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.

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 $2\sqrt{2}$ .
23.

$1$
25.

$9/2$
27.

$1/12(9-2\sqrt{2})\approx 0.514$
29.

1
31.

4

1.

T
3.

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

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: $\pi\int_{1}^{2}(\sqrt[3]{y-1})^{2}\ dy=\frac{3\pi}{5}$ Shell: $2\pi\int_{0}^{1}x(2-(x^{3}+1))\ dx=\frac{3\pi}{5}$ (b) Disk: $\pi\int_{0}^{1}\left[2-(x^{3}+1)\right]^{2}\ dx=\frac{9\pi}{14}$ Shell: $2\pi\int_{1}^{2}(2-y)\sqrt[3]{y-1}\ dy=\frac{9\pi}{14}$ .

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