Applying the approximation (8) to each Rij , we approximate the double integral of f
over R as follows:
m n yy f ͑x, y͒ dA Ϸ ͚ ͚ f ͑x , y ͒ ⌬A
i j i1 j1 R m Ϸ n ͚ ͚ f ͑t͑u , v ͒, h͑u , v ͒͒
i j i j i1 j1 Ϳ Ϳ Ѩ͑x, y͒
⌬u ⌬v
Ѩ͑u, v͒ where the Jacobian is evaluated at ͑ui, vj ͒. Notice that this double sum is a Riemann sum
for the integral
Ѩ͑x, y͒
yy f ͑t͑u, v͒, h͑u, v͒͒ Ѩ͑u, v͒ du dv
S Ϳ Ϳ The foregoing argument suggests that the following theorem is true. (A full proof is
given in books on advanced calculus.)
1
9 Change of Variables in a Double Integral Suppose that T is a C transformation whose Jacobian is nonzero and that maps a region S in the uvplane onto a region
R in the xyplane. Suppose that f is continuous on R and that R and S are type I or
type II plane regions. Suppose also that T is onetoone, except perhaps on the
boundary of S. Then yy f ͑x, y͒ dA yy f ͑x͑u, v͒, y͑u, v͒͒
R S Ϳ Ϳ Ѩ͑x, y͒
du dv
Ѩ͑u, v͒ Theorem 9 says that we change from an integral in x and y to an integral in u and v by
expressing x and y in terms of u and v and writing
dA Ϳ Ϳ Ѩ͑x, y͒
du dv
Ѩ͑u, v͒ 5E16(pp 10741083) 1/18/06 4:44 PM Page 1081 SECTION 16.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS Խ
¨=∫
r=a Խ r=b S x t͑r, ͒ r cos å 1081 Notice the similarity between Theorem 9 and the onedimensional formula in Equation 2.
Instead of the derivative dx͞du, we have the absolute value of the Jacobian, that is,
Ѩ͑x, y͒͞Ѩ͑u, v͒ .
As a ﬁrst illustration of Theorem 9, we show that the formula for integration in polar
coordinates is just a special case. Here the transformation T from the r plane to the
xyplane is given by ¨
∫ ❙❙❙❙ y h͑r, ͒ r sin ¨=å 0 a r b Խ Խ Ѩx
Ѩ͑x, y͒
Ѩr
Ѩ͑r, ͒
Ѩy
Ѩr T
y r=b ¨=∫
R
r=a
∫
0 and the geometry of the transformation is shown in Figure 7. T maps an ordinary rectangle
in the r plane to a polar rectangle in the xyplane. The Jacobian of T is
Ѩx
Ѩ
cos
Ѩy
sin
Ѩ Ϳ Ϳ Ϫr sin
r cos2 ϩ r sin2 r Ͼ 0
r cos Thus, Theorem 9 gives
¨=å yy f ͑x, y͒ dx dy yy f ͑r cos , r sin ͒ å R S x y
FIGURE 7 The polar coordinate transformation  ␣ y b a Ϳ Ϳ Ѩ͑x, y͒
dr d
Ѩ͑r, ͒ f ͑r cos , r sin ͒ r dr d which is the same as Formula 16.4.2.
EXAMPLE 2 Use the change of variables x u 2 Ϫ v 2, y 2uv to evaluate the integral xxR y dA, where R is the region bounded by the xaxis and the parabolas y 2 4 Ϫ 4x and y 2 4 ϩ 4x. SOLUTION The region R is pictured in Figure 2. In Example 1 we discovered that
T͑S͒ R, where S is the square ͓0, 1͔ ϫ ͓0, 1͔. Indeed, the reason for making the
change of variables to evaluate the integral is that S is a much simpler region than R.
First we need to compute the Jacobian: Խ Խ Ѩx
Ѩ͑x, y͒
Ѩu
Ѩ͑u, v͒
Ѩy
Ѩu
Therefore, by Theorem 9, yy y dA yy 2uv
R S 8y 1 0 1 y 1 0 Ϳ Ѩx
Ѩv
2u
Ѩy
2v
Ѩv Ϳ Ϳ Ѩ͑x, y͒
1 1
dA y y ͑2uv͒4͑u2 ϩ v 2 ͒ du dv
0 0
Ѩ͑u, v͒
͑u3v ϩ uv 3 ͒ du dv 8 y 1 0 [ y ͑2v ϩ 4v 3 ͒ dv v 2 ϩ v 4
0 Ϳ Ϫ2v
4u 2 ϩ 4v 2 Ͼ 0
2u 1
0 [ 1 4
4 v 2 u ϩ 1 u2v 3
2 u1 u0 dv 5E16(pp 10741083) 1082 ❙❙❙❙ 1/18/06 4:45 PM Page 1082 CHAPTER 16 MULTIPLE INTEGRALS NOTE
Example 2 was not a very difﬁcult problem to solve because we were given a
suitable change of variables. If we are not supplied with a transformation, then the ﬁrst step
is to think of an appropriate change of variables. If f ͑x, y͒ is difﬁcult to integrate, then the
form of f ͑x, y͒ may suggest a transformation. If the region of integration R is awkward,
then the transformation should be chosen so that the corresponding region S in the uvplane
has a convenient description.
■ EXAMPLE 3 Evaluate the integral xxR e ͑xϩy͒͑͞xϪy͒ dA, where R is the trapezoidal region with vertices ͑1, 0͒, ͑2, 0͒, ͑0, Ϫ2͒, and ͑0, Ϫ1͒. SOLUTION Since it isn’t easy to integrate e ͑xϩy͒͑͞xϪy͒, we make a change of variables sug gested by the form of this function:
uxϩy 10 vxϪy These equations deﬁne a transformation T Ϫ1 from the xyplane to the uvplane.
Theorem 9 talks about a transformation T from the uvplane to the xyplane. It is
obtained by solving Equations 10 for x and y :
x 1 ͑u ϩ v͒
2 11 y 1 ͑u Ϫ v͒
2 The Jacobian of T is Խ Խ Ѩx
Ѩ͑x, y͒
Ѩu
Ѩ͑u, v͒
Ѩy
Ѩu Ѩx
Ѩv
Ѩy
Ѩv Ϳ 1
2
1
2 Ϳ Ϫ1
2
Ϫ1
2
Ϫ1
2 To ﬁnd the region S in the uvplane corresponding to R, we note that the sides of R lie on
the lines
y0 √ √=2 (_2, 2) (2, 2) S u=_√ xϪy2 x0 xϪy1 and, from either Equations 10 or Equations 11, the image lines in the uvplane are u=√ (_1, 1) uv (1, 1) v2 u Ϫv v1 √=1
0 T u Thus, the region S is the trapezoidal region with vertices ͑1, 1͒, ͑2, 2͒, ͑Ϫ2, 2͒, and
͑Ϫ1, 1͒ shown in Figure 8. Since Խ T –! S ͕͑u, v͒ 1 ഛ v ഛ 2, Ϫv ഛ u ഛ v͖ y Theorem 9 gives
xy=1
1 2 0
_1 x R
xy=2 yy e
R ͑xϩy͒͑͞xϪy͒ dA yy e u͞v
S y 2 1 y Ϫv _2 2 FIGURE 8 v Ϳ Ϳ Ѩ͑x, y͒
du dv
Ѩ͑u, v͒
2 1 1 y ͑e Ϫ eϪ1 ͒v dv 3 ͑e Ϫ eϪ1 ͒
2
4
1 uv e u͞v ( 1 ) du dv 1 y [ve u͞v ]uϪv dv
2
2 5E16(pp 10741083) 1/18/06 4:45 PM Page 1083 SECTION 16.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS ❙❙❙❙ 1083 Triple Integrals
There is a similar change of variables formula for triple integrals. Let T be a transformation that maps a region S in u vwspace onto a region R in xyzspace by means of the
equations
x t͑u, v, w͒
y h͑u, v, w͒
z k͑u, v, w͒
The Jacobian of T is the following 3 ϫ 3 determinant: Խ Ѩx
Ѩu
Ѩ͑x, y, z͒
Ѩy
Ѩ͑u, v, w͒
Ѩu
Ѩz
Ѩu 12 Ѩx
Ѩv
Ѩy
Ѩv
Ѩz
Ѩv Ѩx
Ѩw
Ѩy
Ѩw
Ѩz
Ѩw Խ Under hypotheses similar to those in Theorem 9, we have the following formula for triple
integrals: yyy f ͑x, y, z͒ dV yyy f ͑x͑u, v, w͒, y͑u, v, w͒, z͑u, v, w͒͒ 13 R S Ϳ Ϳ Ѩ͑x, y, z͒
du dv dw
Ѩ͑u, v, w͒ EXAMPLE 4 Use Formula 13 to derive the formula for triple integration in spherical
coordinates.
SOLUTION Here the change of variables is given by x sin cos Խ y sin sin We compute the Jacobian as follows:
sin cos Ϫ sin sin cos cos
Ѩ͑x, y, z͒
sin sin Ϫ sin cos cos sin
Ѩ͑ , , ͒
cos
0
Ϫ sin
cos Ϳ Ϳ z cos ԽͿ Ϫ sin sin cos cos
sin cos Ϫ sin sin
Ϫ sin
sin cos
Ϫ sin cos cos sin
sin sin cos ͑Ϫ 2 sin cos sin2 Ϫ 2 sin cos cos2 ͒
Ϫ sin ͑ sin2 cos2 ϩ sin2 sin2 ͒
Ϫ 2 sin cos2 Ϫ 2 sin sin2 Ϫ 2 sin
Since 0 ഛ ഛ , we have sin ജ 0. Therefore Ϳ Ϳ Ѩ͑x, y, z͒
Ϫ 2 sin 2 sin
Ѩ͑ , , ͒ Խ Խ and Formula 13 gives yyy f ͑x, y, z͒ dV yyy f ͑ sin cos , sin sin , cos ͒
R S which is equivalent to Formula 16.8.4. 2 sin d d d Ϳ 5E16(pp 10841087) ❙❙❙❙ 1084 4:47 PM Page 1084 CHAPTER 16 MULTIPLE INTEGRALS  16.9
1– 6 1/18/06 Exercises
2
; 16. xxR y dA, where R is the region bounded by the curves xy 1, Find the Jacobian of the transformation.  1. x u ϩ 4 v,
2. x u 2 Ϫ v 2, xy 2, xy 2 1, xy 2 2; u xy, v xy 2. Illustrate by
using a graphing calculator or computer to draw R. y 3u Ϫ 2v
y u 2 ϩ v2 u
3. x
,
uϩv y 4. x ␣ sin , ■ y ␣ cos  5. x u v, y vw, 6. x e uϪv,
■ ■ 7–10  ■ ■ z e uϩvϩw
■ ■ ■ ■ ■ ■ ■ Find the image of the set S under the given transformation. Խ 10. S is the disk given by u 2 ϩ v 2 ഛ 1;
11–16  ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 19. x au, y b v
■ ■ ■ ■  Evaluate the integral by making an appropriate change of variables. 9. S is the triangular region with vertices ͑0, 0͒, ͑1, 1͒, ͑0, 1͒;
x u 2, y v
■ ■ 18. Evaluate xxxE x 2 y dV , where E is the solid of Exercise 17(a).
19–23 8. S is the square bounded by the lines u 0, u 1, v 0,
v 1; x v, y u͑1 ϩ v 2 ͒ ■ ■ ellipsoid x 2͞a 2 ϩ y 2͞b 2 ϩ z 2͞c 2 1. Use the transformation x au, y b v, z c w.
(b) The Earth is not a perfect sphere; rotation has resulted in
ﬂattening at the poles. So the shape can be approximated by
an ellipsoid with a b 6378 km and c 6356 km. Use
part (a) to estimate the volume of the Earth. z uw 7. S ͕͑u, v͒ 0 ഛ u ഛ 3, 0 ഛ v ഛ 2͖;
x 2u ϩ 3v, y u Ϫ v ■ ■ 17. (a) Evaluate xxxE dV, where E is the solid enclosed by the uϪv y e uϩv, ■ ■ v ■ x Ϫ 2y
dA, where R is the parallelogram enclosed by the
3x Ϫ y
R
lines x Ϫ 2y 0, x Ϫ 2y 4, 3x Ϫ y 1, and 3x Ϫ y 8 yy 2 20. xxR ͑x ϩ y͒e x Ϫy Use the given transformation to evaluate the integral. 2 dA, where R is the rectangle enclosed by the
lines x Ϫ y 0, x Ϫ y 2, x ϩ y 0, and x ϩ y 3 ͩ ͪ yϪx
dA, where R is the trapezoidal region
yϩx
R
with vertices ͑1, 0͒, ͑2, 0͒, ͑0, 2͒, and ͑0, 1͒ 11. xxR ͑x Ϫ 3y͒ dA, where R is the triangular region with
vertices ͑0, 0͒, ͑2, 1͒, and ͑1, 2͒; x 2u ϩ v, y u ϩ 2v 21. yy cos 12. xxR ͑4 x ϩ 8y͒ dA, where R is the parallelogram with
vertices ͑Ϫ1, 3͒, ͑1, Ϫ3͒, ͑3, Ϫ1͒, and ͑1, 5͒;
x 1 ͑u ϩ v͒, y 1 ͑v Ϫ 3u͒
4
4 22. xxR sin͑9x 2 ϩ 4y 2 ͒ dA, xxR x 2 dA, 23. xxR e xϩy dA, 13. where R is the region bounded by the ellipse
9x ϩ 4y 36; x 2u, y 3v
2 2 ■ 14. 15. xxR ͑x 2 Ϫ xy ϩ y 2 ͒ dA, where R is the region bounded
by the ellipse x 2 Ϫ xy ϩ y 2 2;
x s2u Ϫ s2͞3 v, y s2u ϩ s2͞3 v xxR xy dA, where R is the region in the ﬁrst quadrant bounded
by the lines y x and y 3x and the hyperbolas xy 1,
xy 3; x u͞v, y v where R is the region in the ﬁrst quadrant bounded by the ellipse 9x 2 ϩ 4y 2 1 ■ ■ Խ Խ Խ Խ where R is given by the inequality x ϩ y ഛ 1
■ ■ ■ ■ ■ ■ ■ ■ ■ 24. Let f be continuous on ͓0, 1͔ and let R be the triangular region with vertices ͑0, 0͒, ͑1, 0͒, and ͑0, 1͒. Show that yy f ͑x ϩ y͒ dA y 1 0 R u f ͑u͒ du 5E16(pp 10841087) 1/18/06 4:47 PM Page 1085 CHAPTER 16 REVIEW  16 Review ■ CONCEPT CHECK 1. Suppose f is a continuous function deﬁned on a rectangle R ͓a, b͔ ϫ ͓c, d ͔.
(a) Write an expression for a double Riemann sum of f .
If f ͑x, y͒ ജ 0, what does the sum represent?
(b) Write the deﬁnition of xxR f ͑x, y͒ dA as a limit.
(c) What is the geometric interpretation of xxR f ͑x, y͒ dA if
f ͑x, y͒ ജ 0? What if f takes on both positive and negative
values?
(d) How do you evaluate xxR f ͑x, y͒ dA?
(e) What does the Midpoint Rule for double integrals say?
(f) Write an expression for the average value of f . z f ͑x, y͒, ͑x, y͒ ʦ D.
7. (a) Write the deﬁnition of the triple integral of f over a rectangular box B.
(b) How do you evaluate xxxB f ͑x, y, z͒ dV ?
(c) How do you deﬁne xxxE f ͑x, y, z͒ dV if E is a bounded
solid region that is not a box?
(d) What is a type 1 solid region? How do you evaluate
xxxE f ͑x, y, z͒ dV if E is such a region?
(e) What is a type 2 solid region? How do you evaluate
xxxE f ͑x, y, z͒ dV if E is such a region?
(f) What is a type 3 solid region? How do you evaluate
xxxE f ͑x, y, z͒ dV if E is such a region?
8. Suppose a solid object occupies the region E and has density function ͑x, y, z͒. Write expressions for each of the following.
(a) The mass
(b) The moments about the coordinate planes
(c) The coordinates of the center of mass
(d) The moments of inertia about the axes dinates in a double integral? Why would you want to do it?
4. If a lamina occupies a plane region D and has density function 5. Let f be a joint density function of a pair of continuous random variables X and Y .
(a) Write a double integral for the probability that X lies
between a and b and Y lies between c and d. ■ 1. y y
Ϫ1 6 0 x sin͑x Ϫ y͒ dx dy y
2 6 0 y 2 Ϫ1 9. (a) How do you change from rectangular coordinates to cylin drical coordinates in a triple integral?
(b) How do you change from rectangular coordinates to
spherical coordinates in a triple integral?
(c) In what situations would you change to cylindrical or
spherical coordinates?
10. (a) If a transformation T is given by x t͑u, v͒,
y h͑u, v͒, what is the Jacobian of T ? (b) How do you change variables in a double integral?
(c) How do you change variables in a triple integral? TRUEFALSE QUIZ Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement.
2 1 yy
0 x sx ϩ y 2 dy dx y 0 x 0 y 1 0 ■ 5. If D is the disk given by x 2 ϩ y 2 ഛ 4, then yy s4 Ϫ x x sin͑x Ϫ y͒ dy dx
2 sx ϩ y 2 dx dy 2 Ϫ y 2 dA 16
3 D 6.
2. ■ 6. Write an expression for the area of a surface with equation 3. How do you change from rectangular coordinates to polar coor ͑x, y͒, write expressions for each of the following in terms of
double integrals.
(a) The mass
(b) The moments about the axes
(c) The center of mass
(d) The moments of inertia about the axes and the origin 4 1 y y (x
1 0 2 ϩ sy ) sin͑x 2 y 2 ͒ dx dy ഛ 9 7. The integral 2 2 y yy
0 2 4 yy 4. y y 1 1 Ϫ1 4 1 3. 2 3 x 2e y dy dx y x 2 dx y e y dy 3 1 0 ex 2 ϩy 2 sin y dx dy 0 1085 (b) What properties does f possess?
(c) What are the expected values of X and Y ? 2. (a) How do you deﬁne xxD f ͑x, y͒ dA if D is a bounded region that is not a rectangle?
(b) What is a type I region? How do you evaluate xxD f ͑x, y͒ dA
if D is a type I region?
(c) What is a type II region? How do you evaluate
xxD f ͑x, y͒ dA if D is a type II region?
(d) What properties do double integrals have? ❙❙❙❙ 0 2 r dz dr d represents the volume enclosed by the cone z sx 2 ϩ y 2 and
the plane z 2.
8. The integral xxxE kr 3 dz dr d represents the moment of inertia about the z axis of a solid E with constant density k. 5E16(pp 10841087) ❙❙❙❙ 1086 1/18/06 4:48 PM Page 1086 CHAPTER 16 MULTIPLE INTEGRALS ■ EXERCISES ■ 13–14  Calculate the iterated integral by ﬁrst reversing the order
of integration. 1. A contour map is shown for a function f on the square R ͓0, 3͔ ϫ ͓0, 3͔. Use a Riemann sum with nine terms to
estimate the value of xxR f ͑x, y͒ dA. Take the sample points to
be the upper right corners of the squares. 2 13. y
3 1 yy
0 ■  xxD xy dA, yy 3 D 2 ■ ■ 0 1 sy
■ ye x
dx dy
x3
■ ■ ■ ■ Խ 17. 5 ■ 1 yy dA, where R ͕͑x, y͒ 0 ഛ x ഛ 2, 0 ഛ y ഛ 3͖ xxR ye 6 ■ 14. Calculate the value of the multiple integral. 16. 7 2 1 ■ 15. 9
8 4 x cos͑ y 2 ͒ dy dx ■ 15–28 10 1 xy Խ where D ͕͑x, y͒ 0 ഛ y ഛ 1, y 2 ഛ x ഛ y ϩ 2͖ y
dA, where D is bounded by y sx, y 0, x 1
1 ϩ x2 1
dA, where D is the triangular region with vertices
1 ϩ x2
D
͑0, 0͒, ͑1, 1͒, and ͑0, 1͒ 18. yy 19. xxD y dA, where D is the region in the ﬁrst quadrant bounded by
the parabolas x y 2 and x 8 Ϫ y 2 20. xxD y dA, where D is the region in the ﬁrst quadrant that lies
above the hyperbola xy 1 and the line y x and below the
line y 2 21. xxD ͑x 2 ϩ y 2 ͒3͞2 dA, where D is the region in the ﬁrst quadrant bounded by the lines y 0 and y s3x and the circle
x2 ϩ y2 9 22. xxD x dA, where D is the region in the ﬁrst quadrant that lies 1 0 1 3 x 2 2. Use the Midpoint Rule to estimate the integral in Exercise 1.
3–8  3. 2 yy
1 5. Calculate the iterated integral.
2 0 1 yy
0 x 0 1 ͑ y ϩ 2xe ͒ dx dy 4. yy cos͑x 2 ͒ dy dx 6. yy y 1 s1Ϫy 2 0 1 0 1 ex x 0 1 ye xy dx dy y 3xy 2 dy dx between the circles x 2 ϩ y 2 1 and x 2 ϩ y 2 2
1 y yy
0 ■ 0 ■ 0 8. y sin x dz dy dx ■ ■ ■ ■ ■ yyy
0 0 25. ■ y
4 10. y
4 R
R 2
_4
■ 11. _2
■ 0
■ 2 4 x ■ 0 _4 ■ ■ ■ ■ ■ ͞2 sin 2 0 0 ■ ■ ͞2 0 0 y y y
and evaluate the integral. 2 1 xxxH z 3sx 2 ϩ y 2 ϩ z 2 dV , ■ 2 sin d d d Խ where T is the solid tetrahedron with vertices
͑0, 0, 0͒, ( 1 , 0, 0), ͑0, 1, 0͒, and ͑0, 0, 1͒
3
where E is bounded by the paraboloid
x 1 Ϫ y 2 Ϫ z 2 and the plane x 0 where E is bounded by the planes y 0, z 0,
x ϩ y 2 and the cylinder y 2 ϩ z 2 1 in the ﬁrst octant where E lies above the plane z 0, below the
plane z y, and inside the cylinder x 2 ϩ y 2 4 where H is the solid hemisphere that
lies above the xyplane and has center the origin and radius 1 ■ ■ 29–34 r dr d 12. Describe the solid whose volume is given by the integral
͞2 xxxE yz dV , 4 x Describe the region whose area is given by the integral y y xxxE z dV , where
E ͕͑x, y, z͒ 0 ഛ x ഛ 3, 0 ഛ y ഛ x, 0 ഛ z ഛ x ϩ y͖ ■ 9–10  Write xxR f ͑x, y͒ dA as an iterated integral, where R is the
region shown and f is an arbitrary continuous function on R. 9. xxxE y 2z 2 dV , 27. ■ xxxT xy dV , 26. ■ xxxE xy dV , 24. 6xyz dz dx dy x ■ 23. 28. 7. ■  ■ ■ ■ ■ ■ ■ ■ ■ Find the volume of the given solid. 29. Under the paraboloid z x 2 ϩ 4y 2 and above the rectangle R ͓0, 2͔ ϫ ͓1, 4͔ 30. Under the surface z x 2 y and above the triangle in the xyplane with vertices ͑1, 0͒, ͑2, 1͒, and ͑4, 0͒ 31. The solid tetrahedron with vertices ͑0, 0, 0͒, ͑0, 0, 1͒, ͑0, 2, 0͒, and ͑2, 2, 0͒ ■ 5E16(pp 10841087) 1/18/06 4:48 PM Page 1087 CHAPTER 16 REVIEW 32. Bounded by the cylinder x 2 ϩ y 2 4 and the planes z 0
33. One of the wedges cut from the cylinder x ϩ 9y a by the
2 2 planes z 0 and z mx
z sx 2 ϩ y 2
■ ■ 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ 35. Consider a lamina that occupies the region D bounded by the parabola x 1 Ϫ y 2 and the coordinate axes in the ﬁrst quadrant with density function ͑x, y͒ y.
(a) Find the mass of the lamina.
(b) Find the center of mass.
(c) Find the moments of inertia and radii of gyration about
the x and yaxes.
36. A lamina occupies the part of the disk x 2 ϩ y 2 ഛ a 2 that lies in the ﬁrst quadrant.
(a) Find the centroid of the lamina.
(b) Find the center of mass of the lamina if the density function
is ͑x, y͒ xy 2.
37. (a) Find the centroid of a right circular cone with height h and base radius a. (Place the cone so that its base is in the
xyplane with center the origin and its axis along the positive z axis.)
(b) Find the moment of inertia of the cone about its axis
(the z axis).
38. Find the area of the part of the cone z 2 a 2͑x 2 ϩ y 2 ͒ between the planes z 1 and z 2. 39. Find the area of the part of the surface z x 2 ϩ y that lies above the triangle with vertices (0, 0), (1, 0), and (0, 2).
CAS 40. Graph the surface z x sin y, Ϫ3 ഛ x ഛ 3, Ϫ ഛ y ഛ , and ﬁnd its surface area correct to four decimal places.
41. Use polar coordinates to evaluate y 3 0 y s9Ϫx 2 Ϫs9Ϫx 2 ͑x 3 ϩ xy 2 ͒ dy dx. y y
Ϫ2 s4Ϫy 2 y s4Ϫx 2Ϫy 2 Ϫs4Ϫx 2Ϫy 2 0 2
; 43. If D is the region bounded by the curves y 1 Ϫ x and y e , ﬁnd the approximate value of the integral xxD y dA.
(Use a graphing device to estimate the points of intersection
of the curves.) CAS 2 44. Find the center of mass of the solid tetrahedron with vertices ͑0, 0, 0͒, ͑1, 0, 0͒, ͑0, 2, 0͒, ͑0, 0, 3͒ and density function
͑x, y, z͒ x 2 ϩ y 2 ϩ z 2.
45. The joint density function for random variables X and Y is f ͑x, y͒ ͭ C͑x ϩ y͒ if 0 ഛ x ഛ 3, 0 ഛ y ഛ 2
0
otherwise (a) Find the value of the constant C.
(b) Find P͑X ഛ 2, Y ജ 1͒.
(c) Find P͑X ϩ Y ഛ 1͒. 1Ϫy x2 0 f ͑x, y, z͒ dz dy dx as an iterated integral in the order dx dy dz.
48. Give ﬁve other iterated integrals that are equal to
2 y3 yy y
0 0 y2 0 f ͑x, y, z͒ dz dx dy 49. Use the transformation u x Ϫ y, v x ϩ y to evaluate xxR ͑x Ϫ y͒͑͞x ϩ y͒ dA, where R is the square with vertices
͑0, 2͒, ͑1, 1͒, ͑2, 2͒, and ͑1, 3͒. 50. Use the transformation x u 2, y v 2, z w 2 to ﬁnd the volume of the region bounded by the surface
sx ϩ sy ϩ sz 1 and the coordinate planes.
51. Use the change of variables formula and an appropriate trans formation to evaluate xxR xy dA, where R is the square with
vertices ͑0, 0͒, ͑1, 1͒, ͑2, 0͒, and ͑1, Ϫ1͒.
52. The Mean Value Theorem for double integrals says that if f is a continuous function on a plane region D that is of type I
or II, then there exists a point ͑x 0 , y0 ͒ in D such that yy f ͑x, y͒ dA f ͑x , y ͒ A͑D͒
0 0 D Use the Extreme Value Theorem (15.7.8) and Property 16.3.11
of integrals to prove this theorem. (Use the proof of the singlevariable version in Section 6.5 as a guide.) ͑a, b͒. Let Dr be the closed disk with center ͑a, b͒ and radius r.
Use the Mean Value Theorem for double integrals (see Exercise
52) to show that y 2sx 2 ϩ y 2 ϩ z 2 dz dx dy x 1 Ϫ1 y y y 53. Suppose that f is continuous on a disk that contains the point 42. Use spherical coordinates to evaluate
2 800 hours. If we model the probability of failure of the bulbs
by an exponential density function with mean 800, ﬁnd the
probability that all three bulbs fail within a total of 1000 hours.
47. Rewrite the integral 34. Above the paraboloid z x 2 ϩ y 2 and below the halfcone
■ 1087 46. A lamp has three bulbs, each of a type with average lifetime and y ϩ z 3 2 ❙❙❙❙ lim rl0 1
r 2 yy f ͑x, y͒ dA f ͑a, b͒
Dr 1
dA, where n is an integer and D
͑x 2 ϩ y 2 ͒n͞2
D
is the region bounded by the circles with center the origin
and radii r and R, 0 Ͻ r Ͻ R.
(b) For what values of n does the integral in part (a) have a
limit as r l 0 ϩ?
1
(c) Find yyy 2
dV , where E is the region
͑x ϩ y 2 ϩ z 2 ͒n͞2
E
bounded by the spheres with center the origin and radii r
and R, 0 Ͻ r Ͻ R.
(d) For what values of n does the integral in part (c) have a
limit as r l 0 ϩ? 54. (a) Evaluate yy 5E16(pp 10881089) 1/18/06 PROBLEMS
PLUS 4:49 PM Page 1088 1. If ͠x͡ denotes the greatest integer in x, evaluate the integral yy ͠x ϩ y͡ dA
R Խ where R ͕͑x, y͒ 1 ഛ x ഛ 3, 2 ഛ y ഛ 5͖.
2. Evaluate the integral
1 yy
0 1 0 2 2 e max͕x , y ͖ dy dx where max ͕x 2, y 2 ͖ means the larger of the numbers x 2 and y 2.
3. Find the average value of the function f ͑x͒ xx1 cos͑t 2 ͒ dt on the interval [0, 1]. 4. If a, b, and c are constant vectors, r is the position vector x i ϩ y j ϩ z k, and E is given by the inequalities 0 ഛ a ؒ r ഛ ␣, 0 ഛ b ؒ r ഛ , 0 ഛ c ؒ r ഛ ␥, show that
͑␣␥͒2 yyy ͑a ؒ r͒͑b ؒ r͒͑c ؒ r͒ dV 8 a ؒ ͑b ϫ c͒ Խ E Խ 1
dx dy is an improper integral and could be deﬁned as the
5. The double integral y y
0 0 1 Ϫ xy
limit of double integrals over the rectangle ͓0, t͔ ϫ ͓0, t͔ as t l 1Ϫ. But if we expand the
integrand as a geometric series, we can express the integral as the sum of an inﬁnite series.
Show that
1 1 1 y y
0 1 0 ϱ
1
1
dx dy ͚ 2
1 Ϫ xy
n1 n 6. Leonhard Euler was able to ﬁnd the exact sum of the series in Problem 5. In 1736 he proved that
ϱ ͚ n1 1
2
2
n
6 In this problem we ask you to prove this fact by evaluating the double integral in Problem 5.
Start by making the change of variables
uϪv
s2 x y uϩv
s2 This gives a rotation about the origin through the angle ͞4. You will need to sketch the
corresponding region in the uvplane.
[Hint: If, in evaluating the integral, you encounter either of the expressions
͑1 Ϫ sin ͒͞cos or ͑cos ͒͑͞1 ϩ sin ͒, you might like to use the identity
cos sin͑͑͞2͒ Ϫ ͒ and the corresponding identity for sin .]
7. (a) Show that
1 1 yyy
0 0 1 0 ϱ
1
1
dx dy dz ͚ 3
1 Ϫ xyz
n1 n (Nobody has ever been able to ﬁnd the exact value of the sum of this series.) 1088 5E16(pp 10881089) 1/18/06 4:49 PM Page 1089 (b) Show that
1 1 yyy
0 0 1 0 ϱ
1
͑Ϫ1͒ nϪ1
dx dy dz ͚
1 ϩ xyz
n3
n1 Use this equation to evaluate the triple integral correct to two decimal places.
8. Show that y ϱ 0
arctan x Ϫ arctan x
dx
ln
x
2 by ﬁrst expressing the integral as an iterated integral.
9. If f is continuous, show that
x y z 0 0 0 yyy x f ͑t͒ dt dz dy 1 y ͑x Ϫ t͒2 f ͑t͒ dt
2
0 10. (a) A lamina has constant density and takes the shape of a disk with center the origin and radius R. Use Newton’s Law of Gravitation (see Section 14.4) to show that the magnitude
of the force of attraction that the lamina exerts on a body with mass m located at the point
͑0, 0, d ͒ on the positive z axis is ͩ F 2 Gm d 1
1
Ϫ
2
d
sR ϩ d 2 ͪ [Hint: Divide the disk as in Figure 4 in Section 16.4 and ﬁrst compute the vertical component of the force exerted by the polar subrectangle Rij.]
(b) Show that the magnitude of the force of attraction of a lamina with density that occupies
an entire plane on an object with mass m located at a distance d from the plane is
F 2 Gm
Notice that this expression does not depend on d. 1089 ...
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 Linear Algebra, Algebra, rij, y͒ dA

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