Chapter F

T

T

Answers will vary.

Answers may vary; one common answer is when the region has two or more “top” or “bottom” functions when viewing the region with respect to $x$, but has only 1 “top” function and 1 “bottom” function when viewed with respect to $y$. The former area requires multiple integrals to compute, whereas the latter area requires one.

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

$16/3$

$\pi$

$\pi$

$1/2$

$2\sqrt{2}$

$4.5$

$4/3$

$2-\pi /2$

$8$

$1/6$

$37/12$

$\frac{9}{8}$

$\frac{8}{3}$

$\frac{64}{3}$

$9$

$27/2$

$21$

$9/2$

$343/6$

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 $2\sqrt{2}$.

On regions such as $[\pi /6,5\pi /6]$, the area is $3\sqrt{3}/2$. On regions such as $[-\pi /2,\pi /6]$, the area is $3\sqrt{3}/4$.

$1$

$5/3$

$9/2$

$9/4$

$1/12\left(9-2\sqrt{2}\right)\approx 0.514$

$4/3$

1

5

4

21/2

T

Answers will vary.

Recall that “$\mathrm{d}x$” does not just “sit there;” it is multiplied by $A\left(x\right)$ and represents the thickness of a small slice of the solid. Therefore $\mathrm{d}x$ has units of in, giving $A\left(x\right)\mathrm{d}x$ the units of in³.

volume

$48\pi \sqrt{3}/5$ units³

$175\pi /3$ units³

$\pi /6$ units³

${\displaystyle \frac{768\pi }{7}}$

$9\pi /2$ units³

$35\pi /3$ units³

$2\pi /15$ units³

${\displaystyle \frac{96\pi }{5}}$

Placing the tip of the cone at the origin such that the $x$-axis runs through the center of the circular base, we have $A\left(x\right)=\pi x^2/4$. Thus the volume is $250\pi /3$ units³.

The cross-sections of this cone are the same as the cone in 19. Thus they have the same volume of $250\pi /3$ units³.

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 $A\left(x\right)=2x^2/25$, giving a volume of $80/3$ units³.

Orient the solid so that the $x$-axis is parallel to long side of the base. All cross-sections are trapezoids (at the far left, the trapezoid is a square; at the far right, the trapezoid has a top length of 0, making it a triangle). The area of the trapezoid at $x$ is $A\left(x\right)=1/2\left(-1/2x+5+5\right)\left(5\right)=-5/4x+25$. The volume is $187.5$ units³.

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F

F

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$9\pi /2$ units³

$70\pi /3$ units³

${\displaystyle \frac{96\pi }{5}}$

$2\pi /15$ units³

$48\pi \sqrt{3}/5$ units³

$350\pi /3$ units³

${\displaystyle \frac{768\pi }{7}}$

$\pi /6$ units³

$3{\pi }^2$

(a) Disk: $\pi {\int }_0^1\left[1^2-{\left(\sqrt[4]{y}\right)}^2\right]\mathrm{d}y=\frac{\pi }{3}$
Shell: $2\pi {\int }_0^{1}x\cdot x^4\mathrm{d}x=\frac{\pi }{3}$
(b) Disk: $\pi {\int }_0^1{\left(x^4\right)}^2\mathrm{d}x=\frac{\pi }{9}$
Shell: $2\pi {\int }_0^{1}y\left(1-\sqrt[4]{y}\right)\mathrm{d}y=\frac{\pi }{9}$.

(a) Disk: $\pi {\int }_1^2{\left(\sqrt[3]{y-1}\right)}^2\mathrm{d}y=\frac{3\pi }{5}$
Shell: $2\pi {\int }_0^{1}x\left(2-\left(x^3+1\right)\right)\mathrm{d}x=\frac{3\pi }{5}$
(b) Disk: $\pi {\int }_0^1{\left[2-\left(x^3+1\right)\right]}^2\mathrm{d}x=\frac{9\pi }{14}$
Shell: $2\pi {\int }_1^2\left(2-y\right)\sqrt[3]{y-1}\mathrm{d}y=\frac{9\pi }{14}$.

(a) Disk: $\pi {\int }_{-2}^1\left[{\left(-4x+8\right)}^2-{\left(4x^2\right)}^2\right]\mathrm{d}x=\frac{1152\pi }{5}$
Shell: $2\pi {\int }_0^{4}y\sqrt{y}\mathrm{d}y+2\pi {\int }_4^{16}y\left[\left(2-\frac{y}{4}\right)+\frac{\sqrt{y}}{2}\right]\mathrm{d}y=\frac{128\pi }{5}+\frac{1024\pi }{5}$
(b) Disk: $\pi {\int }_0^4{\left[1+\frac{\sqrt{y}}{2}\right]}^2-{\left[1-\frac{\sqrt{y}}{2}\right]}^2\mathrm{d}y+\pi {\int }_4^{16}{\left[1+\frac{\sqrt{y}}{2}\right]}^2-{\left[1-\left(2-\frac{y}{4}\right)\right]}^2\mathrm{d}y=\frac{32\pi }{3}+\frac{130\pi }{3}$
Shell: $2\pi {\int }_{-2}^1\left(1-x\right)\left[\left(-4x+8\right)-4x^2\right]\mathrm{d}x=54\pi$
(c) Disk: $\pi {\int }_{-2}^1\left[{\left(16-4x^2\right)}^2-{\left(16-\left(-4x+8\right)\right)}^2\right]\mathrm{d}x=\frac{1728\pi }{5}$
Shell: $2\pi {\int }_0^4\left(16-y\right)\left[\sqrt{y}\right]\mathrm{d}y+2\pi {\int }_4^{16}\left(16-y\right)\left[\left(2-\frac{y}{4}\right)+\frac{\sqrt{y}}{2}\right]\mathrm{d}y=\frac{2176\pi }{15}+\frac{3008\pi }{15}$.

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.

The same.

Smaller.

force; distance

735 J

11,100 ft-lb

575 ft-lb

125 ft-lb

0.05 J

12.5 ft-lb

5/3 ft-lb

0.2625 = 21/80 J

$f\cdot d/2$ J

45 ft-lb

5 ft-lb

$953,284$ J

192,767 ft-lb. Note that the tank is oriented horizontally. Let the origin be the center of one of the circular ends of the tank. Since the radius is 3.75 ft, the fluid is being pumped to $y=4.75$; thus the distance the gas travels is $h\left(y\right)=4.75-y$. A differential element of water is a rectangle, with length 20 and width $2\sqrt{{3.75}^2-y^2}$. Thus the force required to move that slab of gas is $F\left(y\right)=40\cdot 45.93\cdot \sqrt{{3.75}^2-y^2}\mathrm{d}y$. Total work is ${\int }_{-3.75}^{3.75}40\cdot 45.93\cdot \left(4.75-y\right)\sqrt{{3.75}^2-y^2}\mathrm{d}y$. This can be evaluated without actual integration; split the integral into ${\int }_{-3.75}^{3.75}40\cdot 45.93\cdot \left(4.75\right)\sqrt{{3.75}^2-y^2}\mathrm{d}y+{\int }_{-3.75}^{3.75}40\cdot 45.93\cdot \left(-y\right)\sqrt{{3.75}^2-y^2}\mathrm{d}y$. The first integral can be evaluated as measuring half the area of a circle; the latter integral can be shown to be 0 without much difficulty. (Use substitution and realize the bounds are both 0.)

212,135 ft-lb

187,214 ft-lb

617,400 J

4,917,150 J

Answers will vary.

Answers will vary.

499.2 lb

249.6 lb

6739.2 lb

5241.6 lb

3920.7 lb

15682.8 lb

2496 lb

2496 lb

602.59 lb

291.2 lb

5.1 ft

4.1 ft