(tp5647) padilla HW05 Gilbert (56650) This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points 2. V = 4 cu. units 3. V = 8 cu. units correct 4. V = 8 cu. units 5. V = 6. V = 16 cu. units 3 16 cu. units 3
1
Determine the volume of the right circular cone generated by rotating the line x = y about the y-axis between y = 0 and y = 3. 1. V = 10 cu.units 2. V = 13 cu.units 3. V = 11 cu.units 4. V = 12 cu.units 5. V = 9 cu.units correct Explanation: The volume, V , of the solid of revolution generated by rotating the graph of x = f (y) about the y-axis between y = a and y = b is given by
b
Explanation: The region rotated about the y-axis is similar to the shaded region in y 4
x (not drawn to scale). Now the volume of the solid of revolution generated by revolving the graph of x = f (y) for a y b about the y-axis is given by
b
V =
a
f (y)2 dy.
When f (y) = y and a = 0, b = 3, therefore,
b
V =
a
1 y dx = y 3 3
2
3 0
.
volume =
a
f (y)2 dy .
Consequently, V = 9 cu.units . 002 10.0 points
To apply this we have rst to express x as a function of y since initially y is dened in terms of x by y = x2 . Thus after taking square roots we see that
4
Find the volume, V , of the solid obtained by rotating the region bounded by y = x , x = 0, about the y-axis. 1. V = 4 cu. units
2
V =
0
1 2 y y dy = 2
4
.
0
Consequently, V = 8 . 003 10.0 points
y = 4
padilla (tp5647) HW05 Gilbert (56650) Find the volume, V , of the solid obtained by rotating the bounded region in the rst quadrant enclosed by the graphs of y = x2 , about the x-axis. 1. V = 2. V = 3. V = 5 cu. units 12 1 cu. units 2 7 cu. units 15 x = y4 x-axis is given by
1
2
V =
0 1
(x1/4 )2 (x2 )2 dx x 2 x4 dx 2 3 1 5 x2 x 3 5
1
1
=
0
= Consequently, V = 2 1 3 5 004
.
0
=
7 cu. units . 15
10.0 points
1 4. V = cu. units 2 5. V = 7 cu. units correct 15
Find the volume, V , of the solid generated by rotating about the x-axis the region enclosed by the graphs of y = sec x, x = 0, y = 0, x = . 3 1. V = ln 1 + 3 2 cu. units
5 cu. units 6. V = 12 Explanation: Since the graphs of y = x2 , x = y4
2. V = cu. units 3 3. V = cu. units 8 cu. units 3 5. V = 3 cu. units correct 4. V = Explanation: The solid is generated by rotating the region
intersect at (0, 0) and at (1, 1) the bounded region in the rst quadrant enclosed by their graphs is the shaded area shown in
1
1
Thus the volume of the solid of revolution generated by rotating this region about the
/3
padilla (tp5647) HW05 Gilbert (56650) about the x-axis. The cross-section perpendicular to the x-axis has area A(x) = y 2 = sec2 x . Consequently,
3
3
Thus the volume generated when rotating this shaded region about the dotted line y = 4 is given by
1
V = sec x dx =
2
V =
0
tan x
3
0 1
(y 4)2 (6 4)2 dx ( x + 2)2 4 dx
1
0
= tan 005
= 3 cu. units 3 10.0 points
.
=
0
=
0
x + 4 x dx .
Find the volume, V , of the solid formed by rotating the region bounded by the graphs of y = x + 6, y = 6, x = 0, x = 1 about the line y = 4. 1. V = 2. V = 17 cu. units 6 10 cu. units 3
Consequently, 1 2 8 3/2 x + x 2 3
1
V =
=
0
19 . 6
006
10.0 points
Let A be the bounded region enclosed by the graphs of f (x) = x , g(x) = x3 .
3. V = 3 cu. units 4. V = 5. V = 19 cu. units correct 6 7 cu. units 2
Find the volume of the solid obtained by rotating the region A about the line x + 3 = 0 . 1. volume = 3 correct 11 3 8 3 7 3 10 3
Explanation: The region bounded by the given curves is shown as the shaded area in y
2. volume =
3. volume = 6 4 x 1 (not drawn to scale) and the line about which it is rotated is shown as the dotted line.
4. volume =
5. volume =
Explanation: The solid is obtained by rotating the shaded region about the line x + 3 = 0 as shown in
padilla (tp5647) HW05 Gilbert (56650) Consequently, x+3=0 volume = 53 37 + = 3 . 30 30
4
1
(Remember: (1)1/3 = 1, (1)4/3 = 1, and (1)5/3 = 1.) 007 10.0 points The base of a solid S is the bounded region enclosed by the graphs of
a margarita glass? (not drawn to scale). To compute the volume of this solid we use the washer method. For this we have to express f and g as functions of y: y = f (x) = x while y = g(x) = x3 = x = y 1/3 . = x = y,
y = x2 ,
y = 4.
Determine the volume of S if cross-sections perpendicular to the y-axis are equilateral triangles. 1. volume = 8 2 cu. units 2. volume = 8 3 cu. units correct 3. volume = 4 3 cu. units 4. volume = 4 2 cu. units 5. volume = 4 cu. units 6. volume = 8 cu. units Explanation: A view of the solid together with a crosssection is shown in the gure z
In this case, the volume V of the glass is given by V = V1 + V2 where
1
V1 =
0
(3 + y 1/3 )2 (3 + y)2 dy
is the volume of the top half of the glass, and
0
V2 =
1
(3 + y)2 (3 + y 1/3 )2 dy
is the volume of the bottom half of the glass. But
1
V1 =
0
6y
1/3
+y
2/3
6y y
2
dy
1
9 3 1 = y 4/3 + y 5/3 3y 2 y 3 2 5 3 while
0
,
0
V2 =
1
6y + y 2 6y 1/3 y 2/3 dy
0
y . x
9 3 1 = 3y + y 3 y 4/3 y 5/3 3 2 5
2
1
padilla (tp5647) HW05 Gilbert (56650) Now the length of the base of an equilateral triangular cross-section is specied by its position on the parabola, so its value is given s = 2x = 2 y . Thus, the cross section has area A(y) = 1 (base height) 2 1 3s = s = 3y . 2 2
4
5
in which case
1
I=
0
1 + 4ex + 4e2x dx = x 4ex 2e2x =
1 0
1 4e1 2e2 +6 .
Consequently, I = 7 2 4 2 . e e
Thus
4
V =
0
A(y) dy =
3
0
y dy , 009
10.0 points
and so the solid S has volume = 8 3 cu. units .
Find the value of f (1) when f (x) = 2xe7x , 36 1 7 e 7 7 36 1 7 e correct 7 7 36 1 7 + e 7 7
2
f = (0) 5.
keywords: integral, volume, cross-section 008 10.0 points
1. f (1) = 2. f (1) = 3. f (1) =
Evaluate the denite integral
1
I =
0
(1 + 2ex )2 dx .
1. I = 7 + 2. I = 7
2 4 + 2 e e 4 2 2 correct e e
1 4. f (1) = e7 7 5. f (1) = 1 7 36 e 7 7
3. I = 5 + 4e + 2e2 4. I = 7 + 2 4 2 e e
1 6. f (1) = e7 7 Explanation: By integration f (x) = 2xe7x dx.
2
5. I = 5 4e + 2e2 6. I = 5 4e 2e2 Explanation: After expansion (1 + 2e
x 2
After substitution u = x2 the integral becomes
x
) = 1 + 4e
+ 4e
2x
,
1 e7u du = e7u + C 7
padilla (tp5647) HW05 Gilbert (56650) with C an arbitrary constant. Thus
2 1 f (x) = e7x + C. 7
6
Thus I = 1 2
2 0
eu du =
1 u e 2
2 0
.
The condition f (0) = 5 determines C since f (0) = 5 i.e., 36 1 7x2 e . 7 7 At x = 1, therefore, f (x) = f (1) = 010 36 1 7 e . 7 7 = 36 C= , 7 Consequently,
I =
1 2 e 1 2
.
011
10.0 points
Find the value of the integral
4
10.0 points 1. I =
2
I =
0
3etan x + 2 dx . cos2 x
Evaluate the denite integral
1
I =
0
(4 6x)e8x6x dx. correct
3 e2 2 3 e 2
2. I =
1. I = 2. I = 3. I = 4. I =
1 2 e 1 2 1 2 e 1 2 1 2 e +1 2 1 2 e +1 2
3. I = 3e + 1 4. I = 3e 5. I = 3e 1 correct 6. I = 3 e+2 2
5. I = e2 1 6. I = e + 1 Explanation: Since the integrand is of the form const f (x) ef (x) , f (x) = 8x 6x2 ,
2
Explanation: Since 1 = sec2 x, cos2 x set u = tan x. Then du = sec2 x dx , while x = 0 x = 4 = = u = 0, u = 1. d tan x = sec2 x , dx
the substitution u = 8x 6x2 is suggested. For then du = (8 12x) dx = 2(4 6x) dx, while x = 0 x = 1 = = u = 0 u = 2.
padilla (tp5647) HW05 Gilbert (56650) In this case
1
7
Consequently, (3eu + 2) dx = 3eu + 2u
1 0
I =
0
.
y() = 3e3 + 10 . 013 10.0 points
Consequently, I = (3e + 2) 3 = 3e 1 .
Find the volume of the solid of revolution generated by rotating the graph of y = 2 ln x about the y-axis between y = 0 and y = 6. 1. V = 408.42 cu.units
keywords: 012 10.0 points
2. V = 406.42 cu.units 3. V = 410.42 cu.units 4. V = 402.42 cu.units correct 5. V = 404.42 cu.units Explanation: The volume, V , of the solid of revolution generated by rotating the graph of x = f (y) about the y-axis between y = a and y = b is given by
b
Find the value of y() when dy = 9e3x + 2 sin x, dx 1. y() = 3e3 10 2. y() = 9e 8 3. y() = 3e3 + 10 4. y() = 3e3 + 10 correct 5. y() = 9e 6. y() = 9e
y(0) = 3 .
8
V =
a
f (y)2 dy.
+8
Explanation: If dy = 9e3x + 2 sin x , dx then y = 3e3x 2 cos x + C with C an arbitrary constant whose value is determined by the condition y(0) = 3. For then y(0) = 5 + C = 3 , in which case y = 3e
3x
To apply this we have rst to express x as a function of y since initially y is given in terms of x by y = 2 ln x. But after exponentiation, 1 x = e 2 y . Thus
6
V =
0
ey dy = ey
6 0
.
Conequently, V = 402.42 cu.units. 014 10.0 points
Evaluate the denite integral
2
2 cos x + 8 .
I =
1
2(1 + ln x) xx dx .
padilla (tp5647) HW05 Gilbert (56650) 1. I = 1 2. I = 6 3. I = 4. I = 3 correct 2 5 4 5. I = ln x(1 2 ln x)2 +C 6. I = 2 ln x(1 2 ln x)2 +C 7. I = ln x(1 2 ln x)2 +C 8. I = 2(1 2 ln x)4 +C Explanation: Set u = 1 2 ln x. Then 2 du = dx, x so I = 4 Consequently, I = (1 2 ln x)4 +C with C an arbitrary constant. 016 3 2 10.0 points u3 du = u4 + C .
8
5. I = 4 6. I = 5 Explanation: Since xx = ex ln x and d (x ln x) = 1 + ln x , dx the substitution u = x ln x is suggested. For then
2 ln 2
I =
0
2eu du =
2eu
ln 4 0
.
Consequently, I = 2(e ln 4 e0 ) = because e
ln 4
Evaluate the integral
e
I = 1 = , 4
1
1 {2f (ln x) + 3} dx x
e = 1.
0
when f (0) = 2 and f (1) = 5. 1. I = 8 2. I = 9 correct 3. I = 7 4. I = 6 5. I = 5 Explanation: Set u = ln x. Then while x = 1 x = e = = u = 0, u = 1.
015
10.0 points 8 (1 2 ln x)3 dx. x
Determine the integral I =
1. I = (1 2 ln x)4 +C 2. I = 2(1 2 ln x) +C 3. I = (1 2 ln x)4 +C correct 4. I = 2 ln x(1 2 ln x)2 +C
4
padilla (tp5647) HW05 Gilbert (56650) In this case,
1
9
I =
0 1
{2f (u) + 3} du
018 10.0 points Evaluate the denite integral
3/4 1 0
=
0
{2f (u) + 3} du =
2f (u) + 3u
.
I =
/4
6 cos x 3 sin x dx . 6 sin x + 3 cos x
Consequently, I = 2{f (2) f (1)} + 3 = 9 017 10.0 points
2
1. I = ln .
11 3
2. I = ln (3) correct 3. none of these 4. I = ln (3) 5. I = ln 11 3
Evaluate the denite integral I =
0
2x dx . 4 + x2
1 1. I = 2 2. I = 1 3. I = 2 ln 2 4. I = 2 5. I = 1 ln 2 2
Explanation: Since the integrand is of the form f (x) , f (x) f (x) = 6 sin x + 3 cos x
this suggests the substitution u = 6 sin x + 3 cos x. For then du = (6 cos x 3 sin x) dx, while x = x = Thus 4 3 4 = = 9 u = , 2 3 u = . 2
6. I = ln 2 correct Explanation: Set u = 4 + x2 . Then du = 2x dx, while x = 0 x = 2 Thus
8
= =
u = 4, u = 8. I =
3/ 2 9/ 2
1 du = u
ln |u|
3/ 2 . 9/ 2
Consequently, 1 du = u ln |u|
8 4
I =
4
. I = ln 019
Consequently, I = ln 2 .
1 3
= ln (3) . 10.0 points
padilla (tp5647) HW05 Gilbert (56650) Evaluate the denite integral
1
10
Find the value of the denite integral
e4
I =
0
2x2 + 4x + 2 dx . 5x2 + 5 1. I = 5 2. I = 8 3. I = 7
I =
e
3 dx . x ln x
1. I = 2. I =
4 + 4 ln 3 5 1 + 2 ln 2 5
4 + 4 ln 2 3. I = 5 4. I = 2 + 2 ln 3 5
4. I = 6 correct 5. I = 4 Explanation: Set u = ln x. Then du = while x = e x = e4 In this case,
4
2 + 2 ln 2 5. I = correct 5 1 + 2 ln 3 6. I = 5 Explanation: After division 2x2 Thus
1
1 dx , x u = 1 u = 4.
= =
x + 4x + 2 2 4 . = + 2+5 2+1 5x 5 5 x
I =
1
3 u1/2
du =
6u1/2
4 1
.
I =
0
2 4 dx + 5 5 = 2 x 5
1 0
x2 + 4 5
x dx +1 x dx. x2 + 1
Consequently, I = 6 .
To evaluate the last integral we use substitution. Set u = x2 + 1. Then du = 2x, dx so 2 2 + I = 5 5 Conequently, I = 020 2 + 2 ln 2 . 5 10.0 points
2 1
2 1 du = + u 5
2 ln u 5
2 1
.