CHAPTER SIXTEEN

16.1 SOLUTIONS

1105

CHAPTER SIXTEEN Solutions for Section 16.1 Exercises 1. Mark the values of the function on the plane, as shown in Figure 16.1, so that you can guess respectively at the smallest and largest values the function takes on each small rectangle.

X

Lower sum =

f (xi , yi )∆x∆y

= 4∆x∆y + 6∆x∆y + 3∆x∆y + 4∆x∆y = 17∆x∆y = 17(0.1)(0.2) = 0.34. Upper sum =

X

f (xi , yi )∆x∆y

= 7∆x∆y + 10∆x∆y + 6∆x∆y + 8∆x∆y = 31∆x∆y = 31(0.1)(0.2) = 0.62. y ↓ 3 2.4

2.2

5

4

4

6

8

5

7

6 ∆y = 0.2

? 2.0

1.0

1.1

∆x = 0.1

-

10 1.2 ← x

Figure 16.1 2. In the subrectangle in the top left in Figure 16.4, it appears that f (x, y) has a maximum value of about 9. In the subrectangle in the top middle, f (x, y) has a maximum value of 10. Continuing in this way, and multiplying by ∆x and ∆y, we have Overestimate = (9 + 10 + 12 + 7 + 8 + 10 + 5 + 7 + 8)(10)(5) = 3800. Similarly, we find Underestimate = (7 + 7 + 8 + 4 + 5 + 7 + 1 + 3 + 6)(10)(5) = 2400. Thus, we expect that 2400 ≤

Z

R

f (x, y)dA ≤ 3800.

3. (a) If we take the partition of R consisting of just R itself, we get Lower bound for integral = minR f · AR = 0 · (4 − 0)(4 − 0) = 0. Similarly, we get Upper bound for integral = maxR f · AR = 4 · (4 − 0)(4 − 0) = 64.

1106

Chapter Sixteen /SOLUTIONS

(b) The estimates asked for are just the upper and lower sums. We partition R into subrectangles R (a,b) of width 2 and height 2, where (a, b) is the lower-left corner of R(a,b) . The subrectangles are then R(0,0) , R(2,0) , R(0,2) , and R(2,2) . (0, 4)

(4, 4)

(0, 0)

(4, 0)

Figure 16.2 Then we find the lower sum Lower sum =

X

(a,b)

AR(a,b) · min f = R(a,b)

X

(a,b)

=4

4 · (Min of f on R(a,b) )

X

(Min of f on R(a,b) )

(a,b)

= 4(f (0, 0) + f (2, 0) + f (0, 2) + f (2, 2)) √ √ √ √ = 4( 0 · 0 + 2 · 0 + 0 · 2 + 2 · 2) = 8.

Similarly, the upper sum is Upper sum = 4

X

(Max of f on R(a,b) )

(a,b)

= 4(f (2, 2) + f (4, 2) + f (2, 4) + f (4, 4)) √ √ √ √ = 4( 2 · 2 + 4 · 2 + 2 · 4 + 4 · 4) √ = 24 + 16 2 ≈ 46.63. The upper sum is an√overestimate and the lower sum is an underestimate, so we can get a better estimate by averaging them to get 16 + 8 2 ≈ 27.3.

4. (a) We first find an over- and underestimate of the integral, using four subrectangles. On the first subrectangle (0 ≤ x ≤ 3, 0 ≤ y ≤ 4), the function f (x, y) appears to have a maximum of 100 and a minimum of 79. Continuing in this way, and using the fact that ∆x = 3 and ∆y = 4, we have Overestimate = (100 + 90 + 85 + 79)(3)(4) = 4248, and Underestimate = (79 + 68 + 61 + 55)(3)(4) = 3156. A better estimate of the integral is the average of the overestimate and the underestimate: Better estimate =

4248 + 3156 = 3702. 2

(b) The average value of f (x, y) on this region is the value of the integral divided by the area of the region. Since the area of R is (6)(8) = 48, we approximate Average value =

1 Area

Z

R

f (x, y)dA ≈

1 · 3702 = 77.125. 48

We see in the table that the values of f (x, y) on this region vary between 55 and 100, so an average value of 77.125 is reasonable. 5. Partition R into subrectangles with the lines x = 0, x = 0.5, x = 1, x = 1.5, and x = 2 and the lines y = 0, y = 1, y = 2, y = 3, and y = 4. Then we have 16 subrectangles, each of which we denote R (a,b) , where (a, b) is the location of the lower-left corner of the subrectangle.

16.1 SOLUTIONS

1107

We want to find a lower bound and an upper bound for the volume above each subrectangle. The lower bound for the volume of R(a,b) is 0.5(Min of f on R(a,b) ) because the area of R(a,b) is 0.5 · 1 = 0.5. The function f (x, y) = 2 + xy increases with both x and y over the whole region R, as shown in Figure 16.3. Thus, Min of f on R(a,b) = f (a, b) = 2 + ab, because the minimum on each subrectangle is at the corner closest to the origin. y

z 4

R(a, b)

(a, b) y

x

x 2

Figure 16.4

Figure 16.3 Similarly, Max of f on R(a,b) = f (a + 0.5, b + 1) = 2 + (a + 0.5)(b + 1). So we have Lower sum =

X

0.5(2 + ab) = 0.5

X

(2 + ab)

(a,b)

(a,b)

= 16 + 0.5

X

ab

(a,b)

Since a = 0, 0.5, 1, 1.5 and b = 0, 1, 2, 3, expanding this sum gives

Lower sum = 16 + 0.5 ( 0 · 0 + 0 · 1 + 0 · 2 + 0 · 3

+ 0.5 · 0 + 0.5 · 1 + 0.5 · 2 + 0.5 · 3

+ 1·0+1·1+1·2+1·3

+ 1.5 · 0 + 1.5 · 1 + 1.5 · 2 + 1.5 · 3) = 25.

Similarly, we can compute the upper sum: Upper sum =

X

0.5(2 + (a + 0.5)(b + 1)) = 0.5

(a,b)

X

(2 + (a + 0.5)(b + 1))

(a,b)

= 16 + 0.5

X

(a + 0.5)(b + 1)

(a,b)

= 41.

6. Since f (x, y) is measured in micrograms per square meter, and we are integrating over an area measured in square meters, the units of the integral are micrograms. The integral represents the total quantity of pollution, in micrograms, in the region R. 7. The exact value of the integral is 40/3. 8. The value of the integral is around −2.4.

Problems 9. The function being integrated is f (x, y) = 1, which is positive everywhere. Thus, its integral over any region is positive.

1108


10. The function being integrated is f (x, y) = 1, which is positive everywhere. Thus, its integral over any region is positive. 11. The function being integrated is f (x, y) = 5x. Since x > 0 in R, f is positive in R and thus the integral is positive. 12. The function being integrated is f (x, y) = 5x, which is an odd function in x. Since B is symmetric with respect to x, the contributions to the integral cancel out, as f (x, y) = −f (−x, y). Thus, the integral is zero.

13. The region D is symmetric both with respect to x and y axes. The function being integrated is f (x, y) = 5x, which is an odd function in x. Since D is symmetric with respect to x, the contributions to the integral cancel out. Thus, the integral of the function over the region D is zero. 14. The function being integrated, f (x, y) = y 3 + y 5 , is an odd function in y while D is symmetric with respect to y. Then, by symmetry, the positive and negative contributions of f will cancel out and thus its integral is zero. 15. In a region such as B in which y < 0, the quantity y 3 + y 5 is less than zero. Thus, its integral is negative. 16. The region R is symmetric with respect to y and the integrand is an odd function in y, so the integral over R is zero. 17. The function being integrated, f (x, y) = y − y 3 is always negative in the region B since in that region −1 < y < 0 and |y 3 | < |y|. Thus, the integral is negative.

18. The function being integrated, f (x, y) = y − y 3 , is an odd function in y while D is symmetric with respect to y. By symmetry, the integral is zero. 19. The region D is symmetric both with respect to x and y axes. The function being integrated is odd with respect to y in the region D. Thus, its integral is zero. 20. Since D is a disk of radius 1, in the region D, we have |y| < 1. Thus, −π/2 < y < π/2. Thus, cos y is always positive in the region D and thus its integral is positive. 21. The function f (x, y) = ex is positive for any value of x. Thus, its integral is always positive for any region, such as D, with nonzero area.

22. The region D is symmetric both with respect to x and y axes. Looking at the contributions to the integral of the function f (x, y) = xex , we can see that any contribution made by the point (x, y), where x > 0, is greater than the corresponding contribution made by (−x, y), since ex > 1 > e−x for x > 0. Thus, the integral of f in the region D is positive. 23. The region D is symmetric both with respect to x and y axes. The function f (x, y) = xy 2 is odd with respect to x, and thus the contributions to the integral from (x, y) and (−x, y) cancel. Thus the integral is zero in the region D. 24. The function f (x, y) is odd with respect to x, and thus the integral is zero in region B, which is symmetric with respect to x. 25. We use four subrectangles to find an overestimate and underestimate of the integral: Overestimate = (15 + 9 + 9 + 5)(4)(3) = 456, Underestimate = (5 + 2 + 3 + 1)(4)(3) = 132. A better estimate of the integral is the average of the two:

Z

R

f (x, y)dA ≈

456 + 132 = 294. 2

The units of the integral are milligrams, and the integral represents the total number of mg of mosquito larvae in this 8 meter by 6 meter section of swamp. 26. The question is asking which graph has more volume under it, and from inspection, it appears that it would be the graph for the mosquitos. 27. Let’s break up the room into 25 sections, each of which is 1 meter by 1 meter and has area ∆A = 1. We shall begin our sum as an upper estimate starting with the lower left corner of the room and continue across the bottom and moving upward using the highest temperature, T i , in each case. So the upper Riemann sum becomes

P25

i=1

Ti ∆A = T1 ∆A + T2 ∆A + T3 ∆A + · · · + T25 ∆A = ∆A(T1 + T2 + T3 + · · · + T25 ) = (1)( 31 +29 + 28 + 27 + 27+ 29 +28 + 27 + 27 + 26+ 27 +27 + 26 + 26 + 26+ 26 +26 + 25 + 25 + 25+ 25 +24 + 24 + 24 + 24) = (1)(659) = 659.

16.1 SOLUTIONS

1109

In the same way, the lower Riemann sum is formed by taking the lowest temperature, t i , in each case:

P25

t ∆A i=1 i

= t1 ∆A + t2 ∆A + t3 ∆A + · · · + t25 ∆A = ∆A(t1 + t2 + t3 + · · · + t25 ) = (1)( 27 +27 + 26 + 26 + 25+ 26 +26 + 25 + 25 + 25+ 25 +24 + 24 + 24 + 24+ 24 +23 + 23 + 23 + 23+ 23 +21 + 20 + 21 + 22) = (1)(602) = 602.

So, averaging the upper and lower sums we get: 630.5. To compute the average temperature, we divide by the area of the room, giving Average temperature =

630.5 ≈ 25.2◦ C. (5)(5)

Alternatively we can use the temperature at the central point of each section ∆A. Then the sum becomes

P25

i=1

P25

Ti0 ∆A = ∆A i=1 Ti0 = (1)( 29 +28 + 27 + 26.5 + 26+ 27 +27 + 26 + 26 + 25.5+ 26 +25.5 + 25 + 25 + 25+ 25 +24 + 24 + 24 + 24+ 24 +23 + 22 + 22.5 + 23) = (1)(630) = 630.

Then we get Average temperature =

P25

Ti0 ∆A 630 = ≈ 25.2◦ C. Area (5)(5)

i=1

28. The total area of the square R is (1.5)(1.5) = 2.25. See Figure 16.5. On a disk of radius ≈ 0.5 the function has a value of 3 or more, giving a total contribution to the integral of at least (3)·(π ·0.5 2 ) ≈ 2.3. On less than half of the rest of the square the function has a value between −2 and 0, giving a contribution to the integral of between (1/2·2.25)(−2) = −2.25 R and 0. Since the positive contribution to the integral is therefore greater in magnitude than the negative contribution, R f dA is positive. y 2.0 1.5 1.0 0.5 −1

2

0

3

−0.5 −1.0 −1.0 −0.5

4

x

0 1 2

0

0.5

Figure 16.5

1.0

1.5

2.0

1110


29. The total number of tornados, per year, in a certain region, is the integral of the frequency of tornados in that region. In order to approximate it, we first subdivide the states into smaller regions of 100 miles 2 , as shown in the figure in the text. We will find the upper and lower bounds for the frequency of tornados, and then take the average of the two. We do this by finding the highest frequency of tornados in each subdivision, and then add up all the frequencies, and then do the same for the low frequencies. (a) For the high frequency in Texas, going left to right, top to bottom, we get: [(9 + 8) + (9 + 8 + 9 + 9 + 8) + (7 + 7 + 9 + 9 + 8) + (0 + 3 + 4 + 6 + 8 + 7 + 7 + 6) + (0 + 1 + 4 + 6 + 7 + 7) + (2 + 4 + 5) + (2 + 3)]. This equals 182 tornados. For the low frequency, we get: [(7 + 7) + (7 + 7 + 7 + 7 + 5) + (3 + 5 + 6 + 7 + 5) + (0 + 0 + 0 + 1 + 4 + 6 + 7 + 6) + (0 + 0 + 0 + 3 + 5 + 5) + (0 + 1 + 3) + (0 + 0)]. This equals 114 tornados. The average of the two equals (182 + 114)/2 = 148 tornados. (b) For the high frequency in Florida, going left to right, top to bottom, we get: [(5 + 5 + 5 + 9) + (9 + 9) + (7 + 7) + (5)]. This equals 61 tornados. For the low frequency, we get: [(5 + 5 + 5 + 5) + (7 + 7) + (7 + 5) + (5)]. This equals 51 tornados. The average of the two is equal to (61 + 51)/2 = 56 tornados. (c) For the high frequency in Arizona, going left to right, top to bottom, we get: [(1 + 1 + 0) + (1 + 1 + 0) + (0 + 0 + 0) + (0 + 0)]. This equals 4 tornados. For the low frequency, we get: [(0 + 0 + 0) + (0 + 0 + 0) + (0 + 0 + 0) + (0 + 0)]. This equals 0 tornados. The average of the two is equal to (4 + 0)/2 = 2 tornados. 30. Let R be the region 0 ≤ x ≤ 60,

0 ≤ y ≤ 8. Then Volume =

Z

w(x, y) dA R

Lower estimate: 10·2(1+4+8+10+10+8+0+3+4+6+6+4+0+1+2+3+3+2+0+0+1+1+1+1) = 1580. Upper estimate: 10·2(8+13+16+17+17+16+4+8+10+11+11+10+3+4+6+7+7+6+1+2+3+4+4+3) = 3820. The average of the two estimates is 2700 cubic feet. 31. (a) The graph of f looks like the graph of g in the xz-plane slid parallel to itself forward and backward in the y direction, since the value of y does not affect the value of f . (b) The solid bounded by the graph of f is a cylinder, hence

Z

f dA = Volume under surface = R

= (d − c) 32. (a) (b) (c) (d) (e)

Z

b

Length in y direction

Cross-sectional area in xz-plane

g(x) dx a

On the xy-plane, z = 0, so of the base is x 2 + y 2 = 15, a circle of radius √the 2equation of the edge 2 The area of the base is π( 15) = 15π meters . √ 2 2 The cross-section at z = 10 has equation √ 2 x + y = 25, a circle of radius 5. The area of the cross-section is π( 5) = 5π meters . At height z, the cross-section is the circle x2 + y 2 = 15 − z. √ This is a circle of radius 15 − z, so √ A(z) = π( 15 − z)2 = (15 − z)π meters2 .

(f) The approximate volume of the slice is A(z)∆z meters3 . (g) We have Volume of pile = 33. Let

Z

15 0

(15 − z)π dz =

z2 15z − 2

f (x, y) =

1 1 + a 1 x2 + y 2

g(x, y) =

1 1 + a 2 x2 + y 2

and let

15 225π π = meters3 . 2 0

√

15.

16.2 SOLUTIONS

1111

where 0 < aR1 < a2 . For all points (x, y) we have f (x, y) ≥ g(x, y) > 0, so in Riemann sum approximations for R f dA and R g dA using the same subdivision of R for both integrals, we have R It follows that

R

R

R

f dA ≥

X

f (xi , yj )∆x∆y ≥

X

g(xi , yj )∆x∆y.

g dA, and so increasing the value of a decreases the value of the integral

R

34. Take a Riemann sum approximation to

Z

R

X

f dA ≈

R

1 R 1+ax2 +y 2

f (xi , yi )∆A.

i,j

Then, using the fact that |a + b| ≤ |a| + |b| repeatedly, we have:

Z X X f dA ≈ | f (x , y )∆A| ≤ |f (xi , yi )∆A|. i i R

i,j

Now |f (xi , yj )∆A| = |f (xi , yj |∆A since ∆A is non-negative, so

Z X X f dA ≤ |f (xi , yj )|∆A. |f (x , y )∆A| = i i R

i,j

i,j

But the last expression on the right is a Riemann sum approximation to the integral

R

R

|f |dA, so we have

Z Z X X f dA ≈ |f (x , )|∆A ≈ ≤ |f |dA. f (x , y )∆A i j i j R

Thus,

R

i,j

i,j

Z Z f dA ≤ |f |dA. R

R

Solutions for Section 16.2 Exercises 1. We evaluate the inside integral first:

Therefore, we have

Z

Z

3 0

Z

4

4

(4x + 3y) dx = (2x2 + 3yx) = 32 + 12y. 0

0

4

(4x + 3y) dxdy = 0

Z

Therefore, we have

Z

2 0

Z

3

(x2 + y 2 ) dy = 0

3 2

2

(x + y ) dydx = 0

3. We evaluate the inside integral first:

Therefore, we have

Z

0

3Z

Z 2

(32 + 12y) dy = (32y + 6y 2 ) = 150. 0

0


Z

Z

0

y3 3

y=3 = 3x2 + 9. y=0

2

2

(3x2 + 9) dx = (x3 + 9x) = 26. 0

2

(6xy) dy = (3xy 2 ) = 12x.

(6xy) dydx = 0

x2 y +

2 0

3

3

0

Z

3

0

3

(12x) dx = (6x2 ) = 54. 0

dA.

1112



Z Therefore, we have

5.

Z

1

4Z

2

f dy dx or 1

Z

1

2Z

Z

1 0

Z

2

(x2 y) dy = 0

2

(x2 y) dydx = 0

Z

x2 y 2 2

1

y=2 = 2x2 . y=0

(2x2 ) dx = 0

2x3 3

4

f dx dy

1 = 2. 3 0

1

6. This region lies between x = 0 and x = 4 and between the lines y = 3x and y = 12, and so the iterated integral is

Z

4 0

Z

12

f (x, y) dydx. 3x

Alternatively, we could have set up the integral as follows:

Z

12 0

7. The line connecting (−1, 1) and (3, −2) is

Z

y/3

f (x, y) dxdy. 0

3x + 4y = 1

or y= So the integral becomes

Z

3 −1

Z

1 − 3x 4

Z

(1−3x)/4

f dy dx or −2

1 −2

Z

(1−4y)/3

f dx dy −1

8. The line on the left (through points (0, 0) and (3, 6)) is the line y = 2x; the line on the right (through points (3, 6) and (5, 0)) is the line y = −3x + 15. See Figure 16.6. One way to set up this iterated integral is:

Z

6 0

Z

(15−y)/3

f (x, y) dxdy. y/2

The other option for setting up this integral requires two separate integrals, as follows:

Z

3 0

Z

2x

f (x, y) dydx + 0

Z

5 3

Z

−3x+15

f (x, y) dydx.

0

y y = 2x y = −3x + 15 R x

Figure 16.6 9. Two of the sides of the triangle have equations x =

Z

3 1

y−5 y−1 and x = . So the integral is 2 −2

Z

1 (y−5) −2

1 (y−1) 2

f dx dy

16.2 SOLUTIONS

10. The line connecting (1, 0) and (4, 1) is y= So the integral is

11.

Z

3 1

Z

4

e

x+y

dxdy =

0

Z

3

4 Z e e dx =

Z

0

y

1

Z

2

f dy dx (x−1)/3

3

ex (e4 − 1) dx = (e4 − 1)(e2 − 1)e. See Figure 16.7.

x y

1

4

1 (x − 1) 3

1

y y=x

y=4

x

x 1

3

2

Figure 16.8

Figure 16.7

12. 13.

Z

2 0

Z

x

2

ex dydx = 0

Z

2 0

x

2

ex y dx = 0

Z

5

1

Z

Z

2

2

xex dx = 0

2x

sin x dy dx = x

=

Z Z

5 1 5 1

2

1 1 x2 e = (e4 − 1). See Figure 16.8. 2 2 0

2x

sin x · y x dx sin x · x dx

5

= (sin x − x cos x) 1

= (sin 5 − 5 cos 5) − (sin 1 − cos 1) ≈ −2.68.

See Figure 16.9.

y y

y = 2x

x=

√

y

y=x y=4

4 y=x

1 x 1

x

5

1

Figure 16.9 14.

Z

Figure 16.10

4 1

2

Z

y √

x2 y 3 dxdy = y

Z

4

y3 1

y

x3 dy 3 √y

4

1113

1114


1 3

=

1 3

=

1 = 3 See Figure 16.10.

Z

4 1

9

(y 6 − y 2 ) dy

y 11/2 y7 − 7 11/2

4 1

47 411/2 × 2 − 7 11

−

1 2 − 7 11

≈ 656.082

15. The region of integration ranges from x = 0 to x = 3 and from y = 0 to y = 2x, as shown in Figure 16.11. To evaluate the integral, we evaluate the inside integral first:

Z

2x

(x2 + y 2 ) dy = 0

Therefore, we have

Z

3 0

Z

x2 y +

y3 3

2x

y=2x (2x)3 8x3 14 3 = 2x3 + = x . = x2 (2x) + 3 3 3 y=0 Z

(x2 + y 2 ) dydx = 0

3

0

14 3 x 3

y 6

dx =

14 4 3 x = 94.5. 12 0

x2 + y 2 = 9

y

3

y = 2x

R −3 −2

x

−3

3

Figure 16.12

Figure 16.11 16. See Figure 16.12.

Z

0 −2

Z

0 −

√

2xy dydx =

9−x2

Z

0 −2

=− = =

Z

Z 0

0

−2

−2

0

x y 2 √ −

Z

√ x + y dA = R

= =

Z

Z

2 0 2 0

2 3

Z

Z

1 0

x(9 − x2 ) dx

(x3 − 9x) dx

x4 9 − x2 4 2

0

√ x + y dx dy

1

3 2 (x + y) 2 dy 3 0

2

0

dx

9−x2

= −4 + 18 = 14 17.

3

3

3

((1 + y) 2 − y 2 ) dy

−2

x

16.2 SOLUTIONS

2

5 5 2 2 · [(1 + y) 2 − y 2 ] 3 5 0

=

5 5 4 ((3 2 − 2 2 ) − (1 − 0)) 15 √ 4 √ (9 3 − 4 2 − 1) = 2.38176 = 15

=

18. In the other order, the integral is

Z

1 0

Z

2

√

x + y dy dx.

0

First we keep x fixed and calculate the inside integral with respect to y:

Z

2 0

y=2

√ 2 x + y dy = (x + y)3/2 3 y=0 =

Then the outside integral becomes

Z

1 0

2 (x + 2)3/2 − x3/2 . 3

i 1 h 2 2 2 2 5/2 3/2 3/2 5/2 (x + 2) −x dx = (x + 2) − x 3 3 5 5 0 =

2 2 5/2 3 − 1 − 25/2 = 2.38176 · 3 5

Note that the answer is the same as the one we got in Exercise 17. 19.

Z

(5x2 + 1) sin 3y dA = R

= = = =

Z

Z

1 −1 1

Z

π/3

(5x2 + 1) sin 3y dy dx 0 2

(5x + 1) −1

2 3

Z

1

1 π/3 − cos 3y 3 0

dx

(5x2 + 1) dx −1

1

2 5 3 ( x + x) 3 3 −1

32 2 10 ( + 2) = 3 3 9

20. The region of integration, R, is shown in Figure 16.13. Integrating first over y, as shown in the diagram, we obtain

Z

xy dA = R

Z

1 0

Z

1−x

xy dy 0

Now integrating with respect to x gives

Z

xy dA = R

dx =

Z

1 0

1−x

xy 2 2 0

dx =

Z

1 0

1 2 1 3 1 4 1 1 x − x + x = . 4 3 8 24 0

1 x(1 − x)2 dx 2

1115

1116

Chapter Sixteen /SOLUTIONS y 1 y 1 y =1−x

y = −x + 1

y =x+1

x

x

−1

1

1

Figure 16.13

Figure 16.14

21. It would be easier to integrate first in the x direction from x = y − 1 to x = −y + 1, because integrating first in the y direction would involve two separate integrals.

Z

2

(2x + 3y) dA = R

= = = =

Z

Z

Z

Z

Z

1 0

Z

1 0

1 0 1 0

−y+1

(2x + 3y)2 dx dy

y−1 −y+1

(4x2 + 12xy + 9y 2 ) dx dy

y−1

4 3 x + 6x2 y + 9xy 2 3

−y+1

dy

y−1

8 [ (−y + 1)3 + 9y 2 (−2y + 2)] dy 3 2 9 (−y + 1)4 − y 4 + 6y 3 3 2

−

9 13 2 = − (−1) − + 6 = 3 2 6

1 0

See Figure 16.14. 22. The region is bounded by x = 1, x = 4, y = 2, and y = 2x. Thus Volume =

Z

4 1

Z

2x

(6x2 y) dydx. 2

To evaluate this integral, we evaluate the inside integral first:

Z

Therefore, we have

Z

4 1

Z

2x

2x 2

(6x2 y) dy = (3x2 y 2 )

2x

(6x2 y) dydx = 2

The volume of this object is 2203.2.

Z

2

= 3x2 (2x)2 − 3x2 (22 ) = 12x4 − 12x2 .

4 1

(12x4 − 12x2 ) dx =

23. To find the average value, we evaluate the integral

Z

3 0

Z

12 5 4 x − 4x3 = 2203.2. 5 1

6

(x2 + 4y) dydx, 0

and then divide by the area of the base region. To evaluate this integral, we evaluate the inside integral first:

Z

6 0

y=6

(x2 + 4y) dy = (x2 y + 2y 2 )

y=0

= 6x2 + 72.

16.2 SOLUTIONS

Therefore, we have

Z

3 0

Z

Z

6 2

(x + 4y)dydx = 0

1117

3

3

(6x2 + 72)dx = (2x3 + 72x) = 270. 0

0

The value of the integral is 270. The area of the base region is 3 · 6 = 18. To find the average value of the function, we divide the value of the integral by the area of the base region: 1 Average value = Area

Z

Z

3 0

6

1 · 270 = 15. 18

(x2 + 4y) dydx = 0

The average value is 15. This is reasonable, since the smallest value of f (x, y) on this region is 0, and the largest value is 32 + 4 · 6 = 33.

24. To find the average value, we first find the value of the integral

Z

We evaluate the inside integral first:

Therefore, we have

Z

4

Z

Z

4 0

Z

3

(xy 2 ) dydx. 0

3 2

(xy ) dy = 0

Z

3

xy 3 3

y=3 = 9x. y=0

4

9x2 4 (xy ) dydx = (9x) dx = = 72. 2 0 0 0 0 The value of the integral is 72. To find the average value, we divide the value of the integral by the area of the region: 2

Average value =

1 Area

Z

4

0

Z

3

72 = 6. 3·4

(xy 2 ) dydx = 0

The average value of f (x, y) on this rectangle is 6. This is reasonable since the smallest value of xy 2 on this region is 0 and the largest value is 4 · 32 = 36.

25. (a) See Figure 16.15.

y

(1/2, 1/2)

y=x

x+y =1

R

x

Figure 16.15 (b) If we integrate with respect to x first, we have

Z

f (x, y) dA = R

Z

1/2 0

Z

1−y

f (x, y) dx dy. y

If we integrate with respect to y first, the integral must be split into two parts, so

(c) If f (x, y) = x,

Z

f (x, y) dA = R

Z

x dA = R

=

Z

1/2 0

Z

1/2 0

1 2

Z

Z

x

f (x, y) dy dx + 0

Z

1−y

x dx dy = y

1/2 0

Z

Z

1/2

=

1 . 8

1/2

1/2 0

(1 − y)2 − y 2 dy =

1 = (y − y 2 ) 2 0

1

1 2

Z

1−x

f (x, y) dy dx. 0

1−y

x2 2 y

Z

dy

1/2

0

1 − 2y dy

1118


Alternatively,

Z

x dA = R

= =

Z

1/2 0

Z

Z

1/2 0

Z

x

x dy dx + 0

x

xy dx + 0

1/2

x2 dx + 0

1/2

x3 = 3 0

+

Z

Z

Z

1/2

1

1/2

1

1/2

1

Z

1−x

x dy dx 0

1−x

xy

dx

0

x(1 − x) dx

x3 x2 − 2 3

1

1/2

1 1 1 1 1 1 + − − + = . = 24 2 3 8 24 8 26. (a)

y 4 x = −(y − 4)/2 or y = −2x + 4

x 2

Figure 16.16 (b)

R 2 R −2x+4 0

0

g(x, y) dy dx.

Problems

27. As given, the region of integration is as shown in Figure 16.17. Reversing the limits gives

Z

1 0

Z

y

x

e 0

x2

dydx =

Z

1

0 x2 1

x Z dx = ye

1

x2

0

e−1 e . = = 2 0 2

2

xex dx 0

y

1 y=x x=y

x=1

x

x 1

1

Figure 16.17

Figure 16.18

16.2 SOLUTIONS

1119

28. The function sin (x2 ) has no elementary antiderivative, so we try integrating with respect to y first. The region of integration is shown in Figure 16.18. Changing the order of integration, we get

Z

Z

1 0

1

sin (x2 ) dx dy = y

= =

Z

Z

Z

Z

1 0

x

sin (x2 ) dy dx 0

x sin (x ) · y dx

1

2

0

0

1

sin (x2 ) · x dx

0

1

cos (x2 ) =− 2 0

=−

1 1 cos 1 + = (1 − cos 1) = 0.23. 2 2 2


Z

1 0

Z

x2

p

2+

0

x3

dydx = = =

Z

Z

1

(y 0 1

x 0

p

2+

x3

x2 ) dx 0

p 2

2 + x3 dx

1

√ 3 2 √ 2 (2 + x3 ) 2 = (3 3 − 2 2). 9 9 0

y y

1

x=

√

3

y

x = y2

x=1

x=9 x

x 9

1

Figure 16.20

Figure 16.19


Z

9 0

Z

√

x 2

y sin (x ) dydx =

0

=

Z

0

1 2

=− =

√x !

9

Z

9

y 2 sin (x2 ) 2 0

x sin (x2 ) dx 0

9

cos (x2 ) 4 0

cos (81) 1 − = 0.056. 4 4

dx

1120


31. The region of the integration is shown in Figure 16.21. To make the integration easier, we want to change the order of the integration and get

Z

1 0

Z

e ey

x dx dy = ln x = =

y

Z

Z

Z

e 1 e 1 e

Z

ln x 0

x dy dx ln x

ln x

x · y ln x 0

x dx = 1

(e, 1)

dx

e

1 x2 = (e2 − 1). 2 1 2 8

y = 2x + 8

R

e

1

32. Order reversed:

8 0

Z

y = −2x + 8

x

x

−4

Figure 16.21

Z

y

4

Figure 16.22

(8−y)/2

f (x, y) dx dy. See Figure 16.22. (y−8)/2

33. (a) We divide the base region into four subrectangles as shown in Figure 16.23. The height of the object at each point (x, y) is given by f (x, y) = xy, we label each corner of the subrectangles with the value of the function at that point. (See Figure 16.23.) Since Volume = Height × Length × Width, and ∆x = 2 and ∆y = 3, we have Overestimate = (12 + 24 + 6 + 12)(2)(3) = 324,

and Underestimate = (0 + 6 + 0 + 0)(2)(3) = 36. We average these to obtain Volume ≈

324 + 36 = 180. 2

y 6

3

f =0

f = 12

f = 24

f =0

f =6

f = 12

f =0

f =0

f =0 x

2

4

Figure 16.23 (b) We have f (x, y) = xy, so Volume =

Z

4 0

Z

6

xy dydx = 0

Z

4 0

xy 2 2

y=6 Z dx = y=0

4 0

4

18x dx = 9x2 = 144. 0

The volume of this object is 144. Notice that 144 is between the over- and underestimates, 324 and 36, found in part (a).

16.2 SOLUTIONS

1121

34. (a) The contour f (x, y) = 1 lies in the xy-plane and has equation 2e−(x−1)

2

−y 2

= 1,

so −(x − 1)2 − y 2 = ln(1/2)

(x − 1)2 + y 2 = ln 2 = 0.69.

This is the equation of a circle centered at (1, 0) in the xy-plane. Other contours are of the form 2e−(x−1)

2

−y 2

2

=c

2

−(x − 1) − y = ln(c/2). Thus, all the contours are circles centered at the point (1, 0). 2 (b) The cross-section has equation z = f (1, y) = e−y . If x = 1, the base region in the xy-plane extends from √ √ y = − 3 to y = 3. See Figure 16.24, which shows the circular region below W in the xy-plane. So Area =

Z

√

3

√

2

e−y dy.

− 3

(c) Slicing parallel to the y-axis, we get Volume =

y=

√

Z

2 −2

Z √4−x2 −

√

e−(x−1)

2

−2

(1,

√

2

√ y = − 4 − x2

−y 2

dy dx.

4−x2

y 4 − x2

2

−2

3)

x

√ (1, − 3)

Figure 16.24: Region beneath W in the xy-plane

35. The intersection of the graph of f (x, y) = 25 − x2 − y 2 and xy-plane is a circle x2 + y 2 = 25. The given solid is shown in Figure 16.25. Thus the volume of the solid is V = =

Z

Z

f (x, y) dA R 5 −5

Z √25−y2 −

√

25−y 2

(25 − x2 − y 2 ) dx dy.

1122

Chapter Sixteen /SOLUTIONS f (x, y) = 25 − x2 − y 2

z

z

f (x, y) = 25 − x2 − y 2

z = 16

y y

x

x

Figure 16.25

Figure 16.26

36. The intersection of the graph of f (x, y) = 25 − x2 − y 2 and the plane z = 16 is a circle, x2 + y 2 = 32 . The given solid is shown in Figure 16.26. Thus, the volume of the solid is

Z

V =

Z

=

R

(f (x, y) − 16) dA

Z √9−y2

3

−

−3

√

9−y 2

(9 − x2 − y 2 ) dx dy.

37. The solid is shown in Figure 16.27, and the base of the integral is the triangle as shown in Figure 16.28. z

y

y=0

4

y−x=4

backside

-

2x + y + z = 4

y−x=4

−4

2x + y = 4

y 4 2

x

x

−4

Figure 16.27 Thus, the volume of the solid is V = = =

2

Figure 16.28

Z

z dA

ZR Z

(4 − 2x − y) dA

R 4 0

Z

(4−y)/2 y−4

(4 − 2x − y) dx dy.

16.2 SOLUTIONS

38. Volume =

Z

2 0

Z

Z

2

xy dy dx = 0

=

Z

2 0

2

2x dx 0

2

1 2 xy dx 2 0

2

= x2 =4

1123

0

39. The region of integration is shown in Figure 16.29. Thus Volume =

Z

1 0

Z

x

(x2 + y 2 ) dy dx = 0

Z

1 0

x2 y +

y3 3

y=x Z dx = y=0

1 0

1

x4 1 4 3 x dx = = . 3 3 0 3

y 1

y x=y

x=1

3

x = y2 x=9

x

x

1

9

Figure 16.30

Figure 16.29

40. The region of integration is shown in Figure 16.30. Thus, Volume = =

Z

Z

9 0 9 0

Z

√

x

(x + y) dy dx = 0

x3/2 +

x 2

dx =

Z

9 0

y2 xy + 2

y=√x dx y=0

9 2 5/2 x 2349 x + = 117.45. = 5 4 0 20 2

41. The plane 2x + y + z = 4 cuts the xy-plane in the line 2x + y = 4, so the region of integration is the triangle shown in Figure 16.31. We want to find the volume under the graph of z = 4 − 2x − y. Thus, Volume = = =

Z

Z

Z

2 0 2 0 2 0

Z

−2x+4 0

(4 − 2x − y) dy dx =

Z

2 0

4y − 2xy −

(−2x + 4)2 4(−2x + 4) − 2x(−2x + 4) − 2

(2x2 − 8x + 8) dx =

y2 2

dx

2 3 2 16 x − 4x2 + 8x = . 3 3 0

−2x+4 dx 0

1124

Chapter Sixteen /SOLUTIONS y 4

x = −(y − 4)/2 or y = −2x + 4

x 2

Figure 16.31 42. Let R be the triangle with vertices (1, 0), (2, 2) and (0, 1). Note that (3x + 2y + 1) − (x + y) = 2x + y + 1 > 0 for x, y > 0, so z = 3x + 2y + 1 is above z = x + y on R. We want to find

Z

Volume =

R

((3x + 2y + 1) − (x + y)) dA =

We need to express this in terms of double integrals.

Z

(2x + y + 1) dA. R

y

(2, 2)

2

y = 1 + 0.5x R2 1

R1 y = 2x − 2 y =1−x x 1

O

2

Figure 16.32 To do this, divide R into two regions with the line x = 1 to make regions R 1 for x ≤ 1 and R2 for x ≥ 1. See Figure 16.32. We want to find

Z

(2x + y + 1) dA =

R

Z

(2x + y + 1) dA + R1

Z

(2x + y + 1) dA. R2

Note that the line connecting (0, 1) and (1, 0) is y = 1 − x, and the line connecting (0, 1) and (2, 2) is y = 1 + 0.5x. So

Z

(2x + y + 1) dA = R1

The line between (1, 0) and (2, 2) is y = 2x − 2, so

Z

(2x + y + 1) dA = R2

Z

Z

1 0

2 1

Z

Z

1+0.5x

(2x + y + 1) dy dx. 1−x

1+0.5x

(2x + y + 1) dy dx. 2x−2

16.2 SOLUTIONS

1125

We can now compute the double integral for R1 :

Z

1 0

Z

1+0.5x

(2x + y + 1) dy dx = 1−x

= = =

Z

Z

1 0 1 0

y2 +y 2xy + 2

21 2 x + 3x 8

1+0.5x dx 1−x

dx

1 7 3 3 2 x + x dx 8 2 0

19 , 8

and the double integral for R2 :

Z

2 1

Z

1+0.5x

(2x + y + 1) dy dx = 2x−2

=

Z

Z

2 1 1

1+0.5x

(2xy + y 2 /2 + y)

0

−

dx

2x−2

3 39 2 x + 9x + 8 2

dx

2 13 3 9 2 3 = − x + x + x 8 2 2 1

29 = . 8

19 29 48 + = = 6. 8 8 8 43. We want to calculate the volume of the tetrahedron shown in Figure 16.33. So, Volume =

z 1/c

1/b 1/a

y

=1 ax + by

x

Figure 16.33 We first find the region in the xy-plane where the graph of ax + by + cz = 1 is above the xy-plane. When z = 0 we have ax + by = 1. So the region over which we want to integrate is bounded by x = 0, y = 0 and ax + by = 1. Integrating with respect to y first, we have Volume =

Z

1/a 0

Z

= = =

(1−ax)/b

z dy dx = 0

Z

Z

1/a 0 1/a 0

1 . 6abc

Z

1/a 0

y by 2 axy − − c 2c c

Z

(1−ax)/b 0

1 − by − ax dy dx c

y=(1−ax)/b dx y=0

1 (1 − 2ax + a2 x2 ) dx 2bc

1126


44. The region bounded by the x-axis and the graph of y = x − x 2 is shown in Figure 16.34. The area of this region is A=

Z

1 0

(x − x2 )dx = (

1 1 1 = − = . 2 3 6

1

x3 x2 − ) 2 3 0

y 1 4

x

0

1

1 2

Figure 16.34 So the average distance to the x-axis for points in the region is

R

Average distance =

Z

y dA = R

= Therefore the average distance is

1/60 1/6

Z

Z

1 0 1 0

Z

x−x2

y dy 0

!

x4 x2 − x3 + 2 2

R

y dA

area(R)

dx

dx =

1 1 1 1 − + = . 6 4 10 60

= 1/10.

45. Assume the length of the two legs of the right triangle are a and b, respectively. See Figure 16.35. The line through (a, 0) and (0, b) is given by yb + xa = 1. So the area of this triangle is A=

1 ab. 2

y b

a

x

Figure 16.35 Thus the average distance from the points in the triangle to the y-axis (one of the legs) is Average distance = =

1 A 2 ab

Z

a 0

Z

a 0

Z

b x+b −a

x dy dx

0

b − x2 + bx a

dx

a b 3 b 2 2 − x + x = ab 3a 2 0 2 =

2 ab

a b 6

=

a . 3

16.2 SOLUTIONS

1127

Similarly, the average distance from the points in the triangle to the x-axis (the other leg) is Average distance = = = 46. (a) We have 1 Average value of f = Area of Square =

1 4

1 = 4

Z Z

2 0 2 0

Z

2

Z

1 A 2 ab

Z bZ

2 ab

0

−a y+a b

Z b 0

ab2 6

a − y 2 + ay b

=

b . 3

dy

f dA Square

(ax2 + bxy + cy 2 ) dydx = 0

y dx dy

0

8 2ax2 + 2bx + c 3

1 16 16 = a + 4b + c 4 3 3 4 4 = a+b+ c 3 3

1 dx = 4

1 4

Z

2 0

ax2 y + bx

y2 y3 +c 2 3

2

8 2 3 ax + bx2 + cx 3 3 0

y=2 dx y=0

The average value will be 20 if and only if (4/3)a + b + (4/3)c = 20. (b) Since (4/3)a + b + (4/3)c = 20, we must have b = 20 − (4/3)a − (4/3)c. Any function f (x, y) = ax 2 + (20 − (4/3)a − (4/3)c)xy + cy 2 where a and c are any real numbers is a correct solution. For example, a = 1, c = 3 leads to the function f (x, y) = x2 + (44/3)xy + 3y 2 , and a = −3, c = 0 leads to the function f (x, y) = −3x2 + 24xy, both of which have average value 20 on the given square. See Figures 16.36 and 16.37. y

y

x

Figure 16.36: f (x, y) = x

2

x

Figure 16.37: f (x, y) = −3x + 24xy 2

+ 44 xy+3y 2 3

47. (a) We have Average value of f = = =

Z

1 f dA Area of Rectangle Rectangle 1 6 1 6

Z Z

2 x=0 2 0

Z

3

(ax + by) dydx = y=0

1 6

Z

2 0

axy + b

2 9 1 3 2 9 3ax + b dx = ax + bx 2 6 2 2 0

1 = (6a + 9b) 6 3 = a + b. 2

y2 2

y=3 dx y=0

1128


The average value will be 20 if and only if a + (3/2)b = 20. This equation can also be expressed as 2a + 3b = 40, which shows that f (x, y) = ax + by has average value of 20 on the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 3 if and only if f (2, 3) = 40. (b) Since 2a + 3b = 40, we must have b = (40/3) − (2/3)a. Any function f (x, y) = ax + ((40/3) − (2/3)a)y where a is any real number is a correct solution. For example, a = 1 leads to the function f (x, y) = x + (38/3)y, and a = −3 leads to the function f (x, y) = −3x + (46/3)y, both of which have average value 20 on the given rectangle. See Figure 16.38 and 16.39. y

y

x

Figure 16.38: f (x, y) = x +

x

Figure 16.39: f (x, y) = −3x +

38 y 3

46 y 3

48. (a) One solution would be to arrange that the minimum values of f on the square occur at the corners, so that the corner values give an underestimate of the average. See Figure 16.40. y 1

y 94 96

1

0.3 0.2

98

0.1

100

0

x

x

1

1

Figure 16.40

Figure 16.41

(b) One solution would be to arrange that the maximum values of f on the square occur at the corners, so that the corner values give an overestimate of the average. See Figure 16.41. 49. The force, ∆F , acting on ∆A, a small piece of area, is given by ∆F ≈ p∆A,

where p is the pressure at that point. Thus, if R is the rectangle, the total force is given by F =

Z

p dA. R

We choose coordinates with the origin at one corner of the plate. See Figure 16.42. y

(a, b)

b

a

Figure 16.42

x

16.3 SOLUTIONS

Suppose p is proportional to the square of the distance from the corner represented by the origin. Then we have p = k(x2 + y 2 ), Thus, we want to compute

F =

Z

Z

for some positive constant k.

k(x2 + y 2 )dA. Rewriting as an iterated integral, we have R 2

2

k(x + y ) dA = R

Z bZ 0

=k

a 2

2

k(x + y ) dxdy = k 0

Z b 0

a3 + ay 2 3

Z b 0

dy = k

b ! a3 y y 3 +a 3 3 0

k = (a3 b + ab3 ). 3

Solutions for Section 16.3 Exercises 1.

Z

f dV = W

= = = =

Z

Z

Z

Z

Z

Z

2 0

Z

2 0

Z

2 0 2

1 −1

Z

3

(x2 + 5y 2 − z) dz dy dx

2

3

1 (x z + 5y z − z 2 ) dy dx 2 −1 2 1

2

2

1

−1

(x2 + 5y 2 −

(x2 y + 0 2

(2x2 + 0

5 ) dy dx 2

1

5 3 5 y − y) dx 3 2 −1

10 − 5) dx 3

2

2 5 = ( x3 − x) 3 3 0

=

10 16 − =2 3 3

2.

Z

f dV = W

= =

Z

Z

Z

1 0 1 0 1

Z

Z

1 0 1

Z

2

(ax + by + cz) dz dy dx 0

(2ax + 2by + 2c) dy dx 0

(2ax + b + 2c) dx 0

= a + b + 2c

a

x3 + xy 2 3 0

dy

1129

1130


3.

Z

f dV = W

= = =

Z

Z

Z

Z

Z bZ

a

0

0

Z bZ

a 0

Z

a 0

Z

a 0

0

c

e−x−y−z dz dy dx 0 c

e−x e−y e−z dz dy dx 0

b

e

−x −y

e

0

c

(−e−z ) dy dx 0

b

e−x e−y (−e−c + 1) dy dx 0

= (1 − e−c ) = (1 − e

−b

Z

a 0

b

e−x (−e−y ) dx

)(1 − e

−c

)

Z

0

a

e

−x

dx

0

= (1 − e−a )(1 − e−b )(1 − e−c ) 4.

Z

f dV = W

= = =

Z

Z

Z

π 0

0

Z

πZ π

Z

π 0 π 0

= −2 = −2

Z

π

sin x cos(y + z) dz dy dx 0

π sin x sin(y + z) dy dx 0

π

Z0 π Z0 π 0

Z

0 π

Z0 π 0

sin x[sin(y + π) − sin y] dy dx

sin x(−2 sin y) dy dx

π

sin x(− cos y) dx 0

2 sin x dx

π

= −4(− cos x)

0

= (−4)(2) = −8

5. The region is the half cylinder in Figure 16.43. 6. The region is the half cylinder in Figure 16.44. 7. The region is the quarter sphere in Figure 16.45. z 1

z z

1

1 1 1 x

y

x

Figure 16.43

1

1

Figure 16.44

y

x

1 1

Figure 16.45

y

1131

16.3 SOLUTIONS

8. The region is the half cylinder in Figure 16.46. 9. The region is the cylinder in Figure 16.47. 10. The region is the hemisphere in Figure 16.48. z 1

z 1 z 1 1

x 1

x

y

1

y

1

x

1

Figure 16.47

Figure 16.46

1

y

Figure 16.48

11. The region is the hemisphere in Figure 16.49. 12. The region is the quarter sphere in Figure 16.50. 13. The region is the quarter sphere in Figure 16.51. z z

1 1

z 1

x 1

x

y

1

1

y

1 1 y

x 1

Figure 16.49

Figure 16.51

Figure 16.50

Problems 14. A slice through W for a fixed value of x is a semi-circle the boundary of which is y 2 = r2 − x2 − z 2 , so the inner integral is √

Z

r 2 −x2 −z 2

f (x, y, z) dy.

0

√ √ Lining up these stacks parallel to y-axis gives a slice from z = − r2 − x2 to z = r2 − x2 giving Z √ Z √ r 2 −x2

−

√

r 2 −x2

r 2 −x2 −z 2

f (x, y, z) dy dz.

0

Finally, there is a slice for each x between −r and r, so the integral we want is Z Z √ Z √ r 2 −x2

r

−r

−

√

r 2 −x2

r 2 −x2 −z 2

0

f (x, y, z) dy dz dx.

1132


15. A slice through W for a fixed value of x is a semi-circle the boundary of which is y 2 = r2 − x2 − z 2 , so the inner integral is √

Z

r − x2 −z 2

−

f (x, y, z) dy.

√

r − x2 −z 2

Lining up these stacks parallel to y-axis gives a slice from z = 0 to z = Z √ Z √ r−x2

r−x2 −z 2

−

0

√

√

r2 − x2 giving

f (x, y, z) dy dz.

r−x2 −z 2

Finally, there is a slice for each x between 0 and r, so the integral we want is Z √ Z Z √ r−x2

r

0

r−x2 −z 2

−

0

√


r−x2 −z 2

16. A slice through W for a fixed value of x is a semi-circle the boundary of which is y 2 = 4 − z 2 , so the inner integral is Z √ 4−z 2

f (x, y, z) dy.

0

Lining up these stacks parallel to y-axis gives a slice from z = −2 to z = 2 giving Z Z √ 4−z 2

2

f (x, y, z) dy dz.

−2

0

Finally, there is a slice for each x between 0 and 1, so the integral we want is Z Z Z √ 1

4−z 2

2


−2

0

0

17. A slice through W for a fixed value of y is a semi-circle the boundary of which is z 2 = r2 − x2 , so the inner integral is Z √ r 2 −x2

f (x, y, z) dz.

0

Lining up these stacks parallel to z-axis gives a slice from x = −r to x = r giving Z Z √ r 2 −x2

r

−r

f (x, y, z) dz dx.

0

Finally, there is a slice for each y between 0 and 1, so the integral we want is Z Z Z √ 1

0

18. The required volume, V , is given by V = = = =

r 2 −x2

r

−r

Z

Z

Z

Z

10 0 10 0 10

f (x, y, z) dz dx dy.

0

Z

Z

h

10−x 0 10−x 0

Z

10

dzdydx x+y

(10 − (x + y)) dydx

10y − xy −

0 10 0

500 = 3

1 2 y 2

1 (10 − x)2 dx 2

iy=10−x y=0

dx

16.3 SOLUTIONS

1133

19. The pyramid is shown in Figure 16.52. The planes y = 0, and y − x = 4, and 2x + y + z = 4 intersect the plane z = −6 in the lines y = 0, y − x = 4, 2x + y = 10 on the z = −6 plane as shown in Figure 16.53. z

y

y=0

-

(2, 6, −6)

y − x = 4 backside

z = −6 plane

2x + y + z = 4

−4

y 4

2

y−x=4

x

(−4, 0, −6)

2x + y − 6 = 4

(2, 6, −6) z = −6 bottom

(5, 0, −6)

x (5, 0, −6)

(−4, 0, −6)

Figure 16.52

Figure 16.53

These three lines intersect at the points (−4, 0, −6), (5, 0, −6), and (2, 6, −6). Let R be the triangle in the planes z = −6 with the above three points as vertices. Then, the volume of the solid is V = = = =

Z

Z

Z

Z

6 0 6 0 6

Z

Z

(10−y)/2 y−4 (10−y)/2

Z

4−2x−y

dz dx dy −6

(10 − 2x − y) dx dy = 162

y−4

(10−y)/2 dy (10x − x − xy) 2

0

y−4

6 0

9y 2 ( − 27y + 81) dy 4

= 162 20. Figure 16.54 shows a slice through the region for a fixed x. 3

z

2 z = 3y

1

z=y y

0

1

Figure 16.54

1134


The required volume, V , is given by

V = = = = =

Z

Z

Z

Z

Z

Z

2 1

Z

2 1

Z

2 1 2

1 0 1

Z

3y

dz dy dx y

z|3y y dy dx 0 1

2y dy dx 0

1

y 2 0 dx

1 2

dx

1

= 1.

21. Figure 16.55 shows a slice through the region for a fixed value of y. z

1

z=x

z = x2

x

0

1

Figure 16.55 We break the region into small cubes of volume ∆V = ∆x∆y∆z. A stack of cubes vertically above the point (x, z) in the xz-plane gives the strip shown in Figure 16.55 and so the inner integral is

Z

x

dz x2

The plane and the surface meet when x = x2 , giving x(1 − x) = 0, so x = 0 or x = 1. Lining up the stacks parallel to the z-axis gives a slice from x = 0 to x = 1. Thus, the limits on the middle integral are

Z

1 0

Z

x

dz dx. x2

Finally, there is a slice for each y between 0 and 3, so the integral we want is

The required volume, V , is given by

Z

V =

3 0

Z

Z 3

0

1 0

Z

Z 1

0

x

dz dx dy. x2

Z

x

dz dx dy x2

16.3 SOLUTIONS

Z

=

Z

=

Z

=

3 0 3

Z

0 3 0

Z

1

1135

x − x2 dx dy

0

1

x3 x2 dy − 2 3 0

1 dy 6 3

1 = dy 6 0 1 = ·3 6 1 = . 2 22. The required volume, V , is given by

V = = = = =

Z

Z

Z

Z

Z

5 0

Z

5 0 5

0 5−x

0

dz dy dx 0

(x + y) dy dx

5

x+y

0

1 2 y=5−x y dx 2 y=0

xy + 0

Z

5−x

x(5 − x) +

1 (5 − x)2 2

125 . 3

dx

23. The required volume, V , is given by

V = = = =

Z

Z

Z

Z

5 0 5 0 5 0 5

Z

Z

3 0 3

Z

x2

dz dy dx 0

x2 dy dx 0

y=3

x2 y

dx

y=0

3x2 dx

0

= 125.

24. (a) The vectors ~ u = ~i − ~j and ~v = ~i − ~k lie in the required plane so p ~ =~ u × ~v = ~i + ~j + ~k is perpendicular ~ ~ ~ to this plane. Let (x, y, z) be a point in the plane, then (x − 1)i + y j + z k is perpendicular to p ~ , so ((x − 1)~i + ~ ~ ~ ~ ~ y j + z k ) · (i + j + j ) = 0 and so (x − 1) + y + z = 0. Therefore, the equation of the required plane is x + y + z = 1. (b) The required volume, V , is given by

V =

Z

1 0

Z

0

1−x

Z

1−x−y

dz dy dx 0

1136


= = = =

Z

Z

Z Z

=

1 0

Z

1−x

(1 − x − y) dy dx

0

1

y − xy −

0 1 0 1 0

1 2 1−x y dx 2 0

1 − x − x(1 − x) −

1 (1 − x)2 dx 2

1 (1 − x)2 dx 2

1 . 6

25. (a) The equation of the surface of the whole cylinder along the y-axis is x 2 + z 2 = 1. The part we want is z= See Figure 16.56.

p

1 − x2

0 ≤ y ≤ 10.

z

x

1 10

y

Figure 16.56 (b) The integral is

Z

f (x, y, z) dV = D

Z

10 0

Z

1 −1

Z √1−x2

f (x, y, z) dzdxdy.

0

26. The region of integration is shown in Figure 16.57, and the mass of the given solid is given by z 6

+ y2 + z6 = 1 or z = −2x − 3y + 6 x 3

2

y x 3

x

Figure 16.57

Z

δ dV R

y 2

=1

2 x+2 or y = − 3

3

mass =

+

16.3 SOLUTIONS

=

Z

Z

3 0

Z

2 x+2 −3

0

−2x−3y+6

1137

(x + y) dzdydx

0

−2x−3y+6 Z 3 Z − 2 x+2 3 (x + y)z = dydx 0 0 0 Z 3 Z − 2 x+2 3

(x + y)(−2x − 3y + 6) dydx

=

= = = = =

Z

Z

Z

0

Z

3

0

x+2 −2 3

0

3 0

3 0

0

(−2x2 − 3y 2 − 5xy + 6x + 6y) dydx

− 3 x+2 5 −2x y − y − xy 2 + 6xy + 3y 2 dx 2 0 2

3

14 3 8 2 x − x + 2x + 4 27 3

2

dx

3 7 4 8 3 x − x + x2 + 4x 54 9 0

8 21 15 7 · 34 − · 33 + 32 + 12 = −3= . 54 9 2 2

27. The pyramid is shown in Figure 16.58. The planes y = 0, and y − x = 4, and 2x + y + z = 4 intersect the plane z = −6 in the lines y = 0, y − x = 4, 2x + y = 10 as shown in Figure 16.59. z

y

y=0

-

(2, 6, −6)

y − x = 4 backside

z = −6 plane

2x + y + z = 4

−4

y 4

2

y−x=4

x

(−4, 0, −6)

2x + y − 6 = 4

(2, 6, −6) z = −6 bottom

(5, 0, −6)

x (5, 0, −6)

(−4, 0, −6)

Figure 16.58

Figure 16.59

These three lines (the edges of the pyramid) intersect the plane z = −6 at the points (−4, 0, −6), (5, 0, −6), and (2, 6, −6). Let R be the triangle in the plane z = −6 with these three points as vertices. Then, the mass of the solid is Mass = = = = =

Z

Z

Z

Z

Z

6

0 6 0 6 0 6 0 6

Z

Z

Z

(10−y)/2

y−4 (10−y)/2 y−4 (10−y)/2 y−4

Z

Z

4−2x−y

δ(x, y, z) dz dx dy −6

4−2x−y

y dz dx dy −6

y(10 − 2x − y) dx dy

x=(10−y)/2

y(10x − x2 − xy)

( 0

= 243.

x=y−4

9y 3 − 27y 2 + 81y) dy 4

dy

1138


28. From the problem, we know that (x, y, z) is in the cube which is bounded by the three coordinate planes, x = 0, y = 0, z = 0 and the planes x = 2, y = 2, z = 2. We can regard the value x 2 + y 2 + z 2 as the density of the cube. The average value of x2 + y 2 + z 2 is given by average value = =

= = = = = = 29. Positive. The function

p

R

V

(x2 + y 2 + z 2 ) dV volume(V )

R2R2R2 0

0

0

(x2 + y 2 + z 2 ) dxdydz

R 2 R 2 x3 0

3

0

R2R2 0

R2 0

R2 0

0

8 + 2y 2 + 2z 2 dydz 8

8 y 3

+

16 3

+

32 z 3

64 3

8 3

8 2 + (y 2 + z 2 )x 0 dydz

2 3 y 3

16 3

2 + 2z 2 y 0 dz

8 + 4z 2 dz

8 2 + 34 z 3 0

+ 8

8

32 3

= 4.

x2 + y 2 is positive, so its integral over the solid W is positive.

30. Positive. If (x, y, z) is any point inside the solid W then its integral over the solid W is positive.

p

x2 + y 2 < z. Thus the integrand z −

p

x2 + y 2 > 0, and so

31. Zero. The value of x is positive above the first and fourth quadrants in the xy-plane, and negative (and of equal absolute value) above the second and third quadrants. The integral of x over the entire solid cone is zero because the integrals over the two halves of the cone cancel. 32. Zero. The value of y is positive on the half of the cone above the second and third quadrants and negative (of equal absolute value) on the half of the cone above the third and fourth quadrants. The integral of y over the entire solid cone is zero because the integrals over the four quadrants cancel. 33. Positive. Since z is positive on W , its integral is positive. 34. Zero. You can see this in several ways. One way is to observe that xy is positive on part of the cone above the first and third quadrants (where x and y are of the same sign) and negative (of equal absolute value) on the part of the cone above the second and fourth quadrants (where x and y have opposite signs). These add up to zero in the integral of xy over all of W . Another way to see that the integral is zero is to write the triple integral as an iterated integral, say integrating first with respect to x. For fixed y and z, the x-integral is over an interval symmetric about 0. The integral of x over such an interval is zero. If any of the inner integrals in an iterated integral is zero, then the triple integral is zero. 35. Zero. Write the triple integral as an iterated integral, say integrating first with respect to x. For fixed y and z, the x-integral is over an interval symmetric about 0. The integral of x over such an interval is zero. If any of the inner integrals in an iterated integral is zero, then the triple integral is zero. 36. Negative. If (x, y, z) is any point inside the cone then z < 2. Hence the function z − 2 is negative on W and so is its integral. 37. Positive. The function e−xyz is a positive function everywhere so its integral over W is positive. 38. Positive. The function

p

x2 + y 2 is positive, so its integral over the solid W is positive.

39. Positive. If (x, y, z) is any point inside the solid W then over the solid W is positive.

p

x2 + y 2 < z. Thus z −

40. Positive. The value of x is positive on the half-cone, so its integral is positive.

p

x2 + y 2 > 0, and so its integral

41. Zero. y is positive on the half of the half-cone above the first quadrant in the xy-plane and negative (of equal absolute value) on the half of the half-cone above the fourth quadrant. The integral of y over W is zero because the integrals over

1139

16.3 SOLUTIONS

each half add up to zero. 42. Positive. Since z is positive on W , its integral is positive. 43. Zero. You can see this in several ways. One way is to observe that xy is positive on part of the cone above the first quadrant (where x and y are of the same sign) and negative (of equal absolute value) on the part of the cone above the fourth quadrant (where x and y have opposite signs). These add up to zero in the integral of xy over all of W . Another way to see that the integral is zero is to write the triple integral as an iterated integral, say integrating first with respect to y. For fixed x and z, the y-integral is over an interval symmetric about 0. The integral of y over such an interval is zero. If any of the inner integrals in an iterated integral is zero, then the triple integral is zero. 44. Zero. Write the triple integral as an iterated integral, say integrating first with respect to y. For fixed x and z, the y-integral is over an interval symmetric about 0. The integral of y over such an interval is zero. If any of the inner integrals in an iterated integral is zero, then the triple integral is zero. 45. Negative. If (x, y, z) is any point inside the cone then z < 2. Hence the function z − 2 is negative on W and so is its integral. 46. Positive. The function e−xyz is a positive function everywhere so its integral over W is positive. 47. Orient the region as shown in Figure 16.60 and use Cartesian coordinates with origin at the center of the sphere. The equation of the sphere is x2 + y 2 + z 2 = 25, and we want the volume between the planes z = 3 and z = 5. The plane z = 3 cuts the sphere in the circle x2 + y 2 + 32 = 25, or x2 + y 2 = 16. Z Z √ Z √ 16−x2

4

Volume =

−4

−

25−x2 −y 2

√

16−x2

dzdydx.

3

z

Sphere is

z 5

6

x2 + y 2 + z 2 = 25

2

? 6

ICircle is

y

x2 + y 2 = 16

3

?

y

I

x

x2 + y 2 = 16

x

Figure 16.61

Figure 16.60

48. The intersection of two cylinders x2 + z 2 = 1 and y 2 + z 2 = 1 is shown in Figure 16.61. This region is bounded by four surfaces: p p p p z = − 1 − x2 , z = 1 − x2 , y = − 1 − z 2 , and y = 1 − z 2 So the volume of the given solid is

V = 49. Set up axes as in Figure 16.62.

Z

1 −1

Z √1−x2 Z √1−z2 −

√

1−x2

−

√

1−z 2

dy dz dx

1140

Chapter Sixteen /SOLUTIONS z 2 y

2

x

2

Figure 16.62 The slanting plane has equation z = 2 − x − y and the line where it intersects the xy-plane has equation y = 2 − x. The mass of the bottom part is δ1 times its volume, VBottom , where VBottom =

Z 2Z 0

= =

Z 2Z Z

0

2

0

2−x 0 2−x

Z

2−x−y

dz dy dx = 0

(2 − x − y) dy dx =

0

Z 2Z

Z

0 2

0 3 2

2−x 0

2−x−y

z

y2 (2 − x)y − 2

−(2 − x) (2 − x) 4 3 dx = =3m . 2 6 0 2

dy dx

0

2−x dx 0

Since the cube has volume 23 = 8 m3 , the upper part has volume VTop = 8 − 4/3 = 20/3 m3 . Thus 4 20 δ1 + δ2 gm. 3 3

Mass = VBottom δ1 + VTop δ2 = 50. The mass m is given by m= = = =

Z

Z

Z

Z

1 dV = W 1 0 1

Z

1

Z

1 0

Z

1 0

Z

x+y+1

1 dz dy dx 0

(x + y + 1) dy dx 0

1

xy + y 2 /2 + y) 0 dx

0 1

(x + 3/2) dx = 2 gm. 0

Then the x-coordinate of the center of mass is given by 1 x ¯= 2 = =

1 2 1 2

1 = 2

Z Z Z Z

W 1 0 1 0 1

1 x dV = 2

Z

1

Z

1 0

Z

1 0

Z

x+y+1

x dz dy dx 0

x(x + y + 1) dy dx 0

1

x2 y + xy 2 /2 + xy) 0 dx

(x2 + 3/2x) dx = 13/24 cm. 0

An essentially identical calculation (since the region is symmetric in x and y) gives y¯ = 13/24 cm.

16.3 SOLUTIONS

Finally, we compute z¯:

Z

1 2

z¯ =

Z

1 2

=

Z

1 = 2

1 = 12

z dV = W 1

Z

0 1

1

Z

1 0

Z

1 0

Z

x+y+1

z dz dy dx 0

(x + y + 1)2 /2 dy dx 0

1

(x + y + 1)3 /6 0 dx

0

Z

1 2

1

((x + 2)3 − (x + 1)3 ) dx = 25/24 cm.

0

So (¯ x, y¯, z¯) = (13/24, 13/24, 25/24). 51. The mass m is given by m= =

Z

Z

W 1 0

1 = 3 1 = 3 = =

Z

1 dV =

1 3 1 3

Z

1

0 (1−x)/2

Z

Z

(1−x)/2 0

(1−x−2y)/3

1 dz dy dx 0

1 − x − 2y dy dx 3

0

(1−x)/2 dx (y − xy − y ) 0 0 Z 1 Z Z

1

2

1−x 1−x −x − 2 2

0 1

0

1−x 2

(1 − x)2 (1 − x)2 − 2 4

(1 − x)3 12

2

dx

dx

1 = 1/36 gm. 0

Then the coordinates of the center of mass are given by x ¯ = 36 and y¯ = 36 and z¯ = 36

Z

x dV = 36 W

Z

Z

y dV = 36 W

z dV = 36 W

Z Z

Z

1 0

1 0 1

0

Z Z

Z

(1−x)/2 0

Z

(1−x−2y)/3

x dz dy dx = 1/4 cm.

0

(1−x)/2 0 (1−x)/2

0

Z

Z

(1−x−2y)/3

y dz dy dx = 1/8 cm. 0 (1−x−2y)/3

z dz dy dx = 1/12 cm. 0

52. The volume V of the solid is 1 · 2 · 3 = 6. We need to compute m 6

Z

m x + y dV = 6 2

W

2

m = 6 m = 2 m = 2

Z Z Z Z

1 0 1 0 1 0 1

Z Z

2 0 2

Z

3

x2 + y 2 dz dy dx 0

3(x2 + y 2 ) dy dx 0

2

(x2 y + y 3 /3) 0 dx

(2x2 + 8/3) dx = 5m/3 0

1141

1142


53. The volume of the solid is 8abc, so we need to evaluate m 8abc

Z

(y 2 + z 2 ) dV = W

m 8abc

m = 8abc m = 4bc m = 2c

Z Z

Z

Z

c −c c

−c c

−c c

Z Z

b −b b

Z

a

(y 2 + z 2 ) dx dy dz −a

2a(y 2 + z 2 ) dy dz −b

b

(y 3 /3 + yz 2 ) −b dz

(b2 /3 + z 2 ) dz −c

= m(b2 + c2 )/3 54. By the definition, we have that m a+b = V =

m V

m = V

Z Z

m (y + z ) dV + V W 2

Solutions for Section 16.4 Exercises Z 2π Z

√

2. 3. 4.

Z

Z

0

0

ZW W

π/2 3π/4 π/4

(x2 + y 2 ) dV +

Z

m V

Z

(2z 2 ) dV W

(2z 2 ) dV W

(2z 2 ) dV will be positive, thus a + b > c. W

f rdr dθ 0

π/2 Z

3π/2

(x2 + z 2 ) dV W

2

1.

Z

R

Z

(x2 + y 2 + 2z 2 ) dV

m = c+ V Since z 2 is always positive, the integral

2

1/2

f rdr dθ 0

Z

Z

2

f rdr dθ 1 2

f rdr dθ 0

5. See Figure 16.63. y

y r=4

θ = π/2

r=1

1

R

x

x −1

θ = −π/2

Figure 16.63 6. See Figure 16.64.

Figure 16.64

16.4 SOLUTIONS

1143

7. See Figure 16.65. 8. See Figure 16.66.

y θ = 3π/4

y

r=2

y

θ = π/3

r=1

1

2

r=1

x

x

θ = π/6 r=4

θ = 3π/2

r=3

x

Figure 16.66

Figure 16.65

Figure 16.67

9. See Figure 16.67. 10. See Figure 16.68. y

y θ = π/4

θ = π/4 r = 2/ sin θ

or r sin θ = 2 or y = 2

x

x

r = 1/ cos θ or r cos θ = 1 or x = 1

Figure 16.69

Figure 16.68 11. See Figure 16.69. 12. By using polar coordinates, we get

Z

sin(x2 + y 2 )dA = R

=

Z

Z

2π 0 2π 0

Z

2

sin(r 2 )r dr dθ 0

2

1 − cos(r 2 ) dθ 2 0

Z

2π

1 (cos 4 − cos 0) dθ 2 0 1 = − (cos 4 − 1) · 2π = π(1 − cos 4) 2

=−

13. The region is pictured in Figure 16.70. y 2

1

1

Figure 16.70

2

x

1144


Using polar coordinates, we get

Z

2

R

2

(x − y )dA =

Z

Z

π/2 0

Z

2 2

1

2

2

r (cos θ − sin θ)rdr dθ = 15 = 4 15 = 4 =

Z Z

π/2 0

π/2

2

1 (cos θ − sin θ) · r4 dθ 4 1 2

2

(cos2 θ − sin2 θ) dθ

0 π/2

cos 2θ dθ 0

π/2

15 1 · sin 2θ 4 2 0

= 0.

14. The presence of the term x2 +y 2 suggests that we should convert the integral into polar coordinates. Since the integral becomes

Z p

x2 + y 2 dxdy =

R

15. (a)

Z

2π 0

Z

3

r2 drdθ = 2

Z

2π 0

y

3

r3 dθ = 3 2

Z

2π 0

p

x2 + y 2 = r,

38π 19 dθ = . 3 3

y = x/3 1

x 3

Figure 16.71

(b)

Z

1 0

Z

3y

f (x, y) dx dy. 0

(c) For polar coordinates, on the line y = x/3, tan θ = y/x = 1/3, so θ = tan −1 (1/3). On the y-axis, θ = π/2. The quantity r goes from 0 to the line y = 1, or r sin θ = 1, giving r = 1/ sin θ and f (x, y) = f (r cos θ, r sin θ). Thus the integral is

Z

π/2 tan−1 (1/3)

Z

1/ sin θ

f (r cos θ, r sin θ)r dr dθ. 0

Problems √ √ 16. By the given limits 0 ≤ x ≤ −1, and − 1 − x2 ≤ y ≤ 1 − x2 , the region of integration is in Figure 16.72. In polar coordinates, we have

Z

3π/2 π/2

Z

1

r cos θ r dr dθ = 0

Z

3π/2

cos θ π/2

1 = 3

Z

3π/2

cos θ

1 1 3 r 3 0

dθ

dθ

π/2

3π/2

1 2 1 = (−1 − 1) = − = sin θ 3 3 3 π/2

1145

16.4 SOLUTIONS y

√

y 6 y=x

x

√x 6

−1 y = −x

√ − 6

Figure 16.72

Figure 16.73

17. From the given limits, the region of integration is in√ Figure 16.73. √ In polar coordinates, −π/4 ≤ θ ≤ π/4. Also, 6 = x = r cos θ. Hence, 0 ≤ r ≤ 6/ cos θ. The integral becomes

Z

√

0

6

Z

x

dy dx = −x

=

Z

Z

π/4 −π/4

Z

π/4 −π/4

√

r dr dθ 0

√6/cos θ !

r2 2 0

π/4

= 3 tan θ

6/cos θ

−π/4

dθ =

Z

π/4 −π/4

6 dθ 2 cos2 θ

= 3 · (1 − (−1)) = 6.

Notice that we can check this answer because the integral gives the area of the shaded triangular region which is √ (2 6) = 6. 18. From the given limits, the region of integration is in Figure 16.74. y 2 x=y √

2

π/4 √

x 2

2

Figure 16.74 So, in polar coordinates, we have,

Z

π/4 0

Z

2

(r2 cos θ sin θ)r dr dθ = 0

Z

=4

π/4

cos θ sin θ 0

Z

π/4 0

sin(2θ) 2

2 1 4 r dθ 4 0

dθ

π/4 = 0 − (−1) = 1. = − cos(2θ) 0

1 2

·

√

6·

1146


19. The graph of f (x, y) = 25−x2 −y 2 is an upside down bowl, and the region whose volume we want is contained between the bowl (above) and the xy-plane (below). We must first find the region in the xy-plane where f (x, y) is positive. To do that, we set f (x, y) ≥ 0 and get x2 + y 2 ≤ 25. The disk x2 + y 2 ≤ 25 is the region R over which we integrate. Volume =

Z

2

R

Z

2

(25 − x − y ) dA =

=

Z

2π 0

Z

625 4 625π = 2 =

0

5 0

(25 − r 2 ) rdr dθ

5 25 2 1 4 r − r dθ 2 4 0 2π

dθ

0

20. First, let’s find where the two surfaces intersect.

p

Z

2π

p

8 − x2 − y 2 =

x2 + y 2

8 − x 2 − y 2 = x2 + y 2 x2 + y 2 = 4

So z = 2 at the intersection. See Figure 16.75. z

2

y

x2 + y 2 = 4

2

R

x

Figure 16.75 The volume of the ice cream cone has two parts. The first part (which is the volume of the cone) Z is the volume of the

solid bounded by the plane z = 2 and the cone z =

p

x2 + y 2 . Hence, this volume is given by

R

(2 −

where R is the disk of radius 2 centered at the origin, in the xy-plane. Using polar coordinates, we have:

Z R

2−

p

x2 + y 2

dA = =

Z

Z

Z

2π 0

0

Z

0

"

2π

4 = 3

2

(2 − r) · r dr dθ r3 r − 3 2

2π

dθ 0

= 8π/3

2 # dθ 0

p

x2 + y 2 ) dA,

16.4 SOLUTIONS

1147

TheZsecond part is the volume of the region above the plane z = 2 but inside the sphere x 2 + y 2 + z 2 = 8, which is given by

(

R

p

8 − x2 − y 2 − 2) dA where R is the same disk as before. Now

Z p (

R

8 − x2 − y 2 − 2) dA = = =

Z

Z

Z

2π 0 2π 0 2π 0

Z

Z

Z

2

( 0 2

p

r 0

8 − r 2 − 2)rdr dθ

p

8 − r 2 dr dθ −

2 !

1 − (8 − r 2 )3/2 3 0

2π

21. (a)

Total Population =

Z

3π/2 π/2

Z

2π 0

dθ −

1 (43/2 − 83/2 ) dθ − 3 0 √ 1 = − · 2π(8 − 16 2) − 8π 3 √ 2π = (16 2 − 8) − 8π 3 √ 8π(4 2 − 5) = 3 √ Thus, the total volume is the sum of the two volumes, which is 32π( 2 − 1)/3. =−

Z

Z

Z

Z

2

2r dr dθ 0 2π

2 r dθ 2

0

0

2π

4 dθ 0

4

δ(r, θ) rdr dθ. 1

(b) We know that δ(r, θ) decreases as r increases, so that eliminates (iii). We also know that δ(r, θ) decreases as the x-coordinate decreases, but x = r cos θ. With a fixed r, x is proportional to cos θ. So as the x-coordinate decreases, cos θ decreases and (i) δ(r, θ) = (4 − r)(2 + cos θ) best describes this situation. (c)

Z

3π/2 π/2

Z

4

(4 − r)(2 + cos θ) rdr dθ =

1

Z

=9

3π/2 π/2

Z

(2 + cos θ)(2r 2 −

3π/2

(2 + cos θ) dθ π/2

= 9 2θ + sin θ

4

1 3 r ) dθ 3 1

3π/2 π/2

= 18(π − 1) ≈ 39 Thus, the population is around 39,000. 22. (a) The volume, V , is given by V = Converting to polar coordinates gives V =

Z

2π 0

Z

a

e

−r 2

0 2

Z

e−(x

2

+y 2 )

dA.

x2 +y 2 ≤a2

2π a 2 1 −r2 1 1 −a2 − e r dr dθ = θ − e = 2π = π(1 − e−a ). 2 2 2 0 0

(b) As a → ∞, the value of e−a → 0, so the volume tends to π.

23. The density function is given by

ρ(r) = 10 − 2r

where r is the distance from the center of the disk. So the mass of the disk in grams is

Z

ρ(r) dA = R

Z

2π 0

Z

5 0

(10 − 2r)rdr dθ

1148


= =

Z

Z

2π 0 2π 0

5r2 −

2 3 r 3

5

dθ

0

125 250π dθ = (grams) 3 3

24. A rough graph of the base of the spring is in Figure 16.76, where the coil is roughly of width 0.01 inches. The volume is equal to the product of the base area and the height. To calculate the area we use polar coordinates, taking the following integral: Area = = =

Z

Z

4π 0

1 2 1 2

Z Z

0.26+0.04θ

rdrdθ

0.25+0.04θ 4π

(0.26 − 0.04θ)2 − (0.25 − 0.04θ)2 dθ

0 4π

0.01 · (0.51 + 0.08θ)dθ

0

= 0.0051 · 2π + Therefore, the volume= 0.0636 · 0.2 = 0.0127 in3 .

4π

1 (0.0008θ 2 ) = 0.0636 4 0

Figure 16.76 25. The charge density is δ = k/r, where k is a constant. Total charge =

Z

δ dA = Disk

Z

R 0

Z

2π 0

k r dθ dr = k r

Z

R 0

Z

2π

dθ dr = k 0

Thus the total charge is proportional to R with constant of proportionality 2kπ.

Z

R

2π dr = 2kπR. 0

26. (a) We must first decide where to put the origin. We locate the origin at the center of one disk and locate the center of the second disk at the point (1, 0). See Figure 16.77. (Other choices of origin are possible.) y √

x2 + y 2 = 1

(x − 1)2 + y 2 = 1

y

r=1

r = 2 cos θ

3/2

1 2

1

x

π/3

x

√ − 3/2

Figure 16.77

Figure 16.78

By symmetry, the points of intersection of the circles are half-way between the centers, at x = 1/2. The y-values at these points are given by r √ 2 p 1 3 y = ± 1 − x2 = ± 1 − =± . 2 2

16.5 SOLUTIONS

1149

We integrate in the x-direction first, so that it is not necessary to set up two integrals. The right-side of the circle x2 + y 2 = 1 is given by p x = 1 − y2 . The left side of the circle (x − 1)2 + y 2 = 1 is given by x=1−

Thus the area of overlap is given by Area =

Z

√

3/2

√ − 3/2

p

1 − y2 .

Z √1−y2 1−

dxdy.

√

1−y 2

(b) In polar coordinates, the circle centered at the origin has equation r = 1. See Figure 16.78. The other circle, (x − 1)2 + y 2 = 1, can be written as x2 − 2x + 1 + y 2 = 1

x2 + y 2 = 2x,

so its equation in polar coordinates is

r2 = 2r cos θ,

and, since r 6= 0,

r = 2 cos θ. √ √ At the top point of intersection of the two circles, x = 1/2, y = 3/2, so tan θ = 3, giving θ = π/3. Figure 16.78 shows that if we integrate with respect to r first, we have to write the integral as the sum of two integrals. Thus, we integrate with respect to θ first. To do this, we rewrite r r = 2 cos θ as θ = arccos . 2 This gives the top half of the circle; the bottom half is given by r θ = − arccos . 2 Thus the area is given by Area =

Z

1 0

Z

arccos(r/2)

r dθdr. − arccos(r/2)

Solutions for Section 16.5 Exercises 1. (a) (b) (c) (d) (e) (f) 2.

A vertical plane perpendicular to the x-axis: x = 2. A cylinder: r =√3. A sphere: ρ = 3. A cone: φ = π/4. A horizontal plane: z = −5. A vertical half-plane: θ = π/4.

Z

f dV = W

= =

Z

Z

Z

1 −1 1

−1 1

−1

Z

Z

3π/4 π/4 3π/4

Z

4

(r2 + z 2 ) rdr dθ dz 0

(64 + 8z 2 ) dθ dz π/4

π (64 + 8z 2 ) dz 2

= 64π +

8 200 π= π 3 3

1150 3.


Z

f dV = W

=

Z

Z

3 −1 3

−1

1 =− 2 = −π 4.

Z

Z

Z

Z

Z

2π 0 2π 0 3

−1 3

−1

Z

1

(sin (r 2 )) rdr dθ dz 0

1

1 (− cos r 2 ) dθ dz 2 0

Z

2π

(cos 1 − cos 0) dθ dz

0

(cos 1 − 1) dz = −4π(cos 1 − 1) = 4π(1 − cos 1)

Z

f dV = W

Z

=

Z

=

0

Z

f dV = W

= = =

Z

Z

0

0

0

Z Z Z Z

7 6

Z

2π 0

Z

2π 0 2π

π π/2

1 2 · ρ sin φ dφ dθ dρ ρ

π

ρ sin φ dφ dθ dρ π/2

ρ dθ dρ 0 5

ρ dρ = 25π 0

Z

Z

2π

7 = 6 =

Z

0

7 3

Z

5

2π

7 3

Z

5

= 2π 5.

Z

5

Z

π/4 0 π/4

0 2π

Z

0

Z

2π 0 2π 0 2π

π 4

Z

2

(sin φ)ρ2 sin φ dρ dφ dθ 1 2

ρ2 sin2 φ dρ dφ dθ 1

sin2 φ dφ dθ

0 π/4 0

1 − cos 2φ dφ dθ 2

π/4

1 (φ − sin 2φ) 2 0 (

0

dθ

1 π − ) dθ 4 2

7π(π − 2) π 1 7 = · 2π( − ) = 6 4 2 12 6. Using Cartesian coordinates, we get:

Z

7. Using cylindrical coordinates, we get:

8. Using cylindrical coordinates, we get:

Z Z

3 0

1 0

4 0

Z Z

Z

1 0

2π 0

π/2 0

Z Z

5

f dz dy dx 0

4

f · rdr dθ dz

0

Z

2 0

f · rdr dθ dz

16.5 SOLUTIONS

1151

9. Using spherical coordinates, we get:

Z

π 0

Z

Z

π 0

3

f · ρ2 sin φ dρ dφ dθ

2

10. Using spherical coordinates, we get:

Z

2π 0

Z

π/6 0

Z

3

f · ρ2 sin φ dρ dφ dθ

0

11. We use Cartesian coordinates, oriented as shown in Figure 16.79. The slanted top has equation z = mx, where m is the slope in the x-direction, so m = 1/5. Then if f is an arbitrary funtion, the triple integral is

Z

Other answers are possible.

5 0

Z

2 0

Z

x/5

f dzdydx. 0

z (5, 0, 1)

1

y

2

?

(5, 2, 1) (5, 2, 0)

5

x

Figure 16.79 12. We choose cylindrical coordinates oriented as in Figure 16.80. The cone has equation z = r. Since we have a half cone scooped out of a half cylinder, θ varies between 0 and π. Thus, if f is an arbitrary function, the integral is

Other answers are possible.

Z

π 0

Z

2 0

Z

r

f r dzdrdθ. 0

z z=r 2

π/4 x 2

Figure 16.80 13.

y z

x

Figure 16.81 R is one eighth of a sphere of radius 1, below the xy-plane and under the first quadrant.

1152


14. (a) The region of integration is the region between the cone z = r, the xy-plane and the cylinder r = 3. In spherical coordinates, r = 3 becomes ρ sin φ = 3, so ρ = 3/ sin φ. The cone is φ = π/4 and the xy-plane is φ = π/2. See Figure 16.82. Thus, the integral becomes

Z

2π 0

Z

Z

π/2 π/4

3/ sin φ

ρ2 sin φ dρdφdθ. 0

z z=r

r=3

π 4

x 3

Figure 16.82: Region of integration is between the cone and the xy-plane (b) The original integral is easier to evaluate, so

Z

2π 0

Z

3 0

Z

r

r dzdrdθ = 0

Z

2π 0

Z

3

Z z=r zr drdθ = z=0

0

Problems

2π 0

Z

3 0

3

r3 = 18π. r drdθ = 2π · 3 0 2

√ 15. In spherical coordinates, the spherical cap is part of the surface ρ = 2. If α is the angle at the vertex of the cone, we have tan(α/2) = 2/2 = 1, so α/2 = π/4. Since the cone is below the xy-plane, the angle φ ranges from 3π/4 to π. Thus, the integral is given by √

Z

2π

0

Z

π

3π/4

Z

2

f (ρ, φ, θ)ρ2 sin φ dρ dφ dθ. 0

√ 16. In cylindrical coordinates, the spherical cap has equation z = − 2 − r 2 . If α is the angle at the vertex of the cone, we have tan(α/2) = 2/2 = 1, so α/2 = π/4. The cone has equation z = −r. Thus, the integral is Z Z Z √ 0

0

2−r 2

−

1

2π

g(r, θ, z)r dz dr dθ.

r

p

17. In rectangular coordinates, the spherical cap has equation z = − 2 − x2 − y 2 . If α is the angle at the vertex of the p cone, we have tan(α/2) = 2/2 = 1, so α/2 = π/4. The cone has equation z = − x2 + y 2 . Thus, the integral is Z √ Z √ Z √ 2−x2

2

√ − 2

−

√

2−x2

x2 +y 2

−

−

√

h(x, y, z) dz dy dx.

2−x2 −y 2

18. Orient the cone as shown in Figure 16.83 and use cylindrical coordinates with the origin at the vertex of the cone. Since√ the angle at the vertex of the cone is a right angle, the angles AOB and COB are both π/4. Thus, OB = 5 cos π/4 = 5/ 2. The curved surface of the cone has equation z = r, so Volume =

=

Z

2π 0

Z

2π 0

Z

5/

√

2

0

Z

√ 5/ 2 0

Z

√ 5/ 2

r dz dr dθ r

z=5/√2 Z 2π Z 5/√2 5 rz dr dθ = r √ − r dr dθ 2 0 0 z=r

16.5 SOLUTIONS

1153

√ 2π 5/ 2 5 52 r3 53 5 r2 √ √ = θ = 2π √ · 2 − − 3 2 2 2 2 2·3· 2 0 0 3 3

5 = 2π · √ 2 2

1 1 − 2 3

5 π = √ = 46.28 cm3 . 6 2 z

C

B

A

π/4 π/4

5

^

√ r 5/ 2

O

Figure 16.83 19. (a) In Cartesian coordinates, the bottom half of the sphere x2 + y 2 + z 2 = 1 is given by z = − Z Z Z Z √ 1−x2

1

0

dV =

W

−

0

0

dz dy dx.

√

1−x2 −y 2

p

1 − x2 − y 2 . Thus

√ (b) In cylindrical coordinates, the sphere is r 2 + z 2 = 1 and the bottom half is given by z = − 1 − r 2 . Thus

Z

dV = W

Z

Z

π/2

0

(c) In spherical coordinates, the sphere is ρ = 1. Thus,

Z

dV = W

Z

π/2 0

Z

1

0

π π/2

Z

0

−

Z

√

r dz dr dθ. 1−r 2

1

ρ2 sin φ dρ dφ dθ. 0

20. (a) Since the cone has a right angle at its vertex, it has equation

p

z=

x2 + y 2 .

The sphere has equation x2 + y 2 + z 2 = 1, so the top half is given by z= The cone and the sphere intersect in the circle

p

1 − x2 − y 2 .

x2 + y 2 = See Figure 16.84. Thus

Z

W

dV =

Z

√ 1/ 2 −1/

√ 2

1 , 2

1 z= √ . 2

Z √(1/2)−x2 Z √1−x2 −y2 −

√

(1/2)−x2

√

dz dy dx.

x2 +y 2

z 1

6

ICircle is

√1 2

π 4

x2 + y 2 =

?

1 2

y

I

x

x2 + y 2 =

Figure 16.84

1 2

1154


(b) In cylindrical coordinates, the cone has equation z = r and the sphere has equation z = Z Z Z √ Z √ 1−r 2

1/ 2

2π

dV =

W

√ 1 − r 2 . Thus

r dz dr dθ.

0

0

r

(c) In spherical coordinates, the cone has equation φ = π/4 and the sphere is ρ = 1. Thus

Z

dV = W

Z

2π 0

Z

π/4 0

Z

1

ρ2 sin φ dρ dφ dθ. 0

21. (a) Since the cone has a right angle at its vertex, it has equation

p

z = x2 + y 2 . √ Figure 16.85 shows the plane with equation z = 1/ 2. The plane and the cone intersect in the circle x2 + y 2 = 1/2. Thus, √ √

Z

dV = W

Z

1/ 2

√ −1/ 2

Z

(1/2)−x2

−

√

(1/2)−x2

Z

1

dz dy dx.

√

x2 +y 2

z √1 2 √1 2

π 4

1 y

x

Figure 16.85 (b) In cylindrical coordinates the cone has equation z = r, so

Z

dV = W

Z

2π 0

Z

1/ 0

√

2

Z

1

r dz dr dθ. r

√ √ (c) In spherical coordinates, the cone has equation φ = π/4 and the plane z = 1/ 2 has equation ρ cos φ = 1/ 2. Thus Z Z 2π Z π/4 Z 1/(√2 cos φ) dV = ρ2 sin φ dρ dφ dθ. W

0

0

0

22. The region is a solid cylinder of height 1, radius 1 with base on the xy-plane and axis on the z-axis. Use cylindrical coordinates: Z 1 Z 2π Z 1 Z 1 Z 1 Z √1−x2 1 1 dy dx dz = r dr dθ dz √ 2 + y 2 )1/2 r (x 2 0 0 0 −1 − 1−x 0 = =

Z

Z

1

0 1 0

Z

Z

2π

0

1

r dθ dz 0

2π

dθ dz = 2π.

0

16.5 SOLUTIONS

1155

23. The region of integration is half of a ball centered at the origin, radius 1, on the x ≥ 0 side. Since the integral is symmetric, we can integrate over the quarter unit ball (x ≥ 0, y ≥ 0) and multiply the result by 2. Use spherical coordinates: Z 1 Z √1−x2 Z √1−x2 −z2 1 dy dz dx √ √ (x2 + y 2 + z 2 )1/2 0 − 1−x2 − 1−x2 −z 2 =2 =2 =

Z

Z

Z

π/2

0 π/2 0

π/2

0

Z

Z

π

0 π 0

Z

1

0

1 2 ρ sin φ dρ dφ dθ ρ

1

ρ2 sin φ dφ dθ 2 0

π

(− cos φ) dθ 0

= (−(−1) − 0) · π = π. 24. Use cylindrical coordinates: when r 2 = x2 + y 2 = 1, then x2 + y 2 + z 2 = 1 + z 2 = 2 so z = ±1. The region W is shown in Figure 16.86. Z Z Z Z √ 2−z 2

2π

1

2

2

(x + y ) dV =

W

= =

Z

−1 1

−1

2π 4

π = 2

0

Z

Z

1

2π 0 1

r2 · r drdθdz

√2−z2 Z 1 Z 2π r4 1 (2 − z 2 )2 − 1 dθdz dθdz = 4 1 4 −1 0

3 − 4z 2 + z 4 dz

−1

z5 4 3z − z 3 + 3 5 z

1 = 28π . 15 −1 Cylinder

x2 + y 2 = 1

Sphere

1

x2 + y 2 + z 2 = 2

y x −1

Figure 16.86 25. (a) The angle φ takes on values in the range 0 ≤ φ ≤ π. Thus, sin φ is nonnegative everywhere in W 1 , and so its integral is positive. (b) The function φ is symmetric across the xy plane, such that for any point (x, y, z) in W 1 , with z 6= 0, the point (x, y, −z) has a cos φ value with the same magnitude but opposite sign of the cos φ value for (x, y, z). Furthermore, if z = 0, then (x, y, z) has a cos φ value of 0. Thus, with cos φ positive on the top half of the sphere and negative on the bottom half, the integral will cancel out and be equal to zero. 26. (a) The integral is negative. In W2 , we have 0 < z < 1. Thus, z 2 − z is negative throughout W2 and thus its integral is negative. (b) On the top half of the sphere, z is nonnegative, but x can be both positive and negative. Thus, since W 2 is symmetric with respect to the yz plane, the contribution of a point (x, y, z) will be canceled out by its reflection (−x, y, z). Thus, the integral is zero.

1156


27. The region whose volume we want is shown in Figure 16.87: z

6

5

2

?

π 6

θ=

θ=

π 3

y

x

Figure 16.87 Using cylindrical coordinates, the volume is given by the integral: V = =

Z

Z

2 0 2 0

Z

Z

25 2

=

Z

Z

π/3 π/6 π/3 π/6 2

0 2

Z

Z

5

r dr dθ dz 0

5

r2 dθ dz 2 0

π/3

dθ dz

π/6

25 π π = − dz 2 0 3 6 25 π 25π = · ·2= . 2 6 6 2 2 2 2 28. (a) In cylindrical coordinates, the cone √ √ is z = r and the sphere is r + z = 4. The surfaces intersect where z + z = 2 2z = 4. So z = 2 and r = 2. Z Z √ Z √ 2π

4−r 2

2

Volume =

r dzdrdθ.

0

0

r

(b) In spherical coordinates, the cone is φ = π/4 and the sphere is ρ = 2. Volume =

29. (a) (b)

Z

Z

2π 0 2π 0

Z

Z

π 0 2 0

Z

Z

Z

2π 0

Z

π/4

0

Z

2

ρ2 sin φ dρdφdθ. 0

2

ρ2 sin φ dρdφdθ. 1√

−

4−r 2

√

4−r 2

r dzdrdθ −

Z

2π 0

Z

1 0

Z √1−r2 −

√

r dzdrdθ.

1−r 2

30. Orient the region as shown in Figure 16.88 and use cylindrical coordinates with the origin at the center of the sphere. The equation of the sphere is x2 + y 2 + z 2 = 25, or r 2 + z 2 = 25. If z = 3, then r 2 + 32 = 25, so r 2 = 16 and r = 4. √ √ Volume =

Z

2π

0

Z

4

0

Z

25−r 2

r dzdrdθ =

3

Z

2π

0

Z

4

0

z= rz

z=3

25−r 2

drdθ

16.5 SOLUTIONS

=

Z

2π 0

= 2π

Z

4

(r 0

p

1157

2π 4 2 3/2 2 (25 − r ) 3r − − 25 − r 2 − 3r) drdθ = θ 3 2 0 0

125 27 − 3 3

− 24 =

52π = 54.45 cm3 . 3

z

Sphere

5 2

r 2 + z 2 = 25

6

? 6 5

3

?

4

r

-

Figure 16.88 31. Orient the region as shown in Figure 16.89 and use spherical coordinates with the origin at the center of the sphere. The equation of the sphere is x2 + y 2 + z 2 = 25, or ρ = 5. The plane z = 3 is the plane ρ cos φ = 3, so ρ = 3/ cos φ. In Figure 16.89, angle AOB is given by cos φ =

3 , 5

so

φ = arccos(3/5).

The volume is given by

2π Z p=5 arccos(3/5) 3 ρ sin φ ρ2 sin φ dρdφdθ = θ dφ V = 3 0 3/ cos φ 0 0 0 p=3/ cos φ Z arccos(3/5) Z

2π

Z

arccos(3/5)

0

Z

5

9 125 − 3 cos3 φ

= 2π

= 2π

Z

arccos(3/5)

0

125 sin φ dφ − 3

sin φ dφ

Z

arccos(3/5)

0

9 sin φ dφ cos3 φ

arccos(3/5) arccos(3/5) 125 1 cos φ φ = 2π − −9 3 2 cos2 0 0

1 125 3 9 −1 −1 − 3 5 2 (3/5)2 2 9 16 125 − − = 2π − 3 5 2 9 52π 50 −8 = ≈ 54.45 cm3 . = 2π 3 3 −

z 2

3

Sphere

ρ=5

6

?A 6 φ

= 2π

B

5

? O

Figure 16.89

x

!

1158


32. The density function can be rewritten as δ(ρ, φ, θ) = ρ. So the mass is

Z

δ(P ) dV = W

=

Z

Z

2π 0 2π

Z

Z

π/4 0 π/4

3 0

ρ · ρ2 sin φ dρ dφ dθ

81 sin φ dφ dθ 4 0 Z 2π √ 2 + 1) dθ (− 2 0 √ √ 2 81 · 2π · (− + 1) = π(− 2 + 2) 2 4

0

81 = 4 81 4

=

Z

33. Using spherical coordinates: M = = =

Z

Z

π 0 π 0

27 4

Z

Z

Z

2π 0 2π 0 π

0

Z

Z

3

(3 − ρ)ρ2 sin φ dρ dθ dφ

0

ρ4 ρ − 4 3

2π

3

sin φ dθ dφ

0

sin φ dθ dφ 0

π

27 27 = · 2π · (− cos φ) = π · [−(−1) + 1] = 27π. 4 2 0

34. (a) We use spherical coordinates. Since δ = 9 where ρ = 6 and δ = 11 where ρ = 7, the density increases at 2 gm/cm 3 for each cm increase in radius. Thus, since density is a linear function of radius, the slope of the linear function is 2. Its equation is δ − 11 = 2(ρ − 7) so δ = 2ρ − 3. (b) Thus,

Mass = (c) Evaluating the integral, we have

π Mass = 2π − cos φ

0

Z

2π 0

Z

π 0

Z

7 6

(2ρ − 3)ρ2 sin φ dρ dφ dθ.

7 !

2ρ4 3ρ3 − 4 3 6

35. The distance from a point (x, y, z) to the origin is given by

R p R

= 2π · 2(425.5) = 1702π gm = 5346.991 gm.

p

x2 + y 2 + z 2 . Thus we want to evaluate

x2 + y 2 + z 2 dV

where R is the region bounded by the hemisphere z = We will use spherical coordinates.

Vol(R)

p

8 − x2 − y 2 and the cone z =

p

x2 + y 2 . See Figure 16.90.

16.5 SOLUTIONS

1159

z

2

y

x2 + y 2 = 4

x

Figure 16.90 √ In spherical coordinates, the quantity ρ goes from 0 to 8, and θ goes from 0 to 2π, and φ goes from 0 to π/4 (because the angle of the cone is π/4). Thus we have

Z p

x2 + y 2 + z 2 dV =

R

= =

Z

Z

Z

2π 0 2π 0 2π 0

Z

Z

Z

π/4 0 π/4 0 π/4

Z

√

8

ρ(ρ2 sin φ) dρdφdθ 0

√8

ρ4 sin φ · dφdθ 4 0 16 sin φ dφdθ

0

π/4 16(− cos φ) dθ = 0 0 Z 2π √ Z

2π

=

0

16 1 −

2 2

dθ

√ 2 π 2 √ From Problem 20 of Section 16.4 we know that Vol(R) = 32π( 2 − 1)/3, therefore = 32 1 −

Average distance =

R p R

x2 + y 2 + z 2 dV Vol(R)

√

32 1 − 22 π 3 √ = = √ . [32( 2 − 1)π/3] 2 36. (a) First we must choose a coordinate system, since none is given. We pick the xy-plane to be the fixed plane and the z-axis to be the line perpendicular to the plane. Then the distance from a point to the plane is |z|, so the density at a point is given by Density = ρ = k|z|. Using cylindrical coordinates for the integral, we find Z Z Z √ a2 −r 2

a

2π

Mass =

0

0

−

√

a2 −r 2

k|z|r dzdrdθ.

1160


(b) By symmetry, we can evaluate this integral over the top half of the sphere, where |z| = z. Then √ Z 2π Z a Z √a2 −r2 Z 2π Z a 2 z= a2 −r2 z Mass = 2 r kzr dzdrdθ = 2k drdθ 2 z=0 0 0 0 0 0 =k

Z

2π

0

= 2πk

Z

a

2

2

r(a − r ) drdθ = k2π

0

a4 a4 − 2 4

πka4 . = 2

r2 2 r4 a − 2 4

a 0

37. (a) We use the axes shown in Figure 16.91. Then the sphere is given by r 2 + z 2 = 25, so Z Z Z √ 2π

25−r 2

5

Volume =

0

−

1

r dzdrdθ.

√

25−r 2

(b) Evaluating gives Volume = 2π

Z

5 1

z=√25−r2 Z 5 p rz 2r 25 − r 2 dr dr = 2π √ 1 z=− 25−r 2 5 2 = 2π − (25 − r 2 )3/2

3 1 √ 4π 3/2 (24) = 64 6π = 492.5 mm3 . = 3

r 2 + z 2 = 25 x2 + y 2 + z 2 = 25

y

z

5

1

5

x

a

b

y

x

Figure 16.92

Figure 16.91

38. We must first decide on coordinates. We pick spherical coordinates with the common center of the two spheres as the origin. We imagine the half-melon with the flat side horizontal and the positive z-axis going through the curved surface. See Figure 16.92. The volume is given by the integral Volume =

Z

2π 0

Z

π/2 0

Z

b

ρ2 sin φ dρdφdθ. a

Evaluating gives Volume =

Z

2π 0

Z

π/2 0

p=b

ρ3 π/2 dφdθ = 2π(− cos φ) sin φ 3 p=a 0

b3 a3 − 3 3

=

2π 3 (b − a3 ). 3

To check our answer, notice that the volume is the difference between the volumes of two half spheres of radius a and b. These half spheres have volumes 2πb3 /3 and 2πa3 /3, respectively.

16.5 SOLUTIONS

1161

39. The total volume of the cone is 31 πr2 h = 13 π · 12 · 1 = 31 π, so the total mass is 13 π (since the density is always 1). The center of mass z-coordinate is given by Z 3 z¯ = z dV π C Using cylindrical coordinates to evaluate this integral gives z¯ = =

Z

3 π

Z

3 π

Z

3 = π 40. (a) The mass m of the cone is given by

R

C

2π 0 2π 0 2π

Z Z

1 0 1

Z

z

zr dr dz dθ 0

z3 dz dθ 2

0

3 1 dθ = 8 4

0

δ dV . In cylindrical coordinates this is

Z

m=

Z

=

Z

= (b) The center of mass z-coordinate is given by

2π 0 2π 0 2π

Z

Z

1 0 1

Z

0

z

z 2 r dr dz dθ 0

z4 dz dθ 2

1 π dθ = 10 5

0

5 z¯ = π

Z

C

z · z 2 dV

Using cylindrical coordinates to evaluate this integral gives 5 π

z¯ =

5 = π 5 = π

Z Z Z

Z

2π 0

Z

2π 0 2π

1 0 1 0

Z

z

z 3 r dr dz dθ 0

z5 dz dθ 2

1 5 dθ = 12 6

0

Comparing this answer with the center of mass in Problem 39, where the density was constant, it makes sense that the center of mass would be higher in this problem, since more mass is concentrated near the top of the cone. 41. We first need to find the mass of the solid, using cylindrical coordinates:

m= = =

Z

Z

Z

2π 0 2π 0 2π

Z

Z

0

1 0

Z √z/a

r dr dz dθ

0

1 0

z dz dθ 2a

π 1 dθ = 4a 2a

It makes sense that the mass would vary inversely with a, since increasing a makes the paraboloid skinnier. Now for the z-coordinate of the center of mass, again using cylindrical coordinates: 2a z¯ = π =

2a π

2a = π

Z Z Z

2π 0 2π 0 2π 0

Z Z

1 0

zr dr dz dθ

0

1 0

Z √z/a

z2 dz dθ 2a

1 2 dθ = 6a 3

1162


42. The volume of the hemisphere is 23 πa3 so its mass is 23 πa3 b. To find the location of the center of mass, we place the base of the hemisphere on the xy-plane with the origin at its center, so we can describe it in spherical coordinates by 0 ≤ ρ ≤ a, 0 ≤ φ ≤ π2 and 0 ≤ θ ≤ 2π. Then the x-coordinate of the center of mass is, integrating using spherical coordinates: x ¯= since the first integral

R 2π 0

3 2πa3 b

Z

Z

a 0

Z

π 2

0

2π

ρ sin(φ) cos(θ) · ρ2 sin(φ) dθ dφ dρ = 0

0

cos(θ) dθ is zero. A similar computation shows that y¯ = 0. Now for the z-coordinate:

Z

3 z¯ = 2πa3 b

a

0

3 = · 2π 2πa3 b = =

Z

3 a3 b

a

π 2

0

Z

a 0

Z

Z

2π

ρ cos(φ) · ρ2 sin(φ) dθ dφ dρ

0 π 2

0

ρ3 cos(φ) sin(φ) dφ dρ

π

sin2 (φ) 2 dρ 2 0

ρ3

Z0

3 2a3 b

Z

a

ρ3 dρ = 0

3a 8b

So the x and y-coordinates are located at the center of the base, while the z-coordinate is located the base.

3a 8b

above the center of

43. The sum of the three moments of inertia I for the ball B will be Z Z Z 3 3 3 2 2 2 2 3I = (y + z ) dV + (x + z ) dV + (x2 + y 2 ) dV 4πa3 B 4πa3 B 4πa3 B =

3 4πa3

Z

(2x2 + 2y 2 + 2z 2 ) dV, B

which, in spherical coordinates is 3 2πa3

Z

(x2 + y 2 + z 2 ) dV = B

3 2πa3

3 = 3 a =

6 a3

Thus 3I = 65 a2 , so I = 25 a2 .

Z

Z

a

Z0 a

a 0

Z

Z

π 0

π

Z

2π 0

ρ2 · ρ2 sin(φ) dθ dφ dρ

4

ρ sin(φ) dφ dρ 0

ρ4 dρ =

0

6 2 a . 5

44. First we need to find the volume of the cone. In spherical coordinates we find: V =

Z

a 0

Z

π 3

0

Z

2π

ρ2 sin(φ) dθ dφ dρ = 0

πa3 3

Now, to find the moment of inertia about the z-axis we need to compute the integral this in spherical coordinates as 3 πa3

Z

x2 + y 2 dV = W

3 πa3

3 = πa3 6 = 3 a

Z

Z Z

a 0 a

0 a

0

6 5 = 3 a 24

Z

Z

Z Z

π 3

0 π 3

0 π 3

Z Z

R

W

x2 + y 2 dV . We can do

2π

(ρ2 sin2 (φ) cos2 (θ) + ρ2 sin2 (φ) sin2 (θ)) · ρ2 sin(φ) dθ dφ dρ

0 2π

ρ4 sin3 (φ) dθ dφ dρ 0

4

ρ sin3 (φ) dφ dρ

0 a

ρ4 dρ = 0

3 πa3

a2 . 4

16.5 SOLUTIONS

1163

45. Assume the base of the cylinder sits on the xy-plane with center at the origin. Because the cylinder is symmetric about the z-axis, the force in the horizontal x or y direction is 0. Thus we need only compute the vertical z component of the force. We are going to use cylindrical coordinates; since the force is G · mass/(distance) 2 , a piece of the cylinder of volume dV located at (r, θ, z) exerts on the unit mass a force with magnitude G(δ dV )/(r 2 + z 2 ). See Figure 16.93. Vertical component G(δ dV ) Gδ dV z Gδz dV · cos φ = 2 ·√ . = 2 = 2 r + z2 r + z2 (r + z 2 )3/2 of force r2 + z 2 Adding up all the contributions of all the dV ’s, we obtain Vertical force = = = =

Z

Z

Z

Z

H 0 H 0 H 0 H

Z

Z

Z

Z

2π 0 2π 0

R

Gδzr drdθdz (r2 + z 2 )3/2

0

1 (Gδz) − √ 2 r + z2

2π 0

(Gδz) ·

2πGδ 0

= 2πGδ(z −

√

−√

1 1 + z R2 + z 2

z 1− √ R2 + z 2

p

= 2πGδ(H −

R2

+

z2)

R dθdz 0

H

dθdz

dz

0

p

R2

+

H2

+ R) = 2πGδ(H + R −

p

R2 + H 2 )

r2 + z 2 z

φ

Figure 16.93 46. In the system used in this book the volume element is dV = ρ 2 sin φ dρ dφ dθ. In the system shown in the problem, φ and θ have been interchanged and ρ changed to r. So the volume element is dV = r 2 sin θ dr dθ dφ. 47. The charge density is δ = kz, where k is a constant. In cylindrical coordinates, Total charge =

Z

δ dV = Cylinder

= kπ

Z

Z hZ 0

h

R 0

Z

R2 z dz = k(πR2 ) 0

2π

kzr dθ dr dz = k 0

Z hZ 0

kπ 2 2 h2 = R h . 2 2

R

2πzr dr dz 0

Thus, the total charge is proportional to R2 h2 with constant of proportionality kπ/2. 48. The charge density is δ = k/ρ. Integrating in spherical coordinates, Total charge =

Z

2π 0

Z πZ 0

R 0

k 2 ρ sin φ dρ dφ dθ = k ρ

R2 = 2πkR2 . = 4πk 2

Z

2π 0

Z

Thus, the total charge is proportional to R2 with constant of proportionality 2πk.

π 0

R2 sin φ dφ dθ 2

1164


49. Using spherical coordinates, Stored energy =

1 2

Z bZ π Z a

q2 = 8π

Z

0 b

2π 0

1 q2 dρ = 2 ρ 8π

a

Z bZ π Z

q2 32π 2

E 2 ρ2 sin φ dθ dφ dρ =

a

1 1 − . a b

0

2π

1 sin φ dθ dφ dρ ρ2

0

50. Use cylindrical coordinates, with the z-axis being the axis of the cable. Consider a piece of cable of length 1. Then 1 Stored energy = 2

Z bZ 1Z 0 b

a

q2 = 4π

Z

a

2π

q2 E r dθ dz dr = 8π 2 2

0

Z bZ 1Z a

0

2π 0

1 dθ dz dr r

1 q2 q2 b dr = (ln b − ln a) = ln . r 4π 4π a

So the stored energy is proportional to ln(b/a) with constant of proportionality q 2 /4π.

Solutions for Section 16.6 Exercises 1. We have p(x, y) = 0 for all points (x, y) satisfying x ≥ 3, since all such points lie outside the region R. Therefore the fraction of the population satisfying x ≥ 3 is 0.

2. The fraction is 0, since

R1 1

xy dx = 0, so

R∞ R1 −∞

p(x, y) dx dy =

1

R1R1 0

1

xy dx dy=0.

3. Since x + y ≤ 3 for all points (x, y) in the region R, the fraction of the population satisfying x + y ≤ 3 is 1.

4. Since p(x, y) = 0 for any (x, y) with x < 0 and also p(x, y) = 0 for any (x, y) with y > 1 or y < 0, the fraction of the population is given by the double integral:

Z

1 0

Z

1

xy dx dy = 0

Z

1 0

1

x2 y dy = 2 0

Z

1 0

1

y 1 y 2 = . dy = 2 4 0 4

5. Since p(x, y) = 0 for all (x, y) outside the rectangle R, the population is given by the volume under the graph of p over the region inside the rectangle R and to the right of the line x = y. Therefore the fraction of the population is given by the double integral:

Z

1 0

Z

2

xy dx dy = y

Z

1 0

2

x2 y dy = 2 y

Z

1 0

2y −

y3 2

dy =

y2 −

y4 8

1 = 7. 8 0

6. Since p(x, y) = 0 for all (x, y) outside the rectangle R, the population is given by the volume under the graph of p over the region inside the rectangle R and below the line x + y = 1. This is the same as the region bounded by the x-axis, the y-axis, and the line x + y = 1. Therefore the fraction of the population is given by the double integral:

Z

1 0

Z

1−y

xy dx dy = 0

Z

1 0

1−y

x2 y 2 0

dy =

Z

1 0

(1 − y)2 y dy = 2

y2 y3 y4 − + 4 3 8

7. The fraction of the population is given by the double integral:

Z

1/2 0

Z

1

xy dx dy = 0

Z

1/2 0

1

x2 y dy = 2 0

Z

1/2 0

1/2

y y 2 dy = 2 4 0

=

1 = 1. 24 0

1 . 16

16.6 SOLUTIONS

1165

8. We are looking for points inside the circle x2 + y 2 = 1 and inside the rectangle R. In the first quadrant, all of the circle and its interior lies inside the rectangle R. Thus the fraction of the population we want is given by the volume under the graph of p over the region inside the circle x2 + y 2 = 1 in the first quadrant. We evaluate this double integral using polar coordinates:

Z

π/2 0

Z

1

(r cos θ)(r sin θ) r dr dθ = 0

Z

π/2 0

Making the substitution w = sin θ, we get:

Z

1

r4 1 cos θ sin θ dθ = 4 4 0

π/2

cos θ sin θ dθ = 0

Z

1

w dw = 0

Z

π/2

cos θ sin θ dθ. 0

1 . 2

Thus the fraction is (1/4)(1/2) = 1/8. 9. (a) (b) (c) (d)

The entries in this table are non-negative and their sum is 1. Add up the values along the row x = 2: 0.2 + 0.1 + 0 = 0.3. Add up the columns with y = 1 and y = 2: 0.3 + 0.2 + 0.1 + 0 + 0.2 + 0.1 + 0 + 0 = 0.9. Add up the values in the grid corresponding x = 1, 2, 3 and y = 1, 2: 0.3 + 0.2 + 0.2 + 0.1 + 0.1 + 0 + 0 + 0 = 0.9.

10. No, p is not a joint density function. Since p(x, y) = 0 outside the region R, the volume under the graph of p is the same as the volume under the graph of p over the region R, which is 2 not 1. 11. Yes, p is a joint density function. The values of p(x, y) are nonnegative, since p(x, y) = 1/2 for all points inside R and p(x, y) = 0 for all other points. The volume under the graph of p over the region R is (1/2)(5 − 4)(0 − (−2)) = 1.

12. No, p is a not joint density function, because p(x, y) < 0 for some points (x, y) in the region R. For example, p(−0.7, 0.1) = −0.6.

13. Yes, p is a joint density function. Since x ≤ y everywhere in the region R, we have p(x, y) = 6(y − x) ≥ 0 for all x and y in R, and p(x, y) = 0 for all other (x, y). To check that p is a joint density function, we check that the total volume under the graph of p over the region R is 1:

Z

p(x, y) dA = R

Z

1 0

Z

y

6(y − x) dx dy =

0

Z

1 0

6 yx −

x2 2

y Z dy = 0

1 0

1

3y 2 dx = y 3 = 1. 0

14. Yes, p is a joint density function. In the region R we have 1 ≥ x 2 + y 2 , so p(x, y) = (2/π)(1 − x2 − y 2 ) ≥ 0 for all x and y in R, and p(x, y) = 0 for all other (x, y). To check that p is a joint density function, we check that the total volume under the graph of p over the region R is 1. Using polar coordinates, we get:

Z

2 p(x, y)dA = π R

Z

2π 0

Z

1

2 (1 − r )r dr dθ = π 2

0

Z

2π 0

r2 r4 − 2 4

1 Z dθ = 2 π 0

2π 0

1 dθ = 1. 4

15. Yes, p is a joint density function. Since e−x−y is always positive, p(x, y) = xye−x−y ≥ 0 for all x and y in R, and hence for all x and y. To check that p is a joint density function, we check that the total volume under the graph of p over the region R is 1. Since e−x−y = e−x e−y , we have

Z

xye−x−y dA = R

Z

∞ 0

Z

∞

xye−x−y dx dy =

0

Z

∞

ye−y

0

Z

∞

xe−x dx

0

Using integration by parts:

Z Thus

Z

∞ 0

xye R

b

xe−x dx = lim (−xe−x − e−x ) = (0 − 0) − (0 − 1) = 1. b→∞

−x−y

dA =

Z

∞

0

ye

0

−y

Z

∞

0

xe

−x

dx

dy =

Z

∞

0

ye−y dy = 1.

dy.

1166


16. (a)

Z

1 0

Z

1 1/3

2 (x + 2y) dx dy = 3 =

Z

Z

1 2 1 2 ( x + 2xy) 1/3 dy 3 2

1 0 1

h

0

Z

1

4 4 2 ( + y) dy 3 0 9 3 2 4 2 2 1 = y+ y 0 3 9 3 20 2 10 = . = 3 9 27 =

i

1 2 2 1 ( + 2y) − ( + y) dy 3 2 18 3

(b) It is easier to calculate the probability that x < (1/3) + y does not happen, that is, the probability that x ≥ (1/3) + y, and subtract it from 1. The probability that x ≥ (1/3) + y is

Z

1

1/3

Z

x−(1/3)

0

2 (x + 2y) dy dx = 3

Z

=

2 3

1

1/3

2 = 3

Z Z

x−(1/3) 2 (xy + y 2 ) 0 dx 3

1

1/3

(x(x −

1 1 ) + (x − )2 ) dx 3 3

1 1/3

(2x2 − x +

1 ) dx 9

2 2 1 1 1 = ( x3 − x2 + x) 1/3 3 3 2 9 i h 1 1 2 1 1 2 2 − + ) ( − + )−( = 3 3 2 9 81 18 27 = 44/243. Thus, the probability that x < (1/3) + y is 1 − (44/243) = 199/243.

x
g, we have

X

f (xi , yj , zk )∆V, where (xi , yj , zk ) is a point inside the ijk-th subbox of volume ∆V,

i,j,k

lim

X

f (xi , yj , zk )∆V > lim

i,j,k

X

g(xi , yj , zk )∆V =

i,j,k

Z

g dV. W

29. False. As a counterexample, let W1 be the solid cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, and let W2 be the solid cube 1 1 1 1 − 1 ) = 1 and volume(W2 ) = 8 . Now if f (x, y, z) = −1, then R 2 ≤ x ≤ 0, − 2 ≤ y ≤ 0, − 2 ≤ z ≤R 0. Then volume(W 1 f dV = 1 · −1 which is less than W f dV = 8 · −1. W 1

2

30. True. If W is the solid region lying under theRgraph of f and above the region RRin the xy-plane, we can compute the volume of W either using the double integral R f dA, or using the triple integral W 1 dV.

PROJECTS FOR CHAPTER SIXTEEN

1. (a) We are integrating over the whole plane, so converting to polar coordinates gives ∞ Z 2π Z 2π Z ∞ Z 2π Z ∞Z ∞ 1 −r2 1 −r 2 −(x2 +y 2 ) − e dθ = e rdrdθ = e dxdy = dθ = π. 2 2 0 0 0 0 −∞ −∞ 0 (b) Rewriting the integrand as a product gives Z Z ∞Z ∞ 2 2 e−(x +y ) dxdy = −∞

−∞

∞ −∞

Z

∞

2

2

e−x e−y dxdy.

−∞

2

Now e−y is a constant as far as the integral with respect to x is concerned, so Z ∞ Z ∞Z ∞ Z ∞ −x2 −y 2 −y 2 −x2 e e dxdy = e e dx dy. −∞

−∞

−∞

−∞

1196


We assume that the integral with respect to x converges, and so is a constant as far as the integral with respect to y is concerned. Thus, we have Z ∞ Z ∞ Z ∞ Z ∞ −y 2 −x2 −x2 −y 2 e dy . e dx e dx dy = e But

R∞

−∞

e−x dx and Z

∞ −∞

Z

∞

e

R∞

−∞

−∞

−∞

−∞

−∞

2

2

e−y dy are the same number, so we can write

−(x2 +y 2 )

dxdy =

−∞

Z

∞

e

−x2

dx

−∞

(c) Using the results of parts (a) and (b), we have 2 Z Z ∞ 2 e−x dx = −∞

∞ −∞

Z

Z ∞

∞

e

−y 2

−∞

e−(x

2

+y 2 )

dy

=

Z

∞

e

−x2

dx

−∞

2

.

dxdy = π.

−∞

Taking square roots and observing that the integral we are looking for is positive, we have Z ∞ √ 2 e−x dx = π. −∞

2. (a) We want to find the average value of |x − y| over the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1: Z 1Z 1 |x − y| dy dx. Average distance between gates = 0

0

Let’s fix x, with 0 ≤ x ≤ 1. Then |x − y| = Z

1 0

|x − y| dy =

Z

x 0

(x − y) dy +

y − x for y ≥ x . Therefore x − y for y ≤ x

Z

1 x

(y − x) dy

x 2 1 2 2 y = x2 − x + 1 − x − x + x2 = + − xy 2 2 2 2 0 x 1 = x2 − x + . 2

y2 xy − 2

So, Average distance between gates =

Z

1

Z

1

|x − y| dy dx Z 1 Z 1 Z 1 1 (x2 − x + ) dx |x − y| dy dx = = 2 0 0 0 1 3 2 1 x x 1 = − + x = . 3 2 2 3 0

0

0

(b) There are (n + 1)2 possible pairs (i, j) of gates, i = 0, . . . , n, j = 0, . . . , n, so the sum given represents the average distances apart of all such gates. The Riemann sum with ∆x = ∆y = 1/n, if we choose the least x and y-values in each subdivision is n−1 X X n−1 i=0

i − j 1 , n n n2 j=0

which for large n is just about the same as the other sum. For n = 5 the sum is about 0.389; for n = 10 the sum is about 0.364.