Solutions To Selected Problems Chapter 5 Chapter 7

Chapter 6

Exercises 6.1

1.

T
3.

Answers will vary.
5.

$4\pi+\pi^{2}\approx 22.436$
7.

$\pi$
9.

$1/2$
11.

$4.5$
13.

$2-\pi/2$
15.

$1/6$
17.

$\frac{9}{8}$
19.

$\frac{64}{3}$
21.

$27/2$
23.

$9/2$
25.

All enclosed regions have the same area, with regions being the reflection of adjacent regions. One region is formed on $[\pi/4,5\pi/4]$ , with area $2\sqrt{2}$ .
27.

$1$
29.

$9/2$
31.

$1/12(9-2\sqrt{2})\approx 0.514$
33.

1
35.

4

Exercises 6.2

1.

T
3.

Recall that “ $\operatorname{d}\!x$ ” does not just “sit there;” it is multiplied by $A(x)$ and represents the thickness of a small slice of the solid. Therefore $\operatorname{d}\!x$ has units of in, giving $A(x)\operatorname{d}\!x$ the units of in ${}^{3}$ .
5.

$48\pi\sqrt{3}/5$ units ${}^{3}$
7.

$\pi/6$ units ${}^{3}$
9.

$9\pi/2$ units ${}^{3}$
11.

$2\pi/15$ units ${}^{3}$
13.
(a) $\pi/2$ (b) $5\pi/6$ (c) $4\pi/5$ (d) $8\pi/15$
15.
(a) $4\pi/3$ (b) $2\pi/3$ (c) $4\pi/3$ (d) $\pi/3$
17.
(a) $8\pi$ (b) $8\pi$ (c) $16\pi/3$ (d) $8\pi/3$
19.

Placing the tip of the cone at the origin such that the $x$ -axis runs through the center of the circular base, we have $A(x)=\pi x^{2}/4$ . Thus the volume is $250\pi/3$ units ${}^{3}$ .
21.

Orient the cone such that the tip is at the origin and the $x$ -axis is perpendicular to the base. The cross-sections of this cone are right, isosceles triangles with side length $2x/5$ ; thus the cross-sectional areas are $A(x)=2x^{2}/25$ , giving a volume of $80/3$ units ${}^{3}$ .

Exercises 6.3

1.

T
3.

F
5.

$9\pi/2$ units ${}^{3}$
7.

$\dfrac{96\pi}{5}$
9.

$48\pi\sqrt{3}/5$ units ${}^{3}$
11.

$\dfrac{768\pi}{7}$
13.
(a) $4\pi/5$ (b) $8\pi/15$ (c) $\pi/2$ (d) $5\pi/6$
15.
(a) $4\pi/3$ (b) $\pi/3$ (c) $4\pi/3$ (d) $2\pi/3$
17.
(a) $16\pi/3$ (b) $8\pi/3$ (c) $8\pi$ (d) $8\pi$
19.
(a) Disk: $\pi\int_{0}^{1}\left[1^{2}-(\sqrt[4]{y})^{2}\right]\operatorname{d}\!y=\frac{% \pi}{3}$ Shell: $2\pi\int_{0}^{1}x\cdot x^{4}\operatorname{d}\!x=\frac{\pi}{3}$ (b) Disk: $\pi\int_{0}^{1}(x^{4})^{2}\operatorname{d}\!x=\frac{\pi}{9}$ Shell: $2\pi\int_{0}^{1}y(1-\sqrt[4]{y})\operatorname{d}\!y=\frac{\pi}{9}$ .
21.
(a) Disk: $\pi\int_{-2}^{1}\left[(-4x+8)^{2}-(4x^{2})^{2}\right]\operatorname{d}\!x=\frac% {1152\pi}{5}$ Shell: $2\pi\int_{0}^{4}y\sqrt{y}\operatorname{d}\!y+2\pi\int_{4}^{16}y\left[\left(2-% \frac{y}{4}\right)+\frac{\sqrt{y}}{2}\right]\operatorname{d}\!y=\frac{128\pi}{% 5}+\frac{1024\pi}{5}$ (b) Disk: $\pi\int_{0}^{4}\left[1+\frac{\sqrt{y}}{2}\right]^{2}-\left[1-\frac{\sqrt{y}}{2% }\right]^{2}\operatorname{d}\!y+\pi\int_{4}^{16}\left[1+\frac{\sqrt{y}}{2}% \right]^{2}-\left[1-\left(2-\frac{y}{4}\right)\right]^{2}\operatorname{d}\!y=% \frac{32\pi}{3}+\frac{130\pi}{3}$ Shell: $2\pi\int_{-2}^{1}(1-x)\left[(-4x+8)-4x^{2}\right]\operatorname{d}\!x=54\pi$ (c) Disk: $\pi\int_{-2}^{1}\left[(16-4x^{2})^{2}-(16-(-4x+8))^{2}\right]\operatorname{d}% \!x=\frac{1728\pi}{5}$ Shell: $2\pi\int_{0}^{4}(16-y)\left[\sqrt{y}\right]\operatorname{d}\!y+2\pi\int_{4}^{1% 6}(16-y)\left[\left(2-\frac{y}{4}\right)+\frac{\sqrt{y}}{2}\right]% \operatorname{d}\!y=\frac{2176\pi}{15}+\frac{3008\pi}{15}$ .

Exercises 6.4

1.

In SI units, it is one joule, i.e., one newton-meter, or kg $\cdot$ m/s ${}^{2}\cdot$ m. In Imperial Units, it is ft-lb.
3.

Smaller.
5.
(a) 500 ft-lb (b) $100-50\sqrt{2}\approx 29.29$ ft
7.
(a) $\frac{1}{2}\cdot d\cdot l^{2}$ ft-lb (b) 75 % (c) $\ell(1-\sqrt{2}/2)\approx 0.2929\ell$
9.
(a) 756 ft-lb (b) 60,000 ft-lb (c) Yes, for the cable accounts for about 1% of the total work.
11.

575 ft-lb
13.

0.05 J
15.

5/3 ft-lb
17.

$f\cdot d/2$ J
19.

5 ft-lb
21.
(a) 52,929.6 ft-lb (b) 18,525.3 ft-lb (c) When 3.83 ft of water have been pumped from the tank, leaving about 2.17 ft in the tank.
23.

212,135 ft-lb
25.

187,214 ft-lb
27.

4,917,150 J

Exercises 6.5

1.

Answers will vary.
3.

499.2 lb
5.

6739.2 lb
7.

3920.7 lb
9.

2496 lb
11.

602.59 lb
13.
(a) 2340 lb (b) 5625 lb
15.
(a) 1597.44 lb (b) 3840 lb
17.
(a) 56.42 lb (b) 135.62 lb
19.

5.1 ft