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Unformatted text preview: 5E-16(pp 1016-1025) 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. M ultiple Integrals 5E-16(pp 1016-1025) 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 x is defined for a x b, we start by dividing the interval a, b into n subb a n and we choose sample points x* in these intervals x i 1, x i of equal width x i subintervals. Then we form the Riemann sum n f x* i 1 x i1 and take the limit of such sums as n l y 2 b a to obtain the definite integral of f from a to b : n f x dx f x* i lim nl x i1 0, the Riemann sum can be interpreted as the sum of the In the special case where f x areas of the approximating rectangles in Figure 1, and xab f x d x represents the area under the curve y f x from a to b. y Îx f(x*) i 0 a x* ¡ FIGURE 1 ⁄ ¤ x* ™ ‹ xi-1 x* £ xi b xn-1 x* i x * xn Volumes and Double Integrals In a similar manner we consider a function f of two variables defined on a closed rectangle R a, b c, d x, y 2 a x b, c y d 1017 5E-16(pp 1016-1025) 1018 ❙❙❙❙ 1/18/06 4:14 PM Page 1018 CHAPTER 16 MULTIPLE INTEGRALS z and we first suppose that f x, y 0. The graph of f is a surface with equation z Let S be the solid that lies above R and under the graph of f, that is, z=f(x, y) 0 c a b 0 z f x, y , x, y R (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, b into m subintervals x i 1, x i of equal width x b a m and dividing c, d into n subintervals yj 1, yj of equal width y d c n. By drawing lines parallel to the coordinate axes through the endpoints of these subintervals as in Figure 3, we form the subrectangles d y x 3 x, y, z S f x, y . R FIGURE 2 Rij x i 1, x i each with area A yj 1, yj x, y x i x 1 x i, yj y 1 yj x y. y R ij d (xi, yj) (x * , y* ) ij ij yj Îy yj_1 › c (x * * £™, y £™) F IGURE 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 z z 0 c a 0 f(x * , y* ) ij ij d y y b x x Rij FIGURE 4 A FIGURE 5 5E-16(pp 1016-1025) 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 n ** f x ij , yij V 3 A i 1j 1 (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 V 4 n ** f x ij , yij lim m, n l A i 1j 1 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, y dA n ** f x ij , yij lim m, n l A i 1j 1 R if this limit exists. The precise meaning of the limit in Definition 5 is that for every number an integer N such that m yy f 0 there is n ** f x ij , yij x, y d A A i 1j 1 R ** 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 right-hand corner of Rij [namely x i, yj , see Figure 3], then the expression for the double integral looks simpler: m 6 yy f x, y dA n lim m, n l R f xi, yj A i 1j 1 By comparing Definitions 4 and 5, we see that a volume can be written as a double integral: If f x, y 0, then the volume V of the solid that lies above the rectangle R and below the surface z f x, y is V yy f R x, y dA 5E-16(pp 1016-1025) 1020 ❙❙❙❙ 1/18/06 4:14 PM Page 1020 CHAPTER 16 MULTIPLE INTEGRALS The sum in Definition 5, m n ** f x ij , yij A i 1j 1 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, 2 0, 2 and below the elliptic paraboloid z 16 x 2 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, y 16 x 2 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 FIGURE 6 2 V f x i, yj A f 1, 1 A f 1, 2 13 1 71 i 1j 1 z 16 z=16-≈-2¥ A 10 1 f 2, 1 41 A f 2, 2 A 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 F IGURE 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 EXAMPLE 2 If R x, y (b) m=n=8, VÅ44.875 1 x 1, 2 yy s1 R y (c) m=n=16, VÅ46.46875 2 , evaluate the integral x 2 dA 5E-16(pp 1016-1025) 1/18/06 4:15 PM Page 1021 S ECTION 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 x 2 0, we can compute the integral by interpreting it as a volume. If z s1 x 2, then x 2 z 2 1 and z 0, so the given double integral represents the volume of the solid S that lies below the circular cylinder x 2 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 FIGURE 9 1 2 x 2 dA 1 2 4 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 , yj of Rij. In other words, xi is the midpoint of x i 1, x i and yj is the midpoint of yj 1, yj . Midpoint Rule for Double Integrals m yy f n x, y dA f xi , yj A i 1j 1 R where xi is the midpoint of x i 1, x i and yj is the midpoint of yj 1, yj . EXAMPLE 3 Use the Midpoint Rule with m xxR y x, y 0 n x 2 to estimate the value of the integral 2, 1 y 2 . SOLUTION In using the Midpoint Rule with m n 2, we evaluate f x, y at the centers of the four subrectangles shown in Figure 10. So x1 1 , x2 2 and y2 7 . The area of each subrectangle is A 1 . Thus 4 2 (2, 2) 2 3 2 3y 2 dA, where R x R¡™ R™™ R¡¡ R™¡ 2 yy 1 x f xi , yj 2 A i 1j 1 R 1 x 3y 2 , y1 5 , 4 2 3y 2 dA f x1, y1 0 3 2 f( , ) ) 15 24 x ( FIGURE 10 f x1, y2 f( , 17 24 A ( 67 1 16 2 95 8 Thus, we have A ) ) 139 1 16 2 A f( , A ( f x2 , y1 35 24 ) 51 1 16 2 ) ( A A f x2 , y2 f( , 37 24 ) A A ) 123 1 16 2 11.875 yy x 3y 2 dA 11.875 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 12. (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, ■ 5E-16(pp 1016-1025) 1022 ❙❙❙❙ 1/18/06 4:15 PM Page 1022 CHAPTER 16 MULTIPLE INTEGRALS Number of subrectangles 1 4 16 64 256 1024 Midpoint Rule approximations 11.5000 11.8750 11.9687 11.9922 11.9980 11.9995 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, 12. Average Value Recall from Section 6.5 that the average value of a function f of one variable defined on an interval a, b is 1 b fave ya f x d x ba 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 AR yy f x, y dA R where A R is the area of R. 0, the equation If f x, y AR fave yy f x, y 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, y 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 5E-16(pp 1016-1025) 1/18/06 4:15 PM Page 1023 S ECTION 16.1 DOUBLE INTEGRALS OVER RECTANGLES ❙❙❙❙ 1023 SOLUTION Let’s place the origin at the southwest corner of the state. Then 0 x 388, 0 y 276, and f x, y 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 1 AR fave yy f x, y dA R where A R 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 16 A 388 276 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 yy f R 4 x, y d A f xi , yj A i 1j 1 A 0.4 1.2 0.1 6.1 1.8 16.5 3.9 0 8.8 3.9 1.8 4.0 8.0 6.5 16.2 9.4 6693 88.6 6693 88.6 5.5 388 276 On December 24, 1982, Colorado received an average of approximately 5 1 inches of 2 snow. Therefore fave 5E-16(pp 1016-1025) 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, y yy f t x, y dA R |||| Double integrals behave this way because the double sums that define them behave this way. 8 yy c f dA where c is a constant R t x, y for all x, y in R, then 9 yy yy t x, y f x, y dA R |||| 16.1 yy t x, y R c yy f x, y dA x, y dA R If f x, y x, y dA R dA R Exercises 1. (a) Estimate the volume of the solid that lies below 1 2 0 3 6 5 1.5 3 1 4 8 6 2.0 4 3 0 5 8 2.5 5 5 3 1 4 3.0 1, 3 0, 2 , use a Riemann sum with m 4, n 2 to estimate the value of xxR y 2 2 x 2 d A. Take the sample points to be the upper left corners of the subrectangles. n 2 to estimate the value of xxR sin x y d A, where R 0, 0, . Take the sample points to be lower left corners. (b) Use the Midpoint Rule to estimate the integral in part (a). 0 7 8 6 3 0 x 2. If R 3. (a) Use a Riemann sum with m 2 3 4 6. A 20-ft-by-30-ft swimming pool is filled with water. The depth is measured at 5-ft 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 0 z x 2 y 2 and above the rectangle R 0, 2 0, 4 . 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). 0 5 10 15 20 5. A table of values is given for a function f x, y defined on R 1, 3 0, 4 . (a) Estimate xxR f x, y 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 1.0 the surface z x y and above the rectangle R x, y 0 x 6, 0 y 4 . 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). 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, y 2x s52 x 2 y 2 and above the rectangle given by 4, 2 y 6. We use the lines x 3 and y 4 to 5E-16(pp 1016-1025) 1/18/06 4:16 PM Page 1025 S ECTION 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, 1 0, 1 . Use them to estimate xxR f x, y dA to the nearest integer. 66 y 68 1 14 13 12 74 74 76 70 70 11 68 10 74 9 0 x 1 9. A contour map is shown for a function f on the square R 0, 4 0, 4 . (a) Use the Midpoint Rule with m n value of xxR f x, y dA. (b) Estimate the average value of f . 2 to estimate the 11–13 |||| Evaluate the double integral by first identifying it as the volume of a solid. 11. 12. y 13. 4 xxR 3 d A, R xxR 5 x d A, xxR 4 2y d A, ■ 10 0 0 10 20 30 ■ ■ x, y 2 R 0, 1 ■ ■ 14. The integral xxR s9 y 3 0, 1 ■ ■ ■ ■ ■ ■ 2 0, 4 0, 2 , y d A, where R represents the volume of a solid. Sketch the solid. command on a CAS) to estimate 10 20 yy e 30 x2 y2 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 6 5, 0 x y 15. Use a programmable calculator or computer (or the sum 2 0 2, 1 x, y 0 R ■ x dA where R 0, 1 0, 1 . 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 cos x 4 y 4 d A. 17. If f is a constant function, f x, y R 18. If R a, b 0, 1 k, and c, d , show that xxR k d A k b 0, 1 , show that 0 xxR sin x ad y dA c. 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. 5E-16(pp 1026-1035) 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, b c, d . We use the notation xcd f x, y d y to mean that x is held fixed and f x, y 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, y d y is a number that depends on the value of x, so it defines a function of x : y Ax d c f x, y d y If we now integrate the function A with respect to x from x y 1 b a b yy A x dx a d a to x b, we get f x, y d y d x c The integral on the right side of Equation 1 is called an iterated integral. Usually the brackets are omitted. Thus b yy 2 a d c b yy f x, y dy dx a d f x, y 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 yy 3 c b a d yy f x, y dx dy c b f x, y 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 1 x2 x2 y2 2 y2 y1 22 2 x2 12 2 3 2 x2 Thus, the function A in the preceding discussion is given by A x We now integrate this function of x from 0 to 3: 3 yy 0 2 1 x 2 y dy dx 3 yy 0 33 2 0 y 2 1 x 2 y dy dx x 2 dx x3 2 3 0 27 2 3 2 x 2 in this example. 5E-16(pp 1026-1035) 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 2 2 yy x y dx dy 1 y 3 0 2 x y dx dy x3 y 3 2 1 y2 9 2 9y d y 1 y 2 2 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). 4 Fubini’s Theorem If f is continuous on the rectangle |||| 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. R x, y a x yy f b, c y d , then b yy x, y dA a d c f x, y d y d x d yy c b a f x, y d x d y 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, y 0. Recall that if f is positive, then we can interpret the double integral xxR f x, y dA as the volume V of the solid S that lies above R and under the surface z f x, y . But we have another formula that we used for volume in Chapter 6, namely, C 0 x x a A(x) y y V b b A x dx a where A x is the area of a cross-section of S in the plane through x perpendicular to the x-axis. From Figure 1 you can see that A x is the area under the curve C whose equation is z f x, y , where x is held constant and c y d. Therefore FIGURE 1 Visual 16.2 illustrates Fubini’s Theorem by showing an animation of Figures 1 and 2. y Ax d f x, y d y c and we have z yy f x, y dA V y b a A x dx b yy a d c f x, y d y d x R 0 c A similar argument, using cross-sections perpendicular to the y-axis as in Figure 2, shows that y d y yy f x, y dA c b a f x, y d x d y R x R x, y 0 x 2, 1 y xxR x 3y 2 dA, where 2 . (Compare with Example 3 in Section 16.1.) EXAMPLE 2 Evaluate the double integral FIGURE 2 d yy 5E-16(pp 1026-1035) 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 2 3y 2 dA x yy 0 2 1 y 2 x dx 2 x2 2 7 dx 7x 12 0 SOLUTION 2 Again applying Fubini’s Theorem, but this time integrating with respect to x first, we have z=x-3¥ 0.5 y2 1 y 3]y 0 0 _4 0 3y 2 d y d x x 1 y [ xy y z _8 2 R R 0 _12 Page 1028 1.5 22 yy 0 1 x x 2 3y 2 dA yy 1 2 0 3y 2 d x d y x R y FIGURE 3 1 y 2 1 EXAMPLE 3 Evaluate xxR y sin x y x2 x2 2 2 3xy 2 dy x0 6y 2 d y 2 dA, where R 2 2y 3]1 2y 1, 2 0, 12 . SOLUTION 1 If we first integrate with respect to x, we get yy y sin x y yy dA 0 2 1 y sin x y d x d y R y[ cos x y y xx cos 2y 0 0 1 2 2 1 dy cos y d y sin y]0 sin 2y 0 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, y dA is a difference of volumes: V1 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.) yy y sin x y 0 FIGURE 4 1 y 1 x 2 32 0 1 y sin x y d y d x To evaluate the inner integral we use integration by parts with u and so z=y sin(xy) 2 yy R y 0 y du 1 z0 _1 dA dy y sin x y d y dv sin x y d y cos x y x v y cos x y x cos x x cos x x y y0 1 x y 0 cos x y d y 1 [sin x y x2 sin x x2 yy 0 5E-16(pp 1026-1035) 1/18/06 4:56 PM Page 1029 S ECTION 16.2 ITERATED INTEGRALS If we now integrate the first term by parts with u du d x x 2, v sin x, and cos x y |||| 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. cos x y Therefore 2 yy and so x 0 1 sin x x dx x 1 x and dv sin x x2 1029 x d x, we get sin x dx x2 y sin x x dx 2 sin x x y sin x y d y d x cos ❙❙❙❙ 1 sin 2 2 sin 0 EXAMPLE 4 Find the volume of the solid S that is bounded by the elliptic paraboloid x2 2y 2 z 16, the planes x 2 and y 2, and the three coordinate planes. 16 x 2 2y 2 and above the square R 0, 2 0, 2 . (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 SOLUTION We first observe that S is the solid that lies under the surface z 16 12 z8 4 0 0 0.5 1 y 1.5 yy V 0 1 0.5 2 2 1.5 x x2 2y 2x]x 2 2y 2 dA 1 3 16 yy 0 2 x2 16 0 2y 2 d x d y R 2 y [16x FIGURE 5 0 2 y( 0 x2 0 x3 4y 2 ) dy 88 3 dy [ 88 y 3 4 3 2 y 3 ]0 48 In the special case where f x, y 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, y t x h y and R a, b c, d . Then Fubini’s Theorem gives yy f d yy x, y dA c b a t x h y dx dy d yy c b a t x h y dx dy R In the inner integral y is a constant, so h y is a constant and we can write d yy c b a t x h y dx dy y d hy c y b a y b a t x dx dy d t x dx y h y dy c since xab t x d x is a constant. Therefore, in this case, the double integral of f can be written as the product of two single integrals: yy t x h y R dA y b a d t x dx y h y dy c where R a, b c, d 5E-16(pp 1026-1035) ❙❙❙❙ 1030 1/18/06 4:57 PM Page 1030 CHAPTER 16 MULTIPLE INTEGRALS EXAMPLE 5 If R 0, 2 0, 2 , then 2 yy sin x cos y dA y 2 sin x d x y 0 0 cos y d y R [ [sin y] 2 cos x 0 2 11 0 1 z sin x cos y in |||| The function f x, y 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, y dx and x04 f x, y dy. |||| 1. f x, y ■ 3x 2 y 2x ■ ■ ■ 16. y 2. f x, y ■ ■ ■ R x ■ x2 dA, y2 1 1 yy R x, y 0 x 1, 0 y 1 2 ■ ■ ■ ■ 17. yy x sin x y dA, R 0, 6 0, 3 R 3–12 Calculate the iterated integral. |||| 3 3. yy 5. 1 yy 7. 1 1 0 2 0 2 0 2 yy 0 1 1 x2 y 8 dx dy yy 6. x sin y dy d x 2x 4 4. 4 xy dx dy y y (x 8. 0 2 4 1 2 1 x yy 1 R 19. sy ) dx dy 0 yy 0 1 18. y 2 dy dx yy x ye 11. 12. 4 yy 1 x y 2 1 ln 2 yy 0 yy 0 2x y 1 sx 0 ■ 13–20 e 0 1 ■ ln 5 y x 2 ■ 10. dy dx 2 1 2 yy 1 xe x dy dx y 20. 1 0 x y 2 ■ yy x R R 0, 1 0, 1 0, 2 dA, ■ R 1, 2 ■ ■ 0, 1 ■ ■ ■ ■ ■ ■ |||| Sketch the solid whose volume is given by the iterated 1 1 ■ ■ ■ ■ ■ ■ ■ ■ 22. yy 0 1 0 1 0 1 0 4 x 2 2y d x d y x2 y 2 dy dx Calculate the double integral. |||| 6 x 2y 3 yy cos x yy y2 ■ yy dy dx 5y 4 dA, R x, y 0 x 3, 0 y 2 y dA, R x, y 0 x ,0 y 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ 3x R 2y z 12 and above the rectangle x, y 0 x 1, 2 y 3 . 24. Find the volume of the solid that lies under the hyperbolic xy 2 x2 ■ 23. Find the volume of the solid that lies under the plane 2 R 15. 2 ■ 21. R 14. 0, 1 integral. ■ 13. R dx dy dx dy ■ dA, x yy 21–22 xy y2 x2y dA, R R 9. xy 1 dA, R x, y 0 x 1, 3 y 3 paraboloid z R 1, 1 4 x2 0, 2 . y 2 and above the square ■ 5E-16(pp 1026-1035) 1/18/06 4:57 PM Page 1031 S ECTION 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS 25. Find the volume of the solid lying under the elliptic paraboloid x 2 4 R 1, 1 y2 9 2, 2 . z C AS e x sin y and the planes x 0. 1, y 0, y y and the planes x , e 33–34 |||| 0, x 1, y 0, y x 2 y, 34. f x, y 1, 28. Find the volume of the solid bounded by the elliptic paraboloid y 4y 2, the planes x 3 and y CAS ■ 9 1 yy y 2 and the plane x 2. 30. (a) Find the volume of the solid bounded by the surface ; CAS z 6 x y and the planes x 2, x and z 0. (b) Use a computer to draw the solid. 2, y 0, y ■ 1 0 x x R 0, 4 ■ ■ 1, 5 , 1, 5 , 1, 0 0, 1 ■ y dy dx y3 ■ ■ ■ ■ ■ 1 yy and 0 1 0 x x y dx dy y3 Do the answers contradict Fubini’s Theorem? Explain what is happening. 36. (a) In what way are the theorems of Fubini and Clairaut 3, similar? (b) If f x, y is continuous on a, b x yy t x, y 31. Use a computer algebra system to find the exact value of the integral xxR x 5y 3e x y d A, where R 0, 1 0, 1 . Then use the CAS to draw the solid whose volume is given by the integral. |||| 16.3 e y, 1, 0 , 35. Use your CAS to compute the iterated integrals 2, and the 29. Find the volume of the solid in the first octant bounded by the R has vertices e sx ■ 0 cylinder z Find the average value of f over the given rectangle. 33. f x, y ■ z1 x 12 coordinate planes. x2 cos x 2 y 2 and z 2 x 2 y 2 for x 1, y 1. Use a computer algebra system to approximate the volume of this solid correct to four decimal places. z 27. Find the volume of the solid bounded by the surface z x sx 2 and z 0. 1031 32. Graph the solid that lies between the surfaces 1 and above the rectangle 26. Find the volume of the solid enclosed by the surface z1 and z ❙❙❙❙ for a x b, c a y y c c, d and f s, t dt ds d, show that txy f x, y . tyx 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 F x, y 1 f x, y 0 y if x, y is in D if x, y is in R but not in D y R D 0 FIGURE 1 D x 0 FIGURE 2 x 5E-16(pp 1026-1035) 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, y dA D yy F x, y dA where F is given by Equation 1 R D x Definition 2 makes sense because R is a rectangle and so xxR F x, y dA has been previously defined in Section 16.1. The procedure that we have used is reasonable because the values of F x, y are 0 when x, y 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. 0 we can still interpret xxD f x, y dA as the volume of the In the case where f x, y solid that lies above D and under the surface z f x, y (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 F x, y 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 F x, y dA exists and therefore xxD f x, y 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 D x, y a x b, t1 x y t2 x where t1 and t 2 are continuous on a, b . 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 x b 0 a b x FIGURE 5 Some type I regions y In order to evaluate xxD f x, y dA when D is a region of type I, we choose a rectangle R a, b c, d 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 yy f D D c y=g¡(x) 0 FIGURE 6 a x, y dA x b x yy F x, y dA b yy a d c F x, y d y d x R Observe that F x, y Therefore 0 if y y d c F x, y d y t1 x or y y t2 x t1 x t 2 x because x, y then lies outside D. F x, y d y y t2 x t1 x f x, y d y 5E-16(pp 1026-1035) 1/18/06 4:58 PM Page 1033 S ECTION 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS ❙❙❙❙ 1033 because F x, y f x, y when t1 x y t 2 x . 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 D x, y a yy f then x b, t1 x b yy x, y dA t2 x t1 x a y t2 x f x, y d y d x D y 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, y but also in the limits of integration, t1 x and t 2 x . We also consider plane regions of type II, which can be expressed as d x=h¡(y) D x=h™(y) c D 4 0 x, y c y d, h1 y x h2 y 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) yy f 5 0 c d yy x, y dA c h2 y h1 y f x, y d x d y D x where D is a type II region given by Equation 4. FIGURE 7 Some type II regions EXAMPLE 1 Evaluate y 1 xxD x x 2. y=1+≈ 1 x 2, that is, x 2 1, so x 1. We 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) D D _1 2y dA, where D is the region bounded by the parabolas SOLUTION The parabolas intersect when 2 x 2 y (_1, 2) 2 x 2 and y x yy x 2y dA 1 1 yy 1 1 x2 y x2 1 2x2 x x 2, Equation 3 1 2y d y d x D 1 FIGURE 8 1, 2 x 2 x 2 x 2 and the upper boundary is y Since the lower boundary is y gives y=2≈ 1 x, y y [ xy 1 y y 1 1 1 1 x1 y 1 x2 2x 2 y 2]y dx x2 1 3x 4 x5 3 5 x3 x4 4 2x 2 x3 2 3 x2 2 x 2x 2 x 1 dx x2 2 1 x 1 2x 2 32 15 2 dx 5E-16(pp 1026-1035) ❙❙❙❙ 1034 1/18/06 4:59 PM Page 1034 CHAPTER 16 MULTIPLE INTEGRALS y N OTE 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 t1 x , which gives the lower limit in the integral, and the arrow ends at the upper boundary y t 2 x , 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 x 2 y 2 and 2 x and the parabola y x 2. EXAMPLE 2 Find the volume of the solid that lies under the paraboloid z 0 1 above the region D in the xy-plane bounded by the line y x 2 SOLUTION 1 From Figure 9 we see that D is a type I region and FIGURE 9 D as a type I region D y x, y 0 x2 Therefore, the volume under z 4 2, x 2 x y 2x y 2 and above D is (2, 4) yy V x= 1 y 2 x2 2 y 2 dA yy 2x x2 0 x2 y 2 dy dx D x=œ„ y y 2 0 D y x 0 x6 3 2 0 FIGURE 10 y 2x y3 3 x2y x y dx 2x 3 x 2 2x 0 y x2 14 x 3 3 4 2 x7 21 dx 3 x2 3 x 2x 2 x5 5 7x 4 6 2 0 3 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 xy-plane, below the paraboloid z x 2 y 2, and between the plane y 2 x and the parabolic cylinder y x 2. y=≈ yy x2 z=≈+¥ y 2 15 FIGURE 11 y=2x 4 y 2 dA y 4, 1 y 2 y yy 0 D y5 2 sy 1 2 y x2 x sy x3 3 4 0 x 0 sy } x Therefore, another expression for V is V z { x, y D y 2x dy x 2 7 y7 2 1 2 13 96 y 4 0 y y 2 dx dy y3 2 3 y5 2 y3 24 y3 2 dy 216 35 y 4 40 EXAMPLE 3 Evaluate xxD x y dA, where D is the region bounded by the line y the parabola y 2 2 x 6. x 1 and 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: D {(x , y) 2 y 4, 1 y 2 2 3 x y 1} 5E-16(pp 1026-1035) 1/18/06 4:59 PM Page 1035 ❙❙❙❙ SECTION 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS y y (5, 4) (5, 4) ¥ x= -3 2 y=œ„„„„„ 2x+6 y=x-1 x=y+1 x 0 _3 x 0 (_1, _2) _2 (_1, _2) y=_ œ„„„„„ 2x+6 FIGURE 12 1035 (a) D as a type I region (b) D as a type II region Then (5) gives 4 yy x y dA y y 1 2 2 D 1 2 1 2 z y1 4 y6 24 x+2y+z=2 T y 4 y2 3 3) 2 dy 2y 2 8y d y 4 y3 2 3 4y 2 36 2 If we had expressed D as a type I region using Figure 12(a), then we would have obtained y 1 yy x y dA y y (0, 1, 0) 0 4y 3 dy 1 2 x ( 1 y2 2 2 y5 4 4 1 2 x=2y 1 2 2 (0, 0, 2) 2 y y[ y y y xy dx dy y2 3 xy1 x2 y 2 4 s2 x 6 3 s2 x 6 xy dy dx 5 yy 1 s2 x 6 x1 xy dy dx D 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 FIGURE 13 x y 1 x+2y=2 x ”or y=1- ’ 2 1 ”1, 2 ’ D x y= 2 0 FIGURE 14 1 x 2y, x 0, and z 2y z 2, 0. 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 2y z 2. Since the plane x 2y z 2 intersects the xy-plane (whose equation is z 0) in the line x 2y 2, we see that T lies above the triangular region D in the xy-plane bounded by the lines x 2y, x 2y 2, and x 0. (See Figure 14.) The plane x 2y z 2 can be written as z 2 x 2y, so the required volume lies under the graph of the function z 2 x 2y and above D { x, y 0 x 1, x 2 y 1 x 2} 5E-16(pp 1036-1043) 1036 ❙❙❙❙ 1/18/06 4:20 PM Page 1036 CHAPTER 16 MULTIPLE INTEGRALS Therefore yy V 2 x 1 yy 2y dA 0 1 x2 2 x2 x 2y d y d x D 1 y [2y 0 y y 1 x2 xy y2 x x1 x 2 1 2 0 y 1 0 x 2 2x dx y x2 EXAMPLE 5 Evaluate the iterated integral y D y=x 1 yy 1 0 x dx 1 3 x 0 x01 xx1 sin x2 4 y 2 d y d x. 1 x sin y 2 d y d x yy sin y 2 dA D where FIGURE 15 D x, y 0 x 1, x y 1 We sketch this region D in Figure 15. Then from Figure 16 we see that an alternative description of D is D x, y 0 y 1, 0 x y D as a type I region 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 x2 2 x SOLUTION If we try to evaluate the integral as it stands, we are faced with the task of first evaluating x sin y 2 d y. But it’s impossible to do so in finite terms since x sin y 2 d y 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 0 2 1 x3 3 1 dx x 2 1 yy D 0 x=y 1 x sin y 2 d y d x yy sin y dA D 1 yy 0 2 0 x y 1 0 FIGURE 16 1 2 D as a type II region y 0 1 y sin y 2 d y 1 xy x0 y [ x sin y ] sin y 2 d x d y 2 0 1 2 cos y 2 dy 1 0 cos 1 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, y t x, y dA D 7 yy f x, y dA yy t x, y D yy c f D x, y dA D c yy f x, y dA D dA 5E-16(pp 1036-1043) 1/18/06 4:20 PM Page 1037 S ECTION 16.3 DOUBLE INTEGRALS OVER GENERAL REGIONS If f x, y y ❙❙❙❙ 1037 t x, y for all x, y in D, then D yy f 8 D¡ D™ yy t x, y x, y dA 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 x d x xac f x d x xcb f x d x. If D D1 D2 , where D1 and D2 don’t overlap except perhaps on their boundaries (see Figure 17), then yy f 9 y dA D yy f x, y dA yy f x, y dA D1 D x, y dA 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, y 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 10 AD D y Figure 19 illustrates why Equation 10 is true: A solid cylinder whose base is D and whose height is 1 has volume A D 1 A D , 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 11 If m f x, y M for all x, y in D, then (b) D=D¡ D™, D¡ is type I, D™ is type II. yy f mA D x, y dA MA D 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 and therefore 1 sin x 1 and 0 e D x FIGURE 19 Cylinder with base D and height 1 y Thus, using m e 1 1 e, M 4 e 1 1 1, we have cos y e sin x cos y e1 D sin x cos y sin x cos y e 2 2 in Property 11, we obtain e, and A D yy e 1 dA 4e 1 5E-16(pp 1036-1043) ❙❙❙❙ 1038 1. 3. 5. Page 1038 Exercises 21. Under the surface z Evaluate the iterated integral. |||| 1 yy x2 x 0 0 1 yy ey 2 yy cos ■ ■ 1 yy 0 ■ ■ ■ yy 0 0 ■ x y dx dy x2 y dy dx x, z 1, y x z y ■ ■ ■ x, y ■ 7. Evaluate the double integral. |||| 32 yy x y dA, D x, y 0 z x 2, x y 4y yy x D 9. 2 2y yy x2 D 10. 3 yy e 1 y2 dA, D x, y 1 x 2, 0 y 2x dA, D { x, y 0 x 1, 0 y sx } D x, y 0 y 1, 0 x yy e xy dA, x, y 1 y 2, y x y3 yy x sy 2 x dA, D x, y 0 y 1, 0 x yy x cos y dA, x 2, x 0, y D is bounded by y 31–32 yy x y dA, D is bounded by y yy y 3 yy x y dA, yy 2 dA, 2x ■ ■ z2 ■ r2 ■ ■ ■ D is enclosed by x 0 and x s1 y CAS y dA, 2 xy dA, ■ ■ ■ ■ ■ ■ ■ Find the volume of the given solid. 19. Under the plane x 2y x and y 20. Under the surface z 2 y and x ■ 2 ■ x 2 and the y ■ ■ ■ ■ 2x y 3 z 0 and above the region x4 y 2 and above the region bounded 2 34. Between the paraboloids z ■ ■ 36. Enclosed by z ■ ■ ■ x2 1 x ■ 2 y 2 and z 2 y and z ■ 2 2x y and z y2 1 and inside the cylinder x 2 35. Enclosed by z ■ bounded by y ■ 3y, z ■ ■ ■ 33–36 |||| Use a computer algebra system to find the exact volume of the solid. D is the triangular region with vertices 0, 0 , ■ by x 2, x 3 y 4 x y 2 and above the region bounded by the curves y x 3 x and y x 2 x for x 1, 2 , and 0, 3 |||| z y x 2, 1 33. Under the surface z D ■ Find the volume of the solid by subtracting two volumes. 2 D is bounded by the circle with center the origin and radius 2 yy |||| planes z D 19–28 ■ r 2 and y 2 32. The solid enclosed by the parabolic cylinder y D ■ ■ 1 and the planes x yx 2 x 2 y z 10 0 ■ 18. ■ 2 D is the triangular region with vertices (0, 2), (1, 1), and 3, 2 17. ■ y2 31. The solid enclosed by the parabolic cylinders y x2 sx and y D 16. z, 1 D 15. 1 and the planes y 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.) y D 14. y2 0 in the first octant ; 30. Find the approximate volume of the solid in the first octant 2 D 13. 2y, x-coordinates of the points of intersection of the curves y x 4 and y 3x x 2. If D is the region bounded by these curves, estimate xxD x dA. D 12. 4 and the planes x ; 29. Use a graphing calculator or computer to estimate the y D ■ x and the planes z2 28. Bounded by the cylinders x 2 ■ 0 0 in the first octant 0, z x D 11. x ,y 27. Bounded by the cylinder x 2 dA, 2, and z y 2 4 0, z x 0, and x, x 2 26. Bounded by the cylinder y 2 x D 8. 0, y 0, z 0, y 25. Enclosed by the cylinders z 7–18 0, 1 24. Bounded by the planes z ■ 3y 2 and the planes x 0 23. Bounded by the planes x v 2 du dv s1 x2 22. Enclosed by the paraboloid z y 2x v x y and above the triangle with vertices 1, 1 , 4, 1 , and 1, 2 x 1 6. 2 y 1 4. e sin dr d 0 0 2 yy 2. 2y dy dx sx dx dy y 0 ■ 4:21 PM CHAPTER 16 MULTIPLE INTEGRALS |||| 16.3 1–6 1/18/06 ■ 8 x 2 0 2y 2 0 2y ■ ■ ■ ■ 37– 42 ■ |||| Sketch the region of integration and change the order of integration. 37. 4 yy 0 sx 0 f x, y d y d x 38. 1 yy 0 4 4x f x, y d y d x ■ 5E-16(pp 1036-1043) 1/18/06 4:22 PM Page 1039 S ECTION 16.4 DOUBLE INTEGRALS IN POLAR COORDINATES 39. 41. 3 yy s9 y 2 s9 y 2 0 2 yy 1 ■ ln x 40. f x, y d x d y 0 42. f x, y d y d x 0 ■ 3 yy ■ ■ ■ ■ s9 y 0 1 0 51. 4 yy arctan x ■ 51–52 f x, y d x d y ■ f x, y d y d x ■ ■ ■ ■ 52. |||| Evaluate the integral by reversing the order of integration. 1 yy 45. yy 47. 3 2 48. 0 9 y2 1 0 yy 46. yy 2 arcsin y 8 yy 2 1 0 1 sy 1 x2 sx 3 yy e x2 y2 ■ yy f ■ ■ ■ ■ ■ ■ 50. dA ■ ■ ■ ■ ■ ■ 1 yy x, y dA 0 2 ■ ■ ■ ■ ■ ■ ■ 2y 0 f x, y d x d y 3 yy 1 3y 0 f x, y d x d y 3x 4y dA, where D is the region bounded by the square with vertices 5, 0 and 0, 5 . y=1+≈ (1, 1) D x=1 57. Compute xxD s1 2 0 1 x x=_1 D CAS x y=_1 ■ |||| 16.4 ■ ■ ■ ■ ■ ■ ■ 2 x y of a solid. x=¥ 0 _1 4 dA, where 56. Use symmetry to evaluate xxD 2 y 1 y3 2 D x, y x y 2. [Hint: Exploit the fact that D is symmetric with respect to both axes.] D ■ D is the disk with center the origin and radius 1 2 dA, 55. Evaluate xxD x 2 tan x yy x y dA y ■ 0, 1 Sketch the region D and express the double integral as an iterated integral with reversed order of integration. D _1 0, 1 iterated integrals was obtained as follows: |||| 2 D D ■ yy x y 3 dA, 54. In evaluating a double integral over a region D, a sum of x 3 sin y 3 d y d x Express D as a union of regions of type I or type II and evaluate the integral. 49. 3 53. Prove Property 11. 1 dx dy cos 2 x d x d y cos x s1 yy s x ■ e x dx dy ■ 49–50 0 Use Property 11 to estimate the value of the integral. D 4 3 sy 0 ■ 44. y cos x 2 d x d y 3y 3 1 e x dx dy yy 0 1039 D 43–48 43. |||| ❙❙❙❙ ■ ■ x 2 y 2 dA, where D is the disk 1, by first identifying the integral as the volume 58. Graph the solid bounded by the plane x y z 1 and the paraboloid z 4 x 2 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, y 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 5E-16(pp 1036-1043) 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, angular coordinates x, y by the equations r2 y ¨ O Page 1040 x x x2 y2 x of a point are related to the rect- r cos y r sin (See Section 11.3.) The regions in Figure 1 are special cases of a polar rectangle R FIGURE 2 r, a r b, which is shown in Figure 3. In order to compute the double integral xxR f x, y dA, where R is a polar rectangle, we divide the interval a, b into m subintervals ri 1, ri of equal width r b a m and we divide the interval , into n subintervals j 1, j of equal width 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 Ψ r=a r=ri ¨=å r=ri _1 ∫ å O O F IGURE 3 Polar rectangle FIGURE 4 Dividing R into polar subrectangles The “center” of the polar subrectangle Rij r, ri r ri , j1 * 1 1 2 j has polar coordinates r i* 1 2 ri ri 1 j j1 j 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 j 1 , we find that the area of Rij is Ai 12 2i 1 2 2i1 r 1 2 ri 1 2 r ri 1 ri ri 1 r2i 2 ri 1 r* r i Although we have defined the double integral xxR f x, y 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 n m f r* cos j*, r* sin j* i i 1 i 1j 1 n f r* cos j*, r* sin j* r* r i i i Ai i 1j 1 5E-16(pp 1036-1043) 1/18/06 4:23 PM Page 1041 S ECTION 16.4 DOUBLE INTEGRALS IN POLAR COORDINATES If we write t r, ten as r f r cos , r sin m ❙❙❙❙ 1041 , then the Riemann sum in Equation 1 can be writn t r*, j* i r i 1j 1 which is a Riemann sum for the double integral yy b a t r, dr d Therefore, we have m yy f x, y dA n f r* cos j*, r* sin j* i i lim m, n l m n t r*, j* i lim m, n l yy Ai i 1j 1 R yy r i 1j 1 b a f r cos , r sin b a t r, dr d r dr d 2 Change to Polar Coordinates in a Double Integral If f is continuous on a polar rect- angle R given by 0 a yy f r b, x, y dA , where 0 yy b a f r cos , r sin 2 , then 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 E XAMPLE 1 Evaluate xxR bounded by the circles x 3x 2 4y 2 dA, where R is the region in the upper half-plane y 1 and x 2 y 2 4. 2 SOLUTION The region R can be described as R x, y y 0, 1 x2 y2 4 It is the half-ring shown in Figure 1(b), and in polar coordinates it is given by 1 r 2, 5E-16(pp 1036-1043) 1042 ❙❙❙❙ 1/18/06 4:24 PM Page 1042 CHAPTER 16 MULTIPLE INTEGRALS 0 . Therefore, by Formula 2, yy 3x 4y 2 dA yy 0 2 4r 2 sin 2 3r cos 1 r dr d R yy 0 2 y [r 3 0 1 2 1 r 4 sin 2 15 2 0 cos 2 1 15 2 7 sin See Section 8.2 for advice on integrating trigonometric functions. 4r 3 sin 2 cos y [7 cos |||| Here we use the trigonometric identity sin2 3r 2 cos 1 rr 2 1 dr d y d z 1 x2 d d cos 2 15 sin 2 4 15 2 0 EXAMPLE 2 Find the volume of the solid bounded by the plane z z 15 sin 2 7 cos 0 0 and the paraboloid y 2. 0 in the equation of the paraboloid, we get x 2 y 2 1. This means that the plane intersects the paraboloid in the circle x 2 y 2 1, so the solid lies under the paraboloid and above the circular disk D given by x 2 y 2 1 [see Figures 6 and 1(a)]. In polar coordinates D is given by 0 r 1, 0 2 . Since 1 x 2 y 2 1 r 2, the volume is SOLUTION If we put z (0, 0, 1) 0 D x yy V y x2 1 2 y 2 dA yy 0 1 0 r 2 r dr d 1 D FIGURE 6 y 2 0 y d 1 0 r 3 r dr r2 2 2 1 r4 4 2 0 If we had used rectangular coordinates instead of polar coordinates, then we would have obtained V yy 1 x2 1 y 2 dA yy 1 s1 x 2 s1 x 2 x2 1 y 2 dy dx D which is not easy to evaluate because it involves finding the following integrals: y s1 r=h™(¨) ¨=∫ D ¨=å ∫ 3 1 x2 If f is continuous on a polar region of the form D r=h¡(¨) FIGURE 7 D=s(r, ¨) | 寨¯∫, h¡(¨)¯r¯h™(¨)d y x 2 dx 32 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 y x s1 x 2 dx then yy f D r, x, y dA , h1 yy h2 h1 r f r cos , r sin h2 r dr d 5E-16(pp 1036-1043) 1/18/06 4:25 PM Page 1043 SECTION 16.4 DOUBLE INTEGRALS IN POLAR COORDINATES In particular, taking f x, y 1, h1 that the area of the region D bounded by 0, and h2 h , , and r yy 1 dA y y AD h ❙❙❙❙ 1043 in this formula, we see h is r dr d 0 D h r2 2 y y d 1 2 2 h d 0 and this agrees with Formula 11.4.3. EXAMPLE 3 Use a double integral to find the area enclosed by one loop of the four-leaved rose r π ¨= 4 cos 2 . SOLUTION From the sketch of the curve in Figure 8 we see that a loop is given by the region { D r, 4 4, 0 r cos 2 } So the area is π ¨=_ 4 yy dA y AD 4 4 y cos 2 r dr d 0 D FIGURE 8 4 y 1 4 4 [ 1 r 2]cos 2 2 0 4 y 1 2 cos 4 1 4 d 4 d y cos 2 2 d 4 [ 1 4 1 4 sin 4 4 4 8 EXAMPLE 4 Find the volume of the solid that lies under the paraboloid z y above the xy-plane, and inside the cylinder x (x-1)@+¥=1 (or r=2 cos ¨) y2 x x 2 2 x. 1 2 y2 1 { r, 2 2, 0 r 2 cos } and, by Formula 3, we have z V x2 yy y 2 dA 2 y 2 y 2 cos 0 r 2 r dr d D r4 4 2 y 2 8y 2 0 2y x 0 y FIGURE 10 y 2, (see Figures 9 and 10). In polar coordinates we have x 2 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 D FIGURE 9 x2 2 x or, after completing the square, D 1 y 2 SOLUTION The solid lies above the disk D whose boundary circle has equation x2 0 2 2[ 3 2 2 2 cos 2 0 cos 4 d 1 2 4y d 8y 0 2 cos 2 sin 2 1 2 1 8 1 2 sin 4 cos 4 d cos 2 2 1 cos 4 0 2 2 3 2 2 d d 2 3 2 5E-16(pp 1044-1053) ❙❙❙❙ 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, y dA as an iterated integral, where f is an arbitrary continuous function on R. |||| y 1. xxR y e x dA, where R is the region in the first quadrant enclosed by the circle x 2 y 2 25 0 xxR arctan y x dA, x, y 1 where R 2x x y 3 2 x2 y2 4, 0 ■ ■ ■ ■ ■ ■ ■ 17. One loop of the rose r R 1 0 0 x 2 ■ 3x 20. The region inside the circle r circle r ■ ■ 21–27 ■ ■ ■ cos 3 4 3 cos 19. The region within both of the circles r 1 x Use a double integral to find the area of the region. |||| 18. The region enclosed by the curve r R y where D is the region in the first quadrant that lies between the circles x 2 y 2 4 and x 2 y 2 2 x 17–20 4. y xxD x dA, ■ y 0 xxD e dA, where D is the region bounded by the semicircle x s4 y 2 and the y-axis 16. 3. 4, x 15. R 2x y2 2 14. R 0 2 y 2 dA, x, y x 2 13. 2 2 x2 where R y 2. xxR s4 cos and r 4 sin sin and outside the 2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ Use polar coordinates to find the volume of the given |||| solid. 5. y 6. 5 disk x 2 R 2 2 y cylinder x 5 0 x 2 2 2 x 2 y 2 and above the 9 22. Inside the sphere x 2 R 0 x2 21. Under the paraboloid z y y 2 y2 z2 23. A sphere of radius a 24. Bounded by the paraboloid z plane z x2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ y2 z2 sx 2 y 2 and below the sphere ■ z 7– 8 |||| Sketch the region whose area is given by the integral and evaluate the integral. 7. yy 7 4 ■ ■ 9–16 8. r dr d ■ ■ ■ ■ 2 yy 0 ■ 4 cos 0 4x r dr d ■ ■ ■ x 2 3x 2 y ■ 2 4y ■ 2 z ■ ■ 2 3y 2 and 2 27. Inside both the cylinder x 2 y2 4 and the ellipsoid 64 ■ ■ ■ ■ ■ ■ ■ ■ 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 ring-shaped 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. 4 ■ ■ 3y 2 and the 1 26. Bounded by the paraboloids z 2 3x 2 10 4 25. Above the cone z ■ 16 and outside the 4 |||| xxD x y dA, where D is the disk with center the origin and radius 3 10. xxR x y dA, where R is the region that lies to the left of the y-axis between the circles x 2 y 2 1 and x 2 y 2 4 29–32 |||| Evaluate the iterated integral by converting to polar coordinates. 11. xxR cos x 2 29. y 2 dA, where R is the region that lies above the x-axis within the circle x 2 y 2 9 1 yy 0 s1 x 2 0 ex 2 y2 dy dx 30. a yy a sa 2 y 2 0 x2 y2 32 dx dy 5E-16(pp 1044-1053) 1/18/06 4:51 PM Page 1045 S ECTION 16.5 APPLICATIONS OF DOUBLE INTEGRALS 31. ■ 2 yy s4 y 2 0 s4 y ■ x 2 y 2 dx dy 32. 2 ■ ■ ■ ■ 2 yy 0 ■ s2x x 2 sx 2 ■ ■ ■ ■ yy 33. A swimming pool is circular with a 40-ft diameter. The depth is constant along east-west 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. yy e r 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. x2 y2 1 s2 x s1 x 2 s2 yy xy dy dx 1 x 0 xy dy dx dA 2 s2 s4 x 2 xy dy dx 0 yy e x2 y2 where Sa is the square with vertices show that y dA x2 e dx y e y2 a, a . Use this to dy (c) Deduce that e x2 dx s (d) By making the change of variable t into one double integral. Then evaluate the double integral. 36. (a) We define the improper integral (over the entire plane dA Sa y yy lim al 2 35. Use polar coordinates to combine the sum 1 x2 y2 e (b) An equivalent definition of the improper integral in part (a) is 34. An agricultural sprinkler distributes water in a circular pattern yy 1045 where Da is the disk with radius a and center the origin. Show that y 2 dy dx 0 ■ ❙❙❙❙ y e x2 2 dx s2 x, show that s2 2 (This is a fundamental result for probability and statistics.) I yy e x2 y2 dA yy e x2 y2 dy dx 2 lim al integrals. yy e x2 y2 dA (a) Da |||| 16.5 37. Use the result of Exercise 36 part (c) to evaluate the following y 0 x 2e x2 (b) dx y 0 sx e x 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 x y-plane and its density (in units of mass per unit area) at a point x, y in D is given by x, y , where is a continuous function on D. This means that y (x, y) D x, y 0 FIGURE 1 lim m A x where m and A are the mass and area of a small rectangle that contains x, y and the limit is taken as the dimensions of the rectangle approach 0. (See Figure 1.) 5E-16(pp 1044-1053) 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, y 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* ij ij m A i 1j 1 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 m 1 l x *, y* ij ij lim k, l l A i 1j 1 yy x, y dA 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, y at a point x, y in D, then the total charge Q is given by yy Q 2 x, y dA 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, y is x, y x y, measured in coulombs per square meter (C m2 ). Find the total charge. (1, 1) D SOLUTION From Equation 2 and Figure 3 we have Q 1 yy x, y dA 1 0 1x dx y x dx 1 2 xy dy dx D y=1-x 0 yy y x y2 x 2 1 0 FIGURE 3 1 2 Thus, the total charge is y 5 24 1 0 2x y1 y1x 2 3 0 x2 1 2 1 2x 3 3 1 x4 4 x 1 0 2 dx 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, y . 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 x-axis by x *, y* ij ij A y* ij If we now add these quantities and take the limit as the number of subrectangles becomes 5E-16(pp 1044-1053) 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 x-axis: m Mx 3 n y* ij lim m, n l x *, y* ij ij yy y A i 1j 1 x, y dA D Similarly, the moment about the y-axis is m (x, y) My 4 n x* ij lim m, n l D x *, y* ij ij yy x A i 1j 1 x, y dA D As before, we define the center of mass x, y 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, y of the center of mass of a lamina occupying the region D and having density function (x, y) are My m x 1 m yy x x, y dA Mx m y D 1 m yy y x, y dA D where the mass m is given by m yy x, y dA D EXAMPLE 2 Find the mass and center of mass of a triangular lamina with vertices 0, 0 , 1, 0 , and 0, 2 if the density function is y (0, 2) x, y 1 3x y. SOLUTION The triangle is shown in Figure 5. (Note that the equation of the upper bound- ary is y 2 2 x.) The mass of the lamina is y=2-2x yy m 3 11 x, y dA 1 yy 0 y D 3x y dy dx y y 2 2x y2 2 1 3xy 0 (1, 0) 1 D ” , ’ 8 16 0 2 2x 0 dx y0 x 1 x 2 dx 4y 1 0 FIGURE 5 x3 3 4x 1 8 3 0 Then the formulas in (5) give x 1 m 3 8 3 2 yy x x, y dA 3 8 1 yy 0 2 2x 0 3x 2 x D y 1 3x 2 y xy 0 y 1 0 x 3 x dx x y2 2 x2 2 y 2 2x dx y0 x4 4 1 0 3 8 xy dy dx 5E-16(pp 1044-1053) 1048 ❙❙❙❙ 1/18/06 4:52 PM Page 1048 CHAPTER 16 MULTIPLE INTEGRALS y 1 m 3 8 yy y 1 3 8 x, y dA yy 0 2 2x y 0 y 2 dy dx 3xy D y 1 0 1 4 y2 2 y2 3x 2 x2 9 2 7x x y 2 2x y3 3 The center of mass is at the point ( , 3 11 8 16 y 1 1 0 y0 x4 5 4 3 1 4 dx 7 3x 2 9x 5x 3 d x 11 16 0 ). 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. y a D _a SOLUTION Let’s place the lamina as the upper half of the circle x 2 y 2 a 2 (see Figure 6). Then the distance from a point x, y to the center of the circle (the origin) is s x 2 y 2. Therefore, the density function is ≈+¥=a@ 3a ”0, ’ 2π 0 K sx 2 x, y a x y2 where K is some constant. Both the density function and the shape of the lamina suggest that we convert to polar coordinates. Then s x 2 y 2 r and the region D is given by 0 r a, 0 . Thus, the mass of the lamina is FIGURE 6 yy m x, y dA D a Kr r dr d 0 K 2 y 2 dA D yy 0 yy K s x r3 3 a Ky d 0 y a 0 r 2 dr K a3 3 0 Both the lamina and the density function are symmetric with respect to the y-axis, so the center of mass must lie on the y-axis, that is, x 0. The y-coordinate is given by y 1 m yy y D 3 a3 |||| 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, 4 a 3 . 3 K a3 x, y dA y 0 3 2a a3 4 sin d 4 y a3 0 r dr yy 0 a 0 3 a3 r sin [ cos r r dr d r4 4 0 a 0 3a 2 Therefore, the center of mass is located at the point 0, 3a 2 . 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, y 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 x-axis, and take the limit of the sum 5E-16(pp 1044-1053) 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 x-axis: m Ix 6 n 2 * yij lim m, n l ** x ij , yij yy y A i 1j 1 2 x, y dA 2 x, y dA D Similarly, the moment of inertia about the y-axis is m Iy 7 n 2 * x ij lim m, n l ** x ij , yij yy x A i 1j 1 D It is also of interest to consider the moment of inertia about the origin, also called the polar moment of inertia: m 8 I0 n m, n l Note that I 0 Ix 2 x* ij lim y* ij 2 x *, y* ij ij yy A i 1j 1 x2 y2 x, y dA D Iy. EXAMPLE 4 Find the moments of inertia I x , I y , and I 0 of a homogeneous disk D with den- sity x, y , center the origin, and radius a. S OLUTION The boundary of D is the circle x 2 described by 0 r x2 2 ,0 I0 yy y 2 a 2 and in polar coordinates D is a. Let’s compute I 0 first: y2 2 yy dA 0 a 0 r 2 r dr d D y 2 0 y d a 0 3 r dr 2 r4 4 a 0 Instead of computing I x and I y directly, we use the facts that I x (from the symmetry of the problem). Thus Ix Iy I0 2 a4 2 Iy I 0 and I x Iy a4 4 In Example 4 notice that the mass of the disk is m density area a2 so the moment of inertia of the disk about the origin (like a wheel about its axle) can be written as I0 1 2 ma 2 Thus, if we increase the mass or the radius of the disk, we thereby increase the moment of 5E-16(pp 1044-1053) 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 9 I 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 x-axis and the radius of gyration x with respect to the y-axis are given by the equations 10 my 2 mx 2 Ix Iy Thus x, y 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 x-axis 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 4 Ix m a4 a2 a2 4 Therefore, the radius of gyration about the x-axis is a 2 y which is half the radius of the disk. Probability In Section 9.5 we considered the probability density function f of a continuous random variable X. This means that f x 0 for all x, x f x dx 1, and the probability that X lies between a and b is found by integrating f from a to b : Pa X b y b a f x dx 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, Y lies in a region D is P X, Y D yy f D x, y dA 5E-16(pp 1044-1053) 1/18/06 4:53 PM Page 1051 S ECTION 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 Pa X b, c Y b yy d a d c f x, y d y d x (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, y yy f 0 x, y dA 1 2 As in Exercise 36 in Section 16.4, the double integral over 2 is an improper integral defined as the limit of double integrals over expanding circles or squares and we can write yy f yy x, y dA f x, y dx dy 1 2 EXAMPLE 6 If the joint density function for X and Y is given by f x, y Cx 0 2y if 0 x otherwise find the value of the constant C. Then find P X 10, 0 7, Y y 10 2. SOLUTION We find the value of C by ensuring that the double integral of f is equal to 1. Because f x, y 0 outside the rectangle 0, 10 0, 10 , we have yy f x, y d y d x 10 yy 0 10 0 Cy 10 0 Therefore, 1500C 1 and so C 1 1500 Cx 2y d y d x 10 x 100 d x . Cy 10 0 1500C [ xy y 10 0 y 2]y dx 5E-16(pp 1044-1053) 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: PX 7, Y 7 yy 2 2 1 1500 7 y [ xy 0 y2 0 868 1500 7 yy f x, y d y d x y 10 y2 10 1 1500 2 1 1500 dx y 7 0 x 2y d y d x 8x 96 d x 0.5787 Suppose X is a random variable with probability density function f1 x and Y is a random variable with density function f2 y . Then X and Y are called independent random variables if their joint density function is the product of their individual density functions: f x, y f1 x f2 y In Section 9.5 we modeled waiting times by using exponential density functions 0 ft 1 e if t if t t 0 0 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 0 f1 x 1 10 e x 10 if x if x 0 0 0 f2 y 1 5 e if y if y y5 0 0 Since X and Y are independent, the joint density function is the product: f x, y y 1 50 f1 x f2 y e x 10 y5 e if x 0, y otherwise 0 We are asked for the probability that X Y 0 20: 20 PX Y 20 P X, Y D x+y=20 where D is the triangular region shown in Figure 8. Thus D PX 0 FIGURE 8 20 x Y 20 yy f x, y dA 20 yy 0 20 x 1 50 0 e x 10 D 1 50 20 y [e 0 x 10 5e y 5 y 20 x y0 dx e y5 dy dx 5E-16(pp 1044-1053) 1/18/06 4:53 PM Page 1053 S ECTION 16.5 APPLICATIONS OF DOUBLE INTEGRALS 1 10 y 1 10 y 20 0 1 x 10 e 20 0 e e x 10 4 ex 1 20 5 ❙❙❙❙ 1053 dx e 4e x 10 d x 2e 2 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 y x f x dx Now if X and Y are random variables with joint density function f, we define the X-mean and Y-mean, also called the expected values of X and Y, to be 11 1 yy x f x, y dA yy y f 2 2 x, y dA 2 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 fx where is the mean and 1 s2 e x 2 2 2 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 and 1 2 f1 x 4.0, 1 0.01. So the individual density functions for X and Y are 1 0.01 s2 e x 4 2 0.0002 f2 y 1 0.01 s2 e 2 y 6 2 0.0002 6.0, 5E-16(pp 1054-1063) 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: 1 0.0002 1500 f x, y 1000 f1 x f2 y e x4 2 0.0002 y6 e 2 0.0002 500 0 5.95 5000 3.95 y 4 6 6.05 e 5000 x 4 2 y6 2 x 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 P 3.98 X 4.02, 5.98 Y 6.02 4.02 yy 3.98 6.02 5.98 5000 f x, y d y d x 4.02 yy 3.98 6.02 5.98 e 5000 x 4 2 y 62 dy dx 0.91 Then the probability that either X or Y differs from its mean by more than 0.02 cm is approximately 1 |||| 16.5 x 11. A lamina occupies the part of the disk x 2 3, 0 y 2 so that the charge density at x, y is x, y 2 x y 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 y 2 4 so that the charge density at x, y is x, y x y x2 y2 (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 . 3. D x, y 0 x 2, 4. D x, y 0 x a, 0 1 y y 1; b; x, y x, y x cxy 2 16. Find the moments of inertia I x , I y , I 0 for the lamina of sx, y 0, x 0, and x 1; x, y x, y x y 2 and the line y x 0, and x 9. D is bounded by the parabola x 1; y x, y 0 y ■ cos x, 0 ■ ■ x ■ 2; ■ ■ Exercise 9. 2; x, y ■ Exercise 12. 17. Find the moments of inertia I x , I y , I 0 for the lamina of 18. Consider a square fan blade with sides of length 2 and the 3 ■ 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 e x, y ■ 13. Find the center of mass of a lamina in the shape of an isosceles 15. Find the moments of inertia I x , I y , I 0 for the lamina of 8. D is bounded by y ■ density at any point is proportional to the square of its distance from the origin. y 2 2y but outside the circle x y 1. Find the center of mass if the density at any point is inversely proportional to its distance from the origin. y 7. D is bounded by y 10. D 12. Find the center of mass of the lamina in Exercise 11 if the 2 6. D is the triangular region with vertices 0, 0 , 1, 1 , 4, 0 ; x, y y 2 1 in the first quadrant. Find its center of mass if the density at any point is proportional to its distance from the x-axis. 14. A lamina occupies the region inside the circle x 2 xy 2 5. D is the triangular region with vertices 0, 0 , 2, 1 , 0, 3 ; x, y 0.09 Exercises 1. Electric charge is distributed over the rectangle 1 x, y 0.91 x ■ ■ lower left corner placed at the origin. If the density of the blade is x, y 1 0.1x, is it more difficult to rotate the blade about the x-axis or the y-axis? 5E-16(pp 1054-1063) 1/18/06 4:29 PM Page 1055 S ECTION 16.6 SURFACE AREA CAS x, y 0 y sin x, 0 x 20. D is enclosed by the cardioid r sx 2 x, y ■ ■ ■ ; 1 x, y xy cos ; y2 ■ CAS ■ ■ ■ ■ ■ ■ ■ ■ 21. A lamina with constant density x, y occupies a square with vertices 0, 0 , a, 0 , a, a , and 0, a . 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, y 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 Cx 1 0 f x, y y if 0 x otherwise 1, 0 y 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 P 40 X 50, 20 Y 25 . (b) Find P 4 X 45 2 100 Y 20 2 2 . 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 f1 x 24. (a) Verify that 4 xy if 0 x 0 otherwise 27. Suppose that X and Y are independent random variables, where 2 (a) Find the value of the constant C. (b) Find P X 1, Y 1 . (c) Find P X Y 1 . f x, y 1, 0 y 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 ❙❙❙❙ 1 e 0 x if x if x 0 0 f2 y 1 50 0 y if 0 y otherwise 10 (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 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 1 ) (ii) P(X 1 , Y 1 ) 2 2 2 (c) Find the expected values of X and Y. 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 A x 0 , y0 , assume that the probability function is given by 25. Suppose X and Y are random variables with joint density fP function f x, y 0.1e 0 0.5x 0.2y if x 0, y otherwise 0 (a) Verify that f is indeed a joint density function. (b) Find the following probabilities. (i) P Y 1 (ii) P X 2, Y 4 (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 1 20 20 d P, A where d P, A 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 single-variable calculus. Here we compute the area of a surface with equation z f x, y , the graph of a function of two variables. 5E-16(pp 1054-1063) 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, y , where f has continuous partial derivatives. For simplicity in deriving the surface area formula, we assume that f x, y 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 , yj 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 AS 1 0 Îx i 1j 1 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 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 fx x i , yj x k Pij a Tij m, n l FIGURE 1 z n lim b ÎTij Îy b y yj fy x i , yj yk and x a FIGURE 2 i x 0 b j k 0 fx xi , yj y fy xi , yj fx x i , yj x yi fx x i , yj i Thus Tij a x y b fy x i , yj fy x i , yj j s fx x i , yj x yj k x yk A 2 fy x i , yj 2 1A 2 fy xi , yj 2 1A From Definition 1 we then have m AS n lim m, n l Tij i 1j 1 m n lim m, n l s fx xi , yj i 1j 1 and by the definition of a double integral we get the following formula. 2 The area of the surface with equation z are continuous, is AS yy s D fx x, y 2 f x, y , x, y fy x, y 2 D, where fx and fy 1 dA 5E-16(pp 1054-1063) 1/18/06 4:31 PM Page 1057 S ECTION 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: yy AS 3 2 z x 1 D z y 2 dA Notice the similarity between the surface area formula in Equation 3 and the arc length formula from Section 9.1: y L 1 a y 2 dy dx b dx x 2 2y that lies above the triangular region T in the x y-plane with vertices 0, 0 , 1, 0 , and 1, 1 . EXAMPLE 1 Find the surface area of the part of the surface z (1, 1) SOLUTION The region T is shown in Figure 3 and is described by y=x T T (0, 0) x2 Using Formula 2 with f x, y x (1, 0) x, y 0 FIGURE 3 yy s 2 x A 2 2 x 1, 0 y x 2y, we get 2 1 yy 1 dA 0 x s4 x 2 0 5 dy dx T z y 1 0 x s4 x 2 1 8 5 dx 2 3 4x 2 5 321 0 1 12 (27 5 s5 ) Figure 4 shows the portion of the surface whose area we have just computed. x2 EXAMPLE 2 Find the area of the part of the paraboloid z y T x plane z y 2 that lies under the 9. SOLUTION The plane intersects the paraboloid in the circle x 2 y 2 9, z 9. Therefore, the given surface lies above the disk D with center the origin and radius 3 (see Figure 5). Using Formula 3, we have FIGURE 4 z yy A 9 z x 1 D yy s1 4 x2 2 z y 2 yy s1 dA 2x 2 2y 2 dA D y 2 dA D Converting to polar coordinates, we obtain D x 3 FIGURE 5 A y 2 yy 0 2 3 0 s1 (1)2 1 83 y 4r 2 r dr d 4r 2 323 0 2 0 6 d y 31 8 0 (37 s37 s1 1) 4r 2 8r dr 5E-16(pp 1054-1063) ❙❙❙❙ 1058 1/18/06 |||| Page 1058 CHAPTER 16 MULTIPLE INTEGRALS |||| 16.6 1–12 4:32 PM Exercises 16. (a) Use the Midpoint Rule for double integrals with Find the area of the surface. 1. The part of the plane z rectangle 0, 5 2 4y that lies above the 1, 4 CAS 2. The part of the plane 2x cylinder x 2 3x y2 5y z 10 that lies inside the 2y z 6 that lies in the 9 3. The part of the plane 3x CAS first octant m n 2 to estimate the area of the surface z x y x 2 y 2, 0 x 2, 0 y 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). 17. Find the exact area of the surface z 1 3 x 2 y 2 that lies above the triangle with vertices 0, 0 , 0, 1 , and 2, 1 4. The part of the surface z 1 CAS 5. The part of the cylinder y 2 6. The part of the paraboloid z x2 4 7. The part of the hyperbolic paraboloid z between the cylinders x 2 3 8. The surface z x 32 y y 9. The part of the surface z x2 y2 32 y 1 and x ,0 x 2 2 x that lies y2 4 1, 0 y C AS 2 y 2 z 2 12. The part of the sphere x 2 paraboloid z 4 that lies above the x ■ 2 y y2 z2 4 z that lies inside the 2 ■ ■ 13. The part of the surface z x y 2 ■ ■ ■ ■ ■ ■ e cos x 2 14. The part of the surface z cylinder x ■ ■ x2 y2 ■ x 1 x 2 y 2 that lies above the disk x 2 1 y 1 y2 1. 20. Find, to four decimal places, the area of the part of the 1 x 2 1 y 2 that lies above the square 1. Illustrate by graphing this part of the surface. y ■ 2 ■ ■ ■ ■ ■ ■ x2 z2 [Hint: Project the surface onto 23. Find the area of the finite part of the paraboloid y 25. 24. The figure shows the surface created when the cylinder y 2 that lies inside the y 2 z 2 1 intersects the cylinder x 2 area of this surface. 1 ■ of the sphere x 2 y 2 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 y 2 a 2. However, the integral can be computed as the limit of the integral over the disk x 2 y 2 t 2 as t l a . Use this method to show that the area of a sphere of radius a is 4 a 2. cut off by the plane y the xz-plane.] that lies above the disk 4 2 2 22. If you attempt to use Formula 2 to find the area of the top half 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. 2 x2 a x by c that projects onto a region D in the x y-plane with area A D is sa 2 b 2 1 A D . y 2 z 2 a 2 that lies within the a x and above the xy-plane y2 ■ y x y that lies within the cylinder 1 ■ 1 surface z x y 1 11. The part of the sphere x 2 ■ x 21. Show that the area of the part of the plane z 10. The part of the sphere x cylinder x 2 1 19. Find, to four decimal places, the area of the part of the surface z 2 1 plane z 1. y Illustrate by graphing the surface. CAS 2 4y 2, 3y y 2 that lies above the xy-plane 2 2x 18. Find the exact area of the surface z z 2 9 that lies above the rectangle with vertices 0, 0 , 4, 0 , 0, 2 , and 4, 2 4, 0 x 1 z2 1. Find the ■ 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 y 2 that lies above the square 0, 1 0, 1 . (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 5E-16(pp 1054-1063) 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: z x, y, z a B 1 x b, c y d, r z s The first step is to divide B into sub-boxes. We do this by dividing the interval a, b into l subintervals x i 1, x i of equal width x, dividing c, d into m subintervals of width y, and dividing r, s 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 sub-boxes B x Bi jk y x i 1, x i zk 1, zk yj 1, yj which are shown in Figure 1. Each sub-box has volume V Then we form the triple Riemann sum x y z. Bijk l Îy m n f x* , y* , z* ijk ijk ijk 2 Îz V i 1j 1k 1 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 yyy f x, y, z dV x n f x i* , y i* , z i* jk jk jk l, m, n l V i 1j 1k 1 B y m lim 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 sub-box, but if we choose it to be the point x i , yj , zk we get a simpler-looking expression for the triple integral: l yyy f x, y, z dV m n f xi , yj , zk lim l, m, n l V i 1j 1k 1 B 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, b r, s , then c, d yyy f B x, y, z dV s d yyy r c b a f x, y, z d x d y dz 5E-16(pp 1054-1063) 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 yyy f b s yyy x, y, z dV a r d c f x, y, z d y dz d x B EXAMPLE 1 Evaluate the triple integral xxxB x yz 2 dV , where B is the rectangular box given by x, y, z 0 B x 1, 1 y z 2, 0 3 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 x yz 2 dV 3 2 yy y 0 1 1 0 3 2 x yz d x d y dz yy 0 B 3 yy 2 0 y 3 0 1 yz 2 d y dz 2 3z 2 dz 4 z3 4 y y 2z 2 4 3 0 3 x 2 yz 2 2 2 1 x1 d y dz x0 y2 dz y 1 27 4 0 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, z dV yyy F x, y, z E dV 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 x, y, z E x, y D, u 1 x, y z u 2 x, y z=u™ (x, y) D FIGURE 2 A type 1 solid region where D is the projection of E onto the xy-plane as shown in Figure 2. Notice that the upper boundary of the solid E is the surface with equation z u 2 x, y , while the lower boundary is the surface z u1 x, y . 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 E x, y, z dV yy y D u 2 x, y u1 x, y f x, y, z dz dA 5E-16(pp 1054-1063) 1/18/06 4:34 PM Page 1061 SECTION 16.7 TRIPLE INTEGRALS z ❙❙❙❙ 1061 The meaning of the inner integral on the right side of Equation 6 is that x and y are held fixed, and therefore u1 x, y and u 2 x, y are regarded as constants, while f x, y, z is integrated with respect to z. In particular, if the projection D of E onto the x y-plane is a type I plane region (as in Figure 3), then z=u™(x, y) E x, y, z a E x b, t1 x y z t2 x , u1 x, y u 2 x, y z=u¡(x, y) a x b and Equation 6 becomes 0 y=g¡(x) D y=g™(x) y yyy f 7 b a E FIGURE 3 t2 x yy y x, y, z dV t1 x u 2 x, y u 1 x, y f x, y, z dz d y d x A type 1 solid region If, on the other hand, D is a type II plane region (as in Figure 4), then z E x, y, z c E z=u™(x, y) z=u¡(x, y) y d, h1 y c yyy f 8 d d yy x, y, z dV c E y x z h2 y , u1 x, y u 2 x, y and Equation 6 becomes x=h¡(y) 0 x h2 y h1 y y u 2 x, y u1 x, y f x, y, z dz d x d y D x=h™(y) EXAMPLE 2 Evaluate xxxE z dV , where E is the solid tetrahedron bounded by the four Another type 1 solid region 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 xy-plane (see Figure 6). The lower boundary of the tetrahedron is the plane z 0 and the upper boundary is the plane x y z 1 (or z 1 x y), so we use u1 x, y 0 and u 2 x, y 1 x y in Formula 7. Notice that the planes x y z 1 and z 0 intersect in the line x y 1 (or y 1 x) in the xy-plane. So the projection of E is the triangular region shown in Figure 6, and we have (0, 0, 1) z=1-x-y E y 1. (0, 1, 0) 0 y (1, 0, 0) x 0, and x z planes x z 0, y 0, z FIGURE 4 9 E x, y, z 0 x 1, 0 y 1 z x, 0 1 x y 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 1 1x yy y 0 0 1xy 0 1 z dz d y d x yy 0 E 1 1 2 y=1-x D 1 2 1 0 y 1x 0 1 0 0 y=0 1 x x 3 y y 1 0 1 0 z1xy dy dx z0 y 2 dy dx x 1 6 FIGURE 6 1 1 yy z2 2 1x x 3 dx 3 y1x dx y0 1 6 1 x 4 4 1 0 1 24 5E-16(pp 1054-1063) 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 x, y, z E D 0 y, z D, u1 y, z u 2 y, z x where, this time, D is the projection of E onto the y z-plane (see Figure 7). The back surface is x u1 y, z , the front surface is x u 2 y, z , and we have y E x x=u¡(y, z) yyy f 10 x, y, z dV yy y E x=u™(y, z) D f x, y, z d x dA Finally, a type 3 region is of the form FIGURE 7 x, y, z E A type 2 region x, z D, u1 x, z u 2 x, z y where D is the projection of E onto the x z-plane, y u1 x, z is the left surface, and y u 2 x, z is the right surface (see Figure 8). For this type of region we have z y=u™(x, z) yyy f 11 x, y, z dV yy y E D u 2 y, z u1 y, z E D f x, y, z d y 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) u 2 x, z u1 x, z y x EXAMPLE 3 Evaluate x 2 2 xxxE s x 2 z 2 dV , where E is the region bounded by the paraboloid 4. z and the plane y FIGURE 8 y A type 3 region 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 xy-plane, which is the parabolic region in Figure 10. (The trace of y x 2 z 2 in the plane z 0 is the parabola y x 2.) y z y=≈+z@ 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 x FIGURE 9 FIGURE 10 Region of integration Projection on xy-plane From y x 2 z 2 we obtain z sy z sy x 2 and the upper surface is z a type 1 region is E { x, y, z 2 x x 2, so the lower boundary surface of E is sy x 2. Therefore, the description of E as 2, x 2 y 4, sy x2 z sy and so we obtain yyy s x E 2 z 2 dV 2 4 y yy 2 x 2 sy x 2 sy x 2 sx 2 z 2 dz d y d x x2} 5E-16(pp 1054-1063) 1/18/06 4:36 PM Page 1063 SECTION 16.7 TRIPLE INTEGRALS z ≈+z@=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 x z-plane is the disk x 2 z 2 4 shown in Figure 11. Then the left boundary of E is the paraboloid y x 2 z 2 and the right boundary is the plane y 4, so taking u1 x, z x 2 z 2 and u 2 x, z 4 in Equation 11, we have yyy sx 2 yy y z 2 dV 4 x2 z2 E z 2 dy dA sx 2 D3 FIGURE 11 yy Projection on x z-plane D3 x2 4 z 2 sx 2 z 2 dA Although this integral could be written as 2 yy 2 | 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. s4 x 2 x2 4 s4 x 2 z2 sx 2 z 2 dz d x it’s easier to convert to polar coordinates in the x z-plane: x gives yyy sx 2 yy z 2 dV E x2 4 z 2 sx 2 y r sin . This z 2 dA r 2 r r dr d r cos , z D3 2 yy 0 2 0 4 4r 3 3 2 r5 5 2 2 0 d y 2 0 4r 2 r 4 dr 128 15 0 Applications of Triple Integrals Recall that if f x 0, then the single integral xab f x dx represents the area under the curve y f x from a to b, and if f x, y 0, then the double integral xxD f x, y dA represents the volume under the surface z f x, y and above D. The corresponding interpretation of a triple integral xxxE f x, y, z dV , where f x, y, z 0, is not very useful because it would be the “hypervolume” of a four-dimensional 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 four-dimensional space.) Nonetheless, the triple integral xxxE f x, y, z dV can be interpreted in different ways in different physical situations, depending on the physical interpretations of x, y, z and f x, y, z . Let’s begin with the special case where f x, y, z 1 for all points in E. Then the triple integral does represent the volume of E: VE 12 yyy dV E For example, you can see this in the case of a type 1 region by putting f x, y, z Formula 6: yyy 1 dV yy y E D u 2 x, y u1 x, y dz dA yy D u 2 x, y u1 x, y dA 1 in 5E-16(pp 1064-1073) 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 u1 x, y and z u 2 x, y . z EXAMPLE 4 Use a triple integral to find the volume of the tetrahedron T bounded by the planes x (0, 0, 2) T 2, x 0. 1 1 x2 0 x2 yyy dV y y VT y y 2 x 2y 0 dz d y d x T (0, 1, 0) 0 0, and z 2y, x SOLUTION The tetrahedron T and its projection D on the xy-plane 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 2y z 2, that is, z 2 x 2y. Therefore, we have x+2y+z=2 x=2y z 2y 1 yy 1 ”1, 2 , 0’ 0 1 x2 2 x2 x 1 3 2y d y d x 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.) F IGURE 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, z , in units of mass per unit volume, at any given point x, y, z , then its mass is 1 ”1, 2 ’ D x y= 2 0 FIGURE 13 1 x m 13 yyy x, y, z dV E and its moments about the three coordinate planes are yyy x My z 14 x, y, z dV yyy y Mx z E x, y, z dV E Mx y yyy z x, y, z dV E The center of mass is located at the point x, y, z , where My z m x 15 y Mx z m Mxy m z 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 y2 z2 x, y, z dV Iy E yyy x2 z2 x, y, z dV E Iz yyy x2 y2 x, y, z dV E As in Section 16.5, the total electric charge on a solid object occupying a region E and 5E-16(pp 1064-1073) 1/18/06 4:42 PM Page 1065 S ECTION 16.7 TRIPLE INTEGRALS ❙❙❙❙ 1065 x, y, z is having charge density yyy Q x, y, z 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, Z lies in E is P X, Y, Z yyy f E x, y, z dV E In particular, Pa X b, c Y d, r Z b d yyy s a c s r f x, y, z dz dy dx The joint density function satisfies f x, y, z yyy 0 f x, y, z dz d y d x 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 xy-plane 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: E x, y, z 1 1, y 2 y x 1, 0 z y z x=¥ z=x E D 0 y 1 F IGURE 14 Then, if the density is x, y, z m , the mass is yyy dV 1 1 y yy 1 y2 x dz d x d y 0 E 1 yy 1 y 2 1 y x dx dy y2 1 y 1 x2 2 1 1 1 y5 5 y 4 dy 1 0 4 5 y 1 0 1 x=1 x 0 x x x1 dy x y2 y 4 dy 5E-16(pp 1064-1073) ❙❙❙❙ 1066 1/18/06 4:42 PM Page 1066 CHAPTER 16 MULTIPLE INTEGRALS Because of the symmetry of E and about the xz-plane, we can immediately say that Mx z 0 and, therefore, y 0. The other moments are My z yyy x 1 1 y yy dV 1 y 2 x 0 x dz d x d y E 1 yy 1 2 3 Mxy 1 y y x dx dy 2 x1 x3 3 1 1 1 2 3 6 1 0 yyy z y 2 y dy 1 1 y yy dV 1 y 2 x 0 dy x y2 1 y7 7 y 4 7 0 z dz d x d y E 1 yy 1 3 y 1 0 zx z2 2 1 y 2 1 z0 1 y 2 x 2 dx dy 2 7 y 6 dy 1 1 yy 2 dx dy Therefore, the center of mass is My z Mx z Mxy , , m m m x, y, z |||| 16.7 Exercises 1. Evaluate the integral in Example 1, integrating first with 10. respect to z , then x, and then y. xxxE y dV , where E is bounded by the planes x 0, and 2 x 2 y z 4 z 2. Evaluate the integral xxxE x z y 3 dV , where 1 x y z 2, 0 3. 5. |||| z 1 yyy 0 0 3 1 0 0 ■ 7–16 xxxE x 2e y dV , 2 ze y dx dz dy ■ ■ ■ 6. ■ ■ 1 2x yy y 0 1 x z y 0 0 yyy 0 y 0 ■ 2 xyz dz dy dx y2 ze 14. dx dy dz ■ ■ 15. ■ 0 2, 0 dV , where x, y, z 0 x 1, 0 xxxE 6 xy dV , xxxE x 2y dV , where E is bounded by the parabolic cylinder x 2 and the planes x z, x y, and z 0 xxxE x dV , where E is bounded by the paraboloid x 16. x s4 y 2, 0 z ■ x, x z ■ 4z 2 ■ ■ ■ ■ ■ ■ ■ 9 and ■ ■ y} 17–20 y 4y 2 4 xxxE z dV , where E is bounded by the cylinder y 2 z 2 the planes x 0, y 3x, and z 0 in the first octant ■ y where E is bounded by the parabolic cylinder y 2 and the planes z 0, x 1, and x 1 1 and the plane x ■ Evaluate the triple integral. { x, y, z where E is the solid tetrahedron with vertices 0, 0, 0 , 0, 1, 0 , 1, 1, 0 , and 0, 1, 1 y xxxE yz cos x 5 E 9. |||| s1 z 0 4. 6xz dy dx dz where E is the solid tetrahedron with vertices 0, 0, 0 , 1, 0, 0 , 0, 2, 0 , and 0, 0, 3 z xxxE 2 x dV , where E 8. xz 0 yyy ■ 7. Evaluate the iterated integral. 0, xxxE x z dV , 1 using three different orders of integration. 3–6 0, y xxxE x y dV , 13. 1, 0 11. 12. x, y, z E 5 ( 5 , 0, 14 ) 7 2x where E lies under the plane z 1 x y and above the region in the xy-plane bounded by the curves y sx, y 0, and x 1 |||| Use a triple integral to find the volume of the given solid. 17. The tetrahedron enclosed by the coordinate planes and the plane 2 x y z 4 18. The solid bounded by the cylinder y z 0, z 4, and y 9 x 2 and the planes 5E-16(pp 1064-1073) 1/18/06 4:42 PM Page 1067 S ECTION 16.7 TRIPLE INTEGRALS 19. The solid enclosed by the cylinder x 2 z y 5 and z y2 ■ ■ 31. The figure shows the region of integration for the integral y2 1 z 2 and the ■ ■ ■ ■ ■ ■ ■ ■ 1 yy y 0 16 ■ ■ sx cut from the cylinder y 2 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). CAS 1 z=1-y 0 x 32. The figure shows the region of integration for the integral 1 yy 0 dV , where ln 1 x y z x, y, z 0 x 4, 0 y 8, 0 B 24. xxxB sin x y 2z 3 ■ ■ x ■ |||| 1 z 4 z z=1-≈ y ■ ■ 2, 0 1 ■ ■ ■ 1 ■ ■ Sketch the solid whose volume is given by the iterated 26. 1 1x yy y 0 0 2 2y yy y 0 ■ 2 2z 0 0 ■ d y dz d x 1 1 y yyy 34. 4 y2 yy y d x dz d y ■ ■ ■ ■ ■ ■ ■ ■ ■ y y=1-x 1 33–34 |||| Write five other iterated integrals that are equal to the given iterated integral. 33. 0 f x, y, z d y dz d x 0 4, 0 ■ 1x 0 z integral. 25. y Rewrite this integral as an equivalent iterated integral in the five other orders. x 25–26 0 dV , where x, y, z 0 B ■ 1 x2 1 xxxB y 1 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 sub-boxes of equal size. 23. f x, y, z dz d y d x z 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, z is evaluated at the center xi , yj , zk of the box Bijk. Use the Midpoint Rule to estimate 2 2 2 xxxB e x y z dV , where B is the cube defined by 0 x 1, 0 y 1, 0 z 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). 1y 0 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 1067 1 20. The solid enclosed by the paraboloid x plane x 9 and the planes ❙❙❙❙ ■ 0 y x2 1 0 0 ■ f x, y, z dz d x d y 0 y 0 ■ f x, y, z dz d y d x ■ ■ ■ ■ Express the integral xxxE f x, y, z dV as an iterated integral in six different ways, where E is the solid bounded by the given surfaces. 27. x 2 ■ 35. E is the solid of Exercise 9; ■ ■ ■ ■ ■ 35–38 27–30 |||| z2 28. z 0, 29. z 30. 9x ■ x z 0, 2 ■ y 4, 4y ■ 2 0, y, z ■ 2 0, y x y 6 z 2, 2 |||| Find the mass and center of mass of the solid E with the given density function . 1 y 2x planes x y z 1, x 0, and z 37. E is the cube given by 0 ■ ■ ■ 2 1 y 2 and the x, y, z 4 36. E is bounded by the parabolic cylinder z 1 ■ x, y, z ■ ■ ■ ■ x, y, z x2 y2 z2 x 0; a, 0 y a, 0 z a; 5E-16(pp 1064-1073) 1068 ❙❙❙❙ 1/18/06 4:42 PM Page 1068 CHAPTER 16 MULTIPLE INTEGRALS 38. E is the tetrahedron bounded by the planes x z ■ 0, x ■ z y ■ x, y, z 1; ■ ■ ■ ■ 0, y 45. The joint density function for random variables X , Y , and Z is 0, f x, y, z Cxyz if 0 x 2, 0 y f x, y, z 0 otherwise. (a) Find the value of the constant C. (b) Find P X 1, Y 1, Z 1 . (c) Find P X Y Z 1 . y ■ ■ ■ ■ ■ 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. 39. The solid of Exercise 19; 40. The hemisphere x x, y, z ■ CAS ■ sx 2 ■ ■ 2 y y2 ■ x, y, z 2 z 2 sx 2 1, z ■ function f x, y, z Ce 0.5x 0.2y 0.1z if x and f x, y, z 0 otherwise. (a) Find the value of the constant C. (b) Find P X 1, Y 1 . (c) Find P X 1, Y 1, Z 1 . 0; ■ ■ ■ ■ ■ 41. Let E be the solid in the first octant bounded by the cylinder 2, and 0, z 0, y 0, 47–48 |||| The average value of a function f x, y, z over a solid region E is defined to be x 2 y 2 1 and the planes y z, x 0, and z 0 with the density function x, y, z 1 x y 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 z-axis CAS z 46. Suppose X , Y , and Z are random variables with joint density y2 z2 ■ 2, 0 1 VE fave yyy f x, y, z dV E where V E is the volume of E. For instance, if tion, then ave is the average density of E. 42. If E is the solid of Exercise 16 with density function is a density func- 47. Find the average value of the function f x, y, z x yz 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. x, y, z x 2 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 z-axis 48. Find the average value of the function f x, y, z over the region enclosed by the paraboloid z 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. ■ ■ ■ ■ ■ ■ ■ ■ x 2z 1 x2 ■ y 2z y2 ■ 49. Find the region E for which the triple integral 44. Find the moments of inertia for a rectangular brick with dimen- yyy 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. 1 x2 2y 2 3 z 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 n-dimensional 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 y 2 z 2 w 2 r 2 in 4. (Use only trigonometric substitution and the reduction formulas for x sinnx dx or x cosnx dx.) 4. Use an n-tuple integral to find the volume enclosed by a hypersphere of radius r in n-dimensional space n . [Hint: The formulas are different for n even and n odd.] ■ 5E-16(pp 1064-1073) 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 z Recall from Section 13.7 that the cylindrical coordinates of a point P are r, , z , where r, , and z are shown in Figure 1. Suppose that E is a type 1 region whose projection D on the xy-plane is conveniently described in polar coordinates (see Figure 2). In particular, suppose that f is continuous and P(r, ¨, z) z 0 ¨ D, u1 x, y x, y z u 2 x, y , h1 x, y, z E r h2 y r where D is given in polar coordinates by x D FIGURE 1 r, 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 1 yy y x, y, z dV E D u 2 x, y u1 x, y f x, y, z 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 E yy h2 h1 y u 2 r cos , r sin u1 r cos , r sin f r cos , r sin , z r dz dr d dz d¨ r r d¨ x, y, z dV dr FIGURE 3 Volume element in cylindrical coordinates: d V=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, z involves the expression x 2 y 2. 5E-16(pp 1064-1073) 1070 ❙❙❙❙ 1/18/06 4:42 PM Page 1070 CHAPTER 16 MULTIPLE INTEGRALS z EXAMPLE 1 A solid E lies within the cylinder x 2 y 2 1, below the plane z 4, and above the paraboloid z 1 x y . (See Figure 4.) The density at any point is proportional to its distance from the axis of the cylinder. Find the mass of E. z=4 2 (0, 0, 4) 2 SOLUTION In cylindrical coordinates the cylinder is r z 1 1 and the paraboloid is r 2, so we can write r, , z 0 E 2 ,0 r r2 1, 1 z 4 z=1-r@ Since the density at x, y, z is proportional to the distance from the z-axis, the density function is f x, y, z K s x 2 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 yyy K s x m FIGURE 4 2 1 y 2 dV 2 y yy 1 4 r 2 dr d 0 0 1 r2 Kr r dz dr d E 2 yy 0 1 0 Kr 2 4 EXAMPLE 2 Evaluate z=2 E 2 z=œ„„„„„ ≈+¥ 2 s4 x 2 2 s4 x 2 y d 1 3r 2 0 r 4 dr 12 K 5 0 y 2 0 2 sx2 y2 x2 y 2 dz d y d x. SOLUTION This iterated integral is a triple integral over the solid region z x 2 yy 1 r5 5 2 K r3 Ky 2 { x, y, z 2 x s4 x2 y x 2, s x 2 s4 y2 z 2} and the projection of E onto the xy-plane is the disk x 2 y 2 4. The lower surface of E is the cone z s x 2 y 2 and its upper surface is the plane z 2. (See Figure 5.) This region has a much simpler description in cylindrical coordinates: r, , z 0 E y 2, 2 ,0 r z 2, r 2 Therefore, we have FIGURE 5 2 yy 2 s4 x 2 s4 x 2 y 2 sx2 y2 x2 y 2 dz d y d x x2 yyy y 2 dV E 2 2 y yy 0 y 2 0 2 0 2 r d y [ r4 1 2 2 0 r 2 r dz dr d r3 2 1 5 r5 r dr 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 5E-16(pp 1064-1073) 1/18/06 4:42 PM Page 1071 ❙❙❙❙ S ECTION 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 P(∏, ¨, ˙) E ∏ ˙ a b, ,c d where a 0, 2 , and d c . 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 , half-planes j , and half-cones 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 z ∏ i sin ˙ k Ψ i i Î˙ l ∏ i Î˙ Ψ ri=∏ i sin ˙ k sin k 2 i sin k *** where i , j , k is some point in Eijk . Let x ijk, y ijk , z ijk be the rectangular coordinates of this point. Then 0 x 2 i k 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∏ Vijk ˙k sin yyy f y x, y, z dV l, m, n l E l FIGURE 7 m l, m, n l n *** f x ijk , y ijk , z ijk Vijk i 1j 1k 1 n lim ri Ψ=∏ i sin ˙ k Ψ m lim f i sin k cos j, i sin k sin j , i cos 2 i k sin k i j k i 1j 1k 1 But this sum is a Riemann sum for the function 2 F,, 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, z dV E d yyy c b a f sin cos , sin sin , cos 2 sin ddd where E is a spherical wedge given by z ∏ sin ˙ d¨ ˙ E d∏ ,, a b, ,c d Formula 4 says that we convert a triple integral from rectangular coordinates to spherical coordinates by writing ∏ ∏ d˙ 0 d¨ x FIGURE 8 Volume element in spherical coordinates: d V=∏@ sin ˙ d∏ d¨ d˙ x y sin cos 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 E ,, ,c d, t1 , t2 , 5E-16(pp 1064-1073) 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 t1 , and t 2 , . 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 xxxB e x 2 y2 z2 3 2 dV, where B is the unit ball: x, y, z x 2 B y2 z2 1 SOLUTION Since the boundary of B is a sphere, we use spherical coordinates: B ,, 0 1, 0 2 ,0 In addition, spherical coordinates are appropriate because x2 y2 z2 2 Thus, (4) gives yyy e x2 y2 z2 32 2 yy y dV 0 0 1 e 0 232 2 sin ddd B y [ NOTE y d cos 0 sin 0 2 0 2 y d 1 0 [e ] 1 3 3 1 0 3 2 ed 4 3 e 1 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 yy s1 x 2 1 s1 x y s1 x 2 y 2 s1 x 2 y 2 2 e x2 y2 z2 3 2 dz dy dx EXAMPLE 4 Use spherical coordinates to find the volume of the solid that lies above the cone z sx 2 y 2 and below the sphere x 2 y2 z2 z. (See Figure 9.) z (0, 0, 1) ≈+¥+z@=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 5E-16(pp 1064-1073) 1/18/06 4:42 PM Page 1073 S ECTION 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 2 s sin 2 This gives sin cos , or spherical coordinates is E ,, cos 2 2 sin 2 sin 2 sin 4. Therefore, the description of the solid E in 0 2 ,0 4, 0 cos Figure 11 shows how E is swept out if we integrate first with respect to , then , and then . The volume of E is F IGURE 10 2 4 yyy dV y y y VE 0 cos 2 0 0 sin ddd E y 2 0 2 3 Visual 16.8 shows an animation of Figure 11. 0 y 4 0 z 2 4 1. 2. y yy 3. yyy 4. ■ yy y yyy 0 2 0 2 9 r2 0 6 0 2 0 0 2 ■ 3 2 1 ■ x y ¨ varies from 0 to 2π. 5–6 |||| Set up the triple integral of an arbitrary continuous function f x, y, z in cylindrical or spherical coordinates over the solid shown. 2 z 5. sin sin ■ ddd 2 ■ y x ddd ■ z 6. 3 r dz dr d 2 0 2 8 0 z y r dz d d r r 0 d 4 cos 4 4 2 3 Exercises |||| Sketch the solid whose volume is given by the integral and evaluate the integral. 4 cos 3 0 ˙ varies from 0 to π/4 while ¨ is constant. 1–4 0 sin d 3 x y ∏ varies from 0 to cos ˙ while ˙ and ¨ are constant. |||| 16.8 cos 3 sin z x FIGURE 11 4 y d 1 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ y 2 x ■ ■ ■ ■ ■ 5E-16(pp 1074-1083) 1/18/06 4:43 PM 1074 ❙❙❙❙ CHAPTER 16 MULTIPLE INTEGRALS 7–16 |||| Page 1074 Use cylindrical coordinates. 22. Evaluate xxxE x y z dV , where E lies between the spheres and 7. Evaluate xxxE sx 2 y 2 dV , where E is the region that lies inside the cylinder x 2 y 2 16 and between the planes z 5 and z 4. and below the sphere x2 y2 z sx 2 y 2, the cylinder x 2 y2 5, and the xy-plane. 10. Evaluate xxxE x dV , where E is enclosed by the planes z and z x y x 2 y 2 9. 3 and by the cylinders x 2 y 2 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. 0 4 and 11. Evaluate xxxE x 2 dV , where E is the solid that lies within the cylinder x 2 cone z 2 4 x 2 y 2 1, above the plane z 4y 2. 27. (a) Find the centroid of a solid homogeneous hemisphere of 0, and below the 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 x2 y2 1 and the sphere x 2 y2 z2 28. Find the mass and center of mass of a solid hemisphere of 4. 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 y 2 and z 36 3x 2 3y 2. (b) Find the centroid of E (the center of mass in the case where the density is constant). ■ a cos 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 paraboloid z 4 x constant density K . 4y and the plane z ■ ■ CAS ■ ■ ■ ■ ■ ■ ■ |||| y2 z2 y2 1. y 2 and below the sphere x 2 sx 2 y2 32. (a) Find the volume enclosed by the torus ■ ■ z2 1. sin . (b) Use a computer to draw the torus. ; 33. 34. ■ ■ 1 yy 1 ■ 1 0 y2 z2 20. Evaluate xxxE e s x 1 and x 2 2 y2 z2 4 in the first octant. ■ ■ ■ ■ ■ ■ ■ ■ y2 z2 35. 21. Evaluate xxxE x 2 dV , where E is bounded by the x z-plane and 36. y z 2 the hemispheres y s9 x2 z 2 and y s16 x2 z 2. y s1 y 2 x 2 y2 y sx 2 y 2 x2 y2 32 ■ ■ ■ ■ dz dy dx xyz dz dx dy x 2 y2 0 ■ 2 x2 y2 ■ ■ ■ ■ ■ ■ ■ ■ 35–36 |||| Evaluate the integral by changing to spherical coordinates. dV , where E is enclosed by the sphere 9 in the first octant. 2 s1 x 2 s1 x 2 yy ■ 19. Evaluate xxxE z dV , where E lies between the spheres x ■ 33–34 |||| Evaluate the integral by changing to cylindrical coordinates. z 2 dV , where B is the unit ball y 2 dV , where H is the hemispherical region that lies above the xy-plane and below the sphere x 2 y 2 z 2 1. 2 ■ z x 2 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. ■ 18. Evaluate xxxH x 2 x2 ■ Use spherical coordinates. 17. Evaluate xxxB x 2 x2 ■ 31. Evaluate xxxE z dV , where E lies above the paraboloid ■ 17–28 ■ radius a by two planes that intersect along a diameter at an angle of 6. y 2 z 2 a 2 if the density at any point is proportional to its distance from the z -axis. ■ ■ 30. Find the volume of the smaller wedge cut from a sphere of 16. Find the mass of a ball B given by x 2 ■ ■ cone z 0 if S has aa ■ 29. Find the volume and centroid of the solid E that lies above the 15. Find the mass and center of mass of the solid S bounded by the 2 ■ 29–32 |||| Use cylindrical or spherical coordinates, whichever seems more appropriate. 14. (a) Find the volume of the solid that the cylinder r ; z 2 4, above the xy-plane, and below the cone 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 x2 3 4 cos . 24. Find the volume of the solid that lies within the sphere x y 2 dV , where E is the solid in the first octant that lies beneath the paraboloid z 1 x 2 y 2. 1 2 3. 23. Find the volume of the solid that lies above the cone 8. Evaluate xxxE x 3 z 4 and above the cone ■ 3 yy 3 3 s9 y 2 y s9 x 2 y 2 z sx 2 z 2 dz d y d x y2 0 s18 x 2 y 2 x2 y2 z 2 dz d x d y sx2 y2 0 ■ y s9 x 2 yy 0 s9 x 2 ■ ■ ■ ■ ■ ■ ■ ■ 5E-16(pp 1074-1083) 1/18/06 4:44 PM Page 1075 ❙❙❙❙ A PPLIED PROJECT ROLLER DERBY C AS 1075 (b) Deduce that the volume of the spherical wedge given by 1 2, 1 2, 1 2 is 37. In the Laboratory Project on page 880 we investigated the family of surfaces 1 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. 3 2 V 3 1 cos 3 cos 1 2 2 1 (c) Use the Mean Value Theorem to show that the volume in part (b) can be written as 2 V where lies between 1, 2 sin 1 and 2 1 lies between 2, , and 2 1 and . 2 , 1 40. When studying the formation of mountain ranges, geologists 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 t P and the height is h P . (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 lb ft3 . How much work was done in forming Mount Fuji if the land was initially at sea level? 38. Show that yyy sx 2 y2 z2 e x 2 y2 z2 d x d y 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.) 39. (a) Use cylindrical coordinates to show that the volume of the solid bounded above by the sphere r 2 z 2 a 2 and below by the cone z r cot 0 (or 0 ), where 2, is 0 0 V 2 a3 1 3 cos P 0 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 m th. 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 m th 1 2 mv2 1 2 I 2 1. Show that v2 1 2th I* where I * I mr 2 5E-16(pp 1074-1083) 1076 ❙❙❙❙ 1/18/06 4:44 PM Page 1076 CHAPTER 16 MULTIPLE INTEGRALS 2. If y t is the vertical distance traveled at time t, then the same reasoning as used in Problem 1 shows that v 2 2ty 1 I * at any time t. Use this result to show that y satisfies the differential equation dy dt where 2t 1 I* sin sy 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 2h 1 I * t sin 2 This shows that the object with the smallest value of I * wins the race. 4. Show that I * 1 2 for a solid cylinder and I * 1 for a hollow cylinder. 5. Calculate I * for a partly hollow ball with inner radius a and outer radius r . Express your answer in terms of b a r. What happens as a l 0 and as a l r ? for a solid ball and I * 2 for a hollow ball. Thus, the objects finish in the 3 following order: solid ball, solid cylinder, hollow ball, hollow cylinder. 6. Show that I * 2 5 D ISCOVERY PROJECT T he 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 y 2 1, x 2 z 2 1, and y z 1. Indicate the positions of the coordinate axes and label the faces with the equations of the corresponding cylinders. 2 2 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 hand-drawn sketch or a computer graph. 5. If the first cylinder is x 2 y 2 a 2, where a integral for the volume of the solid. What if a 1, set up, but do not evaluate, a double 1? 5E-16(pp 1074-1083) 1/18/06 4:44 PM Page 1077 S ECTION 16.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS |||| 16.9 ❙❙❙❙ 1077 Change of Variables in Multiple Integrals In one-dimensional 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 where x t u and a tc,b y 2 y f x dx a b a d c f t u t u du t d . Another way of writing Formula 1 is as follows: y f x dx d c dx du du f xu 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, y dA yy f R r cos , r sin r dr d S where S is the region in the r -plane that corresponds to the region R in the xy-plane. More generally, we consider a change of variables that is given by a transformation T from the uv-plane to the x y-plane: T u, v x, y where x and y are related to u and v by the equations x t u, v y h u, v x 3 x u, v y y u, v or, as we sometimes write, We usually assume that T is a C 1 transformation, which means that t and h have continuous first-order partial derivatives. A transformation T is really just a function whose domain and range are both subsets of 2. If T u1, v1 x 1, y1 , then the point x 1, y1 is called the image of the point u1, v1 . If no two points have the same image, T is called one-to-one. Figure 1 shows the effect of a transformation T on a region S in the uv-plane. T transforms S into a region R in the xy-plane 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 5E-16(pp 1074-1083) 1078 ❙❙❙❙ 1/18/06 4:44 PM Page 1078 CHAPTER 16 MULTIPLE INTEGRALS If T is a one-to-one transformation, then it has an inverse transformation T 1 from the xy-plane to the uv-plane and it may be possible to solve Equations 3 for u and v in terms of x and y : u G x, y v H x, y EXAMPLE 1 A transformation is defined by the equations u2 x Find the image of the square S √ S£ (0, 1) S¢ 0 S™ S¡ (1, 0) u, v 0 u T x 4 1 y ¥ x=1- 4 x 5 R (_1, 0) 0 (1, 0) 2uv 1, 0 u 1. v y2 4 0 x 1 which is part of a parabola. Similarly, S 3 is given by v the parabolic arc (0, 2) ¥ x= -1 4 y 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 u 1 . (See Figure 2.) From the given equations we have x u 2, y 0, and so 0 x 1. Thus, S1 is mapped into the line segment from 0, 0 to 1, 0 in the xy-plane. The second side, S 2, is u 1 0 v 1 and, putting u 1 in the given equations, we get x 1 v2 y 2v Eliminating v, we obtain (1, 1) S v2 x FIGURE 2 y2 4 1 1 10 x u 1 , whose image is 0 Finally, S4 is given by u 0 0 v 1 whose image is x v 2, y 0, that is, 1 x 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 x-axis 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 uv-plane 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 x y-plane, one of whose boundary points is x 0 , y0 T u0 , v0 . The vector r u, v t u, v i h u, v j 5E-16(pp 1074-1083) 1/18/06 4:44 PM Page 1079 S ECTION 16.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS ❙❙❙❙ 1079 is the position vector of the image of the point u, v . The equation of the lower side of S is v v0 , whose image curve is given by the vector function r u, v0 . The tangent vector at x 0 , y0 to this image curve is ru tu u0 , v0 i x i u hu u0 , v0 j y j u Similarly, the tangent vector at x 0 , y0 to the image curve of the left side of S (namely, u u0 ) is rv r (u¸, √¸+Î√) b R r (u¸, √¸) a r u0 u, v0 FIGURE 5 j v b r u0 , v0 r u0 , v0 v r u0 u, v0 u u, v0 lim r u0 , v0 u ru v r u0 , v0 v rv ul0 and so Î u ru y i shown in Figure 4. But ru r (u¸, √¸) v T S by a parallelogram determined by the r u0 , v0 FIGURE 4 Î√ r√ x hv u0 , v0 j We can approximate the image region R secant vectors a r (u¸+Î u, √¸) tv u0 , v0 i r u0 Similarly r u0 , v0 r u0 , v0 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 u ru 6 v rv ru rv uv Computing the cross product, we obtain i rv x u x y u y v ru j k v x u x 0 y u k y v 0 x u y u v x v y k v 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 y h u, v is x, y u, v x u y u x v y v x u y x v v y u t u, v and 5E-16(pp 1074-1083) 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 of R: x, y u, v A 8 A uv where the Jacobian is evaluated at u0 , v0 . Next we divide a region S in the uv-plane into rectangles Sij and call their images in the xy-plane 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 yy f n x, y dA f x i , yj A i 1j 1 R m n f t ui , vj , h ui , vj i 1j 1 x, y u, v uv where the Jacobian is evaluated at ui, vj . Notice that this double sum is a Riemann sum for the integral x, y yy f t u, v , h u, v u, v du dv S 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 uv-plane onto a region R in the xy-plane. 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 one-to-one, except perhaps on the boundary of S. Then yy f R x, y dA yy f x u, v , y u, v S x, y u, v du dv 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 x, y u, v du dv 5E-16(pp 1074-1083) 1/18/06 4:44 PM Page 1081 S ECTION 16.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS ¨=∫ r=a r=b S x å 1081 Notice the similarity between Theorem 9 and the one-dimensional formula in Equation 2. Instead of the derivative d x du, we have the absolute value of the Jacobian, that is, x, y u, v . 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 x y-plane is given by ¨ ∫ ❙❙❙❙ t r, r cos y h r, r sin ¨=å 0 a r b T r=b ¨=∫ R r=a 0 x r y r x, y r, y ∫ 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 xy-plane. The Jacobian of T is x cos sin y r sin r cos r cos2 r sin2 r 0 Thus, Theorem 9 gives ¨=å yy f å yy f x, y d x d y R x, y r, r cos , r sin S dr d x yy FIGURE 7 The polar coordinate transformation b a f r cos , r sin r dr d which is the same as Formula 16.4.2. u 2 v 2, y 2uv to evaluate the integral xxR y dA, where R is the region bounded by the x-axis and the parabolas y 2 4 4 x and y 2 4 4 x. EXAMPLE 2 Use the change of variables x SOLUTION The region R is pictured in Figure 2. In Example 1 we discovered that TS R, where S is the square 0, 1 0, 1 . 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: x u y u x, y u, v x 2u 2v v y 2v 2u 4u 2 4v 2 0 v Therefore, by Theorem 9, yy y dA yy 2uv R S 8y 1 0 y 1 0 y 1 0 2v x, y u, v u3v 1 yy dA 0 1 0 uv 3 du dv 4v 3 dv [v 2 2uv 4 u2 8y 1 0 v4 1 0 [ 14 4v 2 u v 2 du dv 1 2 u2v 3 u1 u0 dv 5E-16(pp 1074-1083) 1082 ❙❙❙❙ 1/18/06 4:45 PM Page 1082 CHAPTER 16 MULTIPLE INTEGRALS N OTE 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, y is difficult to integrate, then the form of f x, y 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 uv-plane has a convenient description. ■ EXAMPLE 3 Evaluate the integral vertices 1, 0 , 2, 0 , 0, xxR e x y 2 , and 0, xy dA, where R is the trapezoidal region with 1. xy SOLUTION Since it isn’t easy to integrate e xy , we make a change of variables sug- gested by the form of this function: u 10 x y x v y These equations define a transformation T 1 from the xy-plane to the uv-plane. Theorem 9 talks about a transformation T from the uv-plane to the xy-plane. It is obtained by solving Equations 10 for x and y : 1 2 x 11 u 1 2 y v u v The Jacobian of T is x u y u x, y u, v x 1 2 1 2 v y 1 2 1 2 1 2 v To find the region S in the uv-plane corresponding to R, we note that the sides of R lie on the lines √ y √=2 (_2, 2) (2, 2) S u=_√ 0 x y 2 x 0 x y 1 and, from either Equations 10 or Equations 11, the image lines in the uv-plane are u=√ (_1, 1) u (1, 1) v 2 v u v 1 v √=1 0 T u Thus, the region S is the trapezoidal region with vertices 1, 1 , 2, 2 , 1, 1 shown in Figure 8. Since T –! S y u, v 1 2, v u v 2, 2 , and v Theorem 9 gives x-y=1 1 2 0 _1 x R x-y=2 yy e R xy xy dA yy e uv S 2 yy 1 v v x, y u, v du dv e u v ( 1 ) du dv 2 1 2 2 uv u v u v y [ve ] 1 _2 FIGURE 8 1 2 y 2 1 e e 1 v dv 3 4 e e 1 dv 5E-16(pp 1074-1083) 1/18/06 4:45 PM Page 1083 SECTION 16.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS ❙❙❙❙ 1083 T riple Integrals There is a similar change of variables formula for triple integrals. Let T be a transformation that maps a region S in u vw-space onto a region R in x y z-space by means of the equations x t u, v, w y h u, v, w z k u, v, w The Jacobian of T is the following 3 3 determinant: x u y u z u x, y, z u, v, w 12 x x v w y y v w z z v w Under hypotheses similar to those in Theorem 9, we have the following formula for triple integrals: yyy f 13 yyy f x, y, z dV R x u, v, w , y u, v, w , z u, v, w S x, y, z u, v, w du dv dw 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 y sin z sin cos We compute the Jacobian as follows: sin cos sin sin cos x, y, z ,, sin sin sin cos 0 sin sin cos 2 cos sin Since 0 sin cos cos sin2 cos cos2 2 sin , we have sin cos sin sin cos sin 2 sin2 cos2 sin 2 sin cos cos cos sin sin sin sin cos cos sin sin sin cos2 sin2 sin2 sin2 2 sin 0. Therefore x, y, z ,, 2 2 sin sin and Formula 13 gives yyy f R x, y, z dV yyy f sin cos , S which is equivalent to Formula 16.8.4. sin sin , cos 2 sin ddd sin cos 5E-16(pp 1084-1087) ❙❙❙❙ 1084 1/18/06 Page 1084 CHAPTER 16 MULTIPLE INTEGRALS |||| 16.9 1– 6 4:47 PM Exercises Find the Jacobian of the transformation. |||| 1. x u 2. x u2 4 v, y 3u y u2 v2 ■ u 3. x y 5. x v u v, e u v, 6. x ■ 7–10 y |||| 7. S u z e u v, ■ eu z ■ ■ vw ■ ■ ■ ■ ■ ■ Find the image of the set S under the given transformation. 2; v 10. S is the disk given by u 2 0, u 1, v 11. v2 1; x a u, y 0, 19. ■ ■ ■ bv x 2y dA, where R is the parallelogram enclosed by the 3x y R lines x 2 y 0, x 2 y 4, 3x y 1, and 3x y 8 2 8y dA, where R is the parallelogram with 1, 3 , 1, 3 , 3, 1 , and 1, 5 ; 1 v, y 3u 4v 22. where R is the region bounded by the ellipse 4y 36; x 2u, y 3v 23. xxR e x 2 xxR sin 9x 2 ■ ■ ■ ■ ■ x 3y dA, where R is the triangular region with vertices 0, 0 , 2, 1 , and 1, 2 ; x 2u v, y u xxR 4 x vertices x 1u 4 13. ■ ■ 2v xxR x 2 dA, 2 ■ 15. ■ yy cos ■ Use the given transformation to evaluate the integral. 12. 14. ■ yy 21. ■ xxR 9x ■ Evaluate the integral by making an appropriate change of xxR ■ 2 |||| 20. ■ |||| ■ variables. 9. S is the triangular region with vertices 0, 0 , 1, 1 , 0, 1 ; x u 2, y v 11–16 ■ 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 v 1; x v, y u 1 v 2 ■ ■ ellipsoid x 2 a 2 y 2 b 2 z 2 c 2 1. Use the transformation x a u, 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. uw u, v 0 u 3, 0 2u 3v, y u v x ■ 17. (a) Evaluate xxxE dV, where E is the solid enclosed by the v cos v w, y ■ ■ v y sin , 4. x ■ , u 1, xy 2, xy 2 1, xy 2 2; u x y, v x y 2. Illustrate by using a graphing calculator or computer to draw R. 2v v 2, 2 ; 16. xxR y dA, where R is the region bounded by the curves xy x y e x y dA, where R is the rectangle enclosed by the lines x y 0, x y 2, x y 0, and x y 3 yx dA, where R is the trapezoidal region yx R with vertices 1, 0 , 2, 0 , 0, 2 , and 0, 1 4y 2 dA, where R is the region in the first quadrant bounded by the ellipse 9x 2 4y 2 1 ■ y dA, where R is given by the inequality x ■ ■ ■ ■ ■ ■ ■ y ■ 1 ■ ■ xxR x 2 x y y 2 dA, where R is the region bounded by the ellipse x 2 x y y 2 2; x s2 u s2 3 v, y s2 u s2 3 v xxR x y 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 u v, y v 24. Let f be continuous on 0, 1 and let R be the triangular region with vertices 0, 0 , 1, 0 , and 0, 1 . Show that yy f R x y dA y 1 0 u f u du 5E-16(pp 1084-1087) 1/18/06 4:47 PM Page 1085 C HAPTER 16 REVIEW |||| 16 Review ■ CONCEPT CHECK 1. Suppose f is a continuous function defined on a rectangle R a, b c, d . (a) Write an expression for a double Riemann sum of f . If f x, y 0, what does the sum represent? (b) Write the definition of xxR f x, y dA as a limit. (c) What is the geometric interpretation of xxR f x, y dA if f x, y 0? What if f takes on both positive and negative values? (d) How do you evaluate xxR f x, y dA? (e) What does the Midpoint Rule for double integrals say? (f) Write an expression for the average value of f . z ■ 8. Suppose a solid object occupies the region E and has density function x, y, z . 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 1. 2 yy 1 6 0 2 x sin x 6 yy y dx dy 0 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 t u, v , y h u, v , 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? TRUE-FALSE 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. yy 0 x yy y 2 dy d x sx 0 x 0 1 0 sx ■ 5. If D is the disk given by x 2 yy s4 2 2 1 x sin x y dy dx y 2 d x dy y2 x2 y 2 dA 4 1 y y (x 1 2 0 sy ) sin x 2 y 2 d x d y 2 0 4 3. yy 4. yy 1 x 2e y d y d x 3 1 1 1 0 ex 2 y2 y 2 1 9 2 0 2 r dz dr d 4 x 2 dx y e y dy sin y d x d y 16 3 7. The integral y yy 2 4, then D 6. 1 D. rectangular box B. (b) How do you evaluate xxxB f x, y, z dV ? (c) How do you define xxxE f x, y, z 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, z dV if E is such a region? (e) What is a type 2 solid region? How do you evaluate xxxE f x, y, z dV if E is such a region? (f) What is a type 3 solid region? How do you evaluate xxxE f x, y, z dV if E is such a region? 4. If a lamina occupies a plane region D and has density function 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. f x, y , x, y 7. (a) Write the definition of the triple integral of f over a dinates in a double integral? Why would you want to do it? 5. Let f be a joint density function of a pair of continuous ■ 6. Write an expression for the area of a surface with equation 3. How do you change from rectangular coordinates to polar coor- x, y , 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 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, y 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, y dA if D is a type I region? (c) What is a type II region? How do you evaluate xxD f x, y dA if D is a type II region? (d) What properties do double integrals have? ❙❙❙❙ 3 0 represents the volume enclosed by the cone z the plane z 2. sx 2 y 2 and 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. 5E-16(pp 1084-1087) ❙❙❙❙ 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, 3 0, 3 . Use a Riemann sum with nine terms to estimate the value of xxR f x, y dA. Take the sample points to be the upper right corners of the squares. 2 13. y 3 1 yy 0 ■ ■ xxR ye xxD x y dA, 17. yy 6 5 3 D 2 14. ■ ■ 1 yy 0 ye x dx dy x3 1 sy ■ ■ ■ ■ ■ ■ Calculate the value of the multiple integral. |||| 16. 7 2 1 ■ 15. 9 8 4 x cos y 2 d y d x ■ 15–28 10 1 xy dA, where R where D y x2 1 x, y 0 x, y 0 2, 0 y 3 1, y 2 x y x y dA, where D is bounded by y sx, y 2 0, x 1 1 dA, where D is the triangular region with vertices 1 x2 D 0, 0 , 1, 1 , and 0, 1 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 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 y 2 3 2 dA, where D is the region in the first quadrant bounded by the lines y 0 and y s3 x and the circle x2 y2 9 22. xxD x dA, where D is the region in the first quadrant that lies 1 0 1 3x 2 2. Use the Midpoint Rule to estimate the integral in Exercise 1. 3–8 |||| 3. 2 Calculate the iterated integral. yy 1 5. 7. 2 y 0 1 yy 0 x 0 yyy ■ 0 ■ s1 y 2 0 ■ ■ yy 6. dx dy yy 0 ■ 1 0 1 1 ye xy dx dy ex x 0 y 3xy 2 dy dx between the circles x 2 yyy 8. y sin x dz dy dx ■ 1 4. cos x 2 dy dx 1 0 2 xe y 0 ■ 0 1 23. 6 xyz dz d x d y x ■ ■ ■ y 4 10. _2 0 4 xxxE y 2z 2 dV , 26. R 4x 27. 0 _4 11. ■ ■ xxxE z dV , ■ ■ ■ ■ ■ ■ ■ ■ Describe the region whose area is given by the integral 2 yy sin 2 0 0 12. Describe the solid whose volume is given by the integral 2 2 yyy 0 0 and evaluate the integral. 2 1 2 sin ddd y x, 0 z x y ■ where E is bounded by the paraboloid z 2 and the plane x 0 where E is bounded by the planes y 0, z 0, 2 and the cylinder y 2 z 2 1 in the first octant xxxE yz dV , where E lies above the plane z 0, below the y, and inside the cylinder x 2 y 2 4 xxxH z 3s x 2 y 2 z 2 dV , where H is the solid hemisphere that lies above the xy-plane and has center the origin and radius 1 ■ 29–34 r dr d y plane z 4x 28. ■ y2 1 x 2 3, 0 where T is the solid tetrahedron with vertices 0, 0, 0 , ( 1 , 0, 0), 0, 1, 0 , and 0, 0, 1 3 x 2 2 xxxT x y dV , 25. y R _4 where x, y, z 0 x 24. |||| 9. y2 ■ Write xxR f x, y dA as an iterated integral, where R is the region shown and f is an arbitrary continuous function on R. 9–10 1 and x 2 xxxE x y dV , E ■ y2 ■ |||| ■ ■ ■ ■ ■ ■ ■ Find the volume of the given solid. 29. Under the paraboloid z R ■ 0, 2 x2 4y 2 and above the rectangle 1, 4 x 2 y and above the triangle in the x y-plane with vertices 1, 0 , 2, 1 , and 4, 0 30. Under the surface z 31. The solid tetrahedron with vertices 0, 0, 0 , 0, 0, 1 , 0, 2, 0 , and 2, 2, 0 ■ 5E-16(pp 1084-1087) 1/18/06 4:48 PM Page 1087 C HAPTER 16 REVIEW 32. Bounded by the cylinder x 2 z and y y2 4 and the planes z 33. One of the wedges cut from the cylinder x planes z 0 and z ■ sx 2 ■ 2 a by the mx x2 34. Above the paraboloid z z 9y 2 47. Rewrite the integral y 2 and below the half-cone 1 ■ ■ ■ ■ ■ ■ ■ ■ parabola x 1 y 2 and the coordinate axes in the first quadrant with density function x, y 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 y-axes. 36. A lamina occupies the part of the disk x 2 y2 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, y x y 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 xy-plane 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 the planes z 1 and z a2 x2 y 2 between x 2 y that lies above the triangle with vertices (0, 0), (1, 0), and (0, 2). 40. Graph the surface z x sin y, 3 x 3, y find its surface area correct to four decimal places. 41. Use polar coordinates to evaluate y 3 0 y s9 x 2 s9 x 2 x3 , and yy 2 0 y s4 x 2 y 2 s4 x 2 y 2 y 2sx 2 y2 z 2 dz d x d y 1 x 2 and y e , find the approximate value of the integral xxD y 2 dA. (Use a graphing device to estimate the points of intersection of the curves.) 44. Find the center of mass of the solid tetrahedron with vertices 0, 0, 0 , 1, 0, 0 , 0, 2, 0 , 0, 0, 3 and density function x, y, z x 2 y 2 z 2. 45. The joint density function for random variables X and Y is Cx 0 as an iterated integral in the order dx d y dz. 48. Give five other iterated integrals that are equal to 2 y3 yy y 0 0 y2 0 f x, y, z dz d x d y 49. Use the transformation u x y, v x y to evaluate x y x y dA, where R is the square with vertices 0, 2 , 1, 1 , 2, 2 , and 1, 3 . xxR u 2, y v 2, z w 2 to find the volume of the region bounded by the surface s x sy sz 1 and the coordinate planes. 50. Use the transformation x 51. Use the change of variables formula and an appropriate trans- formation to evaluate xxR x y dA, where R is the square with vertices 0, 0 , 1, 1 , 2, 0 , and 1, 1 . 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 y if 0 x otherwise (a) Find the value of the constant C. (b) Find P X 2, Y 1 . (c) Find P X Y 1 . 3, 0 yy f x, y dA f x 0 , y0 A D 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, b . Let Dr be the closed disk with center a, b and radius r. Use the Mean Value Theorem for double integrals (see Exercise 52) to show that x f x, y f x, y, z dz d y d x 53. Suppose that f is continuous on a disk that contains the point ; 43. If D is the region bounded by the curves y C AS 1y 0 x y 2 d y d x. 42. Use spherical coordinates to evaluate s4 y 2 1 2. 39. Find the area of the part of the surface z 2 1 x2 yyy ■ 35. Consider a lamina that occupies the region D bounded by the CAS 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. y2 ■ 1087 46. A lamp has three bulbs, each of a type with average lifetime 0 3 2 ❙❙❙❙ y 2 lim rl0 1 r2 yy f x, y dA f a, b Dr 1 dA, where n is an integer and D x2 y2 n 2 D is the region bounded by the circles with center the origin and radii r and R, 0 r R. (b) For what values of n does the integral in part (a) have a limit as r l 0 ? 1 (c) Find yyy 2 dV , where E is the region x y 2 z2 n 2 E bounded by the spheres with center the origin and radii r and R, 0 r R. (d) For what values of n does the integral in part (c) have a limit as r l 0 ? 54. (a) Evaluate yy 5E-16(pp 1088-1089) 1/18/06 P ROBLEMS PLUS 4:49 PM Page 1088 1. If x denotes the greatest integer in x, evaluate the integral yy x y dA R where R x, y 1 x 3, 2 5. y 2. Evaluate the integral 1 yy 0 1 2 2 e max x , y dy dx 0 where max x 2, y 2 means the larger of the numbers x 2 and y 2. xx1 cos t 2 3. Find the average value of the function f x dt on the interval [0, 1]. 4. If a, b, and c are constant vectors, r is the position vector x i the inequalities 0 ,0 ar ,0 br y j z k, and E is given by , show that cr 2 yyy a r b r c r dV 8a E 5. The double integral y 1 y b c 1 1 d x d y is an improper integral and could be defined as the 1 xy limit of double integrals over the rectangle 0, t 0, t as t l 1 . But if we expand the integrand as a geometric series, we can express the integral as the sum of an infinite series. Show that 0 0 1 yy 0 1 1 0 1 xy dx dy n1 1 n2 6. Leonhard Euler was able to find the exact sum of the series in Problem 5. In 1736 he proved that n1 1 n2 2 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 u x v s2 y u v s2 This gives a rotation about the origin through the angle 4. You will need to sketch the corresponding region in the uv-plane. [Hint: If, in evaluating the integral, you encounter either of the expressions 1 sin cos or cos 1 sin , you might like to use the identity and the corresponding identity for sin .] cos sin 2 7. (a) Show that 1 1 yyy 0 0 1 1 0 1 xyz dx dy dz n1 1 n3 (Nobody has ever been able to find the exact value of the sum of this series.) 1088 5E-16(pp 1088-1089) 1/18/06 4:49 PM Page 1089 (b) Show that 1 1 yyy 0 0 1 1 0 1 xyz 1n n3 dx dy dz n1 1 Use this equation to evaluate the triple integral correct to two decimal places. 8. Show that arctan x y arctan x dx x 0 2 ln by first expressing the integral as an iterated integral. 9. If f is continuous, show that x y z 0 0 0 yyy 1 2 f t dt dz dy y x 0 t 2 f t dt x 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 d 1 sR 2 d2 [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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