# Chapter 14

## Exercises 14.1

1. 1.

$C(y)$, meaning that instead of being just a constant, like the number 5, it is a function of $y$, which acts like a constant when taking derivatives with respect to $x$.

2. 3.

curve to curve, then from point to point

3. 5.
(a) $18x^{2}+42x-117$ (b) $-108$
4. 7.
(a) $x^{4}/2-x^{2}+2x-3/2$ (b) $23/15$
5. 9.
(a) $\sin^{2}y$ (b) $\pi/2$
6. 11.
$\displaystyle\int_{1}^{4}\int_{-2}^{1}\operatorname{d}\!y\operatorname{d}\!x$ and $\displaystyle\int_{-2}^{1}\int_{1}^{4}\operatorname{d}\!x\operatorname{d}\!y$. area of $R=9\text{ units}^{2}$
7. 13.
$\displaystyle\int_{2}^{4}\int_{x-1}^{7-x}\operatorname{d}\!y\operatorname{d}\!x$. The order $\operatorname{d}\!x\operatorname{d}\!y$ needs two iterated integrals as $x$ is bounded above by two different functions. This gives: $\int_{1}^{3}\int_{2}^{y+1}\operatorname{d}\!x\operatorname{d}\!y+\int_{3}^{5}% \int_{2}^{7-y}\operatorname{d}\!x\operatorname{d}\!y.$ area of $R=4\text{ units}^{2}$
8. 15.
$\displaystyle\int_{0}^{1}\int_{x^{4}}^{\sqrt{x}}\operatorname{d}\!y% \operatorname{d}\!x$ and $\displaystyle\int_{0}^{1}\int_{y^{2}}^{\sqrt[4]{y}}\operatorname{d}\!x% \operatorname{d}\!y$ area of $R=7/15\text{ units}^{2}$
9. 17.

area of $R=\displaystyle\int_{0}^{4}\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\operatorname{d}\!x% \operatorname{d}\!y$

10. 19.

area of $R=\displaystyle\int_{0}^{4}\int_{-\sqrt{4-x^{2}/4}}^{\sqrt{4-x^{2}/4}}% \operatorname{d}\!y\operatorname{d}\!x$

11. 21.

area of $R=\displaystyle\int_{-1}^{2}\int_{x^{2}}^{x+2}\operatorname{d}\!y\operatorname% {d}\!x$

## Exercises 14.2

1. 1.

volume

2. 3.

The double integral gives the signed volume under the surface. Since the surface is always positive, it is always above the $x$-$y$ plane and hence produces only “positive” volume.

3. 5.

6; $\displaystyle\int_{-1}^{1}\int_{1}^{2}\left(\frac{x}{y}+3\right)\operatorname{% d}\!y\operatorname{d}\!x$

4. 7.

112/3; $\displaystyle\int_{0}^{2}\int_{0}^{4-2y}\left(3x^{2}-y+2\right)\operatorname{d% }\!x\operatorname{d}\!y$

5. 9.

16/5; $\displaystyle\int_{-1}^{1}\int_{0}^{1-x^{2}}\left(x+y+2\right)\operatorname{d}% \!y\operatorname{d}\!x$

6. 11.
(a) (b) $\displaystyle\int_{0}^{1}\int_{x^{2}}^{\sqrt{x}}x^{2}y\operatorname{d}\!y% \operatorname{d}\!x=\int_{0}^{1}\int_{y^{2}}^{\sqrt{y}}x^{2}y\operatorname{d}% \!x\operatorname{d}\!y$. (c) $\frac{3}{56}$
7. 13.
(a) (b) $\displaystyle\int_{-1}^{1}\int_{-1}^{1}x^{2}-y^{2}\operatorname{d}\!y% \operatorname{d}\!x=\int_{-1}^{1}\int_{-1}^{1}x^{2}-y^{2}\operatorname{d}\!x% \operatorname{d}\!y$. (c) 0
8. 15.
(a) (b) (c) $\displaystyle\int_{0}^{2}\int_{0}^{3-3/2x}\big{(}6-3x-2y\big{)}\operatorname{d% }\!y\operatorname{d}\!x=\int_{0}^{3}\int_{0}^{2-2/3y}\big{(}6-3x-2y\big{)}% \operatorname{d}\!x\operatorname{d}\!y$. (d) 6
9. 17.
(a) (b) $\displaystyle\int_{-3}^{3}\int_{0}^{\sqrt{9-x^{2}}}\big{(}x^{3}y-x\big{)}% \operatorname{d}\!y\operatorname{d}\!x=\int_{0}^{3}\int_{-\sqrt{9-y^{2}}}^{% \sqrt{9-y^{2}}}\big{(}x^{3}y-x\big{)}\operatorname{d}\!x\operatorname{d}\!y$. (c) 0
10. 19.

Integrating $e^{x^{2}}$ with respect to $x$ is not possible in terms of elementary functions. $\displaystyle\int_{0}^{2}\int_{0}^{2x}e^{x^{2}}\operatorname{d}\!y% \operatorname{d}\!x=e^{4}-1$.

11. 21.

Integrating $\displaystyle\int_{y}^{1}\frac{2y}{x^{2}+y^{2}}\operatorname{d}\!x$ gives $\tan^{-1}(1/y)-\pi/4$; integrating $\tan^{-1}(1/y)$ is hard.

$\displaystyle\int_{0}^{1}\int_{0}^{x}\frac{2y}{x^{2}+y^{2}}\operatorname{d}\!y% \operatorname{d}\!x=\ln 2$.

12. 23.

average value of $f=6/2=3$

13. 25.

average value of $f=\frac{112/3}{4}=28/3$

## Exercises 14.3

1. 1.

$f\big{(}r\cos\theta,r\sin\theta\big{)}$, $r\operatorname{d}\!r\operatorname{d}\!\theta$

2. 3.

$\displaystyle\int_{0}^{2\pi}\int_{0}^{1}\big{(}3r\cos\theta-r\sin\theta+4\big{% )}r\operatorname{d}\!r\operatorname{d}\!\theta=4\pi$

3. 5.

$\displaystyle\int_{0}^{\pi}\int_{\cos\theta}^{3\cos\theta}\big{(}8-r\sin\theta% \big{)}r\operatorname{d}\!r\operatorname{d}\!\theta=16\pi$

4. 7.

$\displaystyle\int_{0}^{2\pi}\int_{1}^{2}\big{(}\ln(r^{2})\big{)}r\operatorname% {d}\!r\operatorname{d}\!\theta=2\pi\big{(}\ln 16-3/2\big{)}$

5. 9.

$\displaystyle\int_{-\pi/2}^{\pi/2}\int_{0}^{6}\big{(}r^{2}\cos^{2}\theta-r^{2}% \sin^{2}\theta\big{)}r\operatorname{d}\!r\operatorname{d}\!\theta=\int_{-\pi/2% }^{\pi/2}\int_{0}^{6}\big{(}r^{2}\cos(2\theta)\big{)}r\operatorname{d}\!r% \operatorname{d}\!\theta=0$

6. 11.

$\displaystyle\int_{-\pi/2}^{\pi/2}\int_{0}^{5}\big{(}r^{2}\big{)}\operatorname% {d}\!r\operatorname{d}\!\theta=125\pi/3$

7. 13.

$\displaystyle\int_{0}^{\pi/4}\int_{0}^{\sqrt{8}}\big{(}r\cos\theta+r\sin\theta% \big{)}r\operatorname{d}\!r\operatorname{d}\!\theta=16\sqrt{2}/3$

8. 15.
(a) This is impossible to integrate with rectangular coordinates as $e^{-(x^{2}+y^{2})}$ does not have an antiderivative in terms of elementary functions. (b) $\displaystyle\int_{0}^{2\pi}\int_{0}^{a}re^{r^{2}}\operatorname{d}\!r% \operatorname{d}\!\theta=\pi(1-e^{-a^{2}})$. (c) $\displaystyle\lim_{a\to\infty}\pi(1-e^{-a^{2}})=\pi$. This implies that there is a finite volume under the surface $e^{-(x^{2}+y^{2})}$ over the entire $x$-$y$ plane. (d) If $R=\mathbb{R}^{2}$, we can write the original integral as $\displaystyle\left(\int_{-\infty}^{\infty}e^{-t^{2}}\operatorname{d}\!t\right)% ^{2}=\pi$.
9. 17.

$3\pi/4-9\sqrt{3}/16$

10. 19.

$2$

11. 21.

$2(1-a^{3})/3(1-a^{2})$; $2/3$; $1$

## Exercises 14.4

1. 1.

Because they are scalar multiples of each other.

2. 3.

“little masses”

3. 5.

$M_{x}$ measures the moment about the $x$-axis, meaning we need to measure distance from the $x$-axis. Such measurements are measures in the $y$-direction.

4. 7.

$\overline{x}=5.25$

5. 9.

$(\overline{x},\overline{y})=(0,3)$

6. 11.

$M=150$g;

7. 13.

$M=2$lb

8. 15.

$M=16\pi\approx 50.27$kg

9. 17.

$M=54\pi\approx 169.65$lb

10. 19.

$M=150$g; $M_{y}=600$; $M_{x}=-75$; $(\overline{x},\overline{y})=(4,-0.5)$

11. 21.

$M=2$lb; $M_{y}=0$; $M_{x}=2/3$; $(\overline{x},\overline{y})=(0,1/3)$

12. 23.

$M=16\pi\approx 50.27$kg; $M_{y}=4\pi$; $M_{x}=4\pi$; $(\overline{x},\overline{y})=(1/4,1/4)$

13. 25.

$M=54\pi\approx 169.65$lb; $M_{y}=0$; $M_{x}=504$; $(\overline{x},\overline{y})=(0,2.97)$

14. 27.

$I_{x}=64/3$; $I_{y}=64/3$; $I_{O}=128/3$

15. 29.

$I_{x}=16/3$; $I_{y}=64/3$; $I_{O}=80/3$

## Exercises 14.5

1. 1.

arc length

2. 3.

surface areas

3. 5.

Intuitively, adding $h$ to $f$ only shifts $f$ up (i.e., parallel to the $z$-axis) and does not change its shape. Therefore it will not change the surface area over $R$.

Analytically, $f_{x}=g_{x}$ and $f_{y}=g_{y}$; therefore, the surface area of each is computed with identical double integrals.

4. 7.

$\displaystyle S=\int_{0}^{2\pi}\int_{0}^{2\pi}\sqrt{1+\cos^{2}x\cos^{2}y+\sin^% {2}x\sin^{2}y}\operatorname{d}\!x\operatorname{d}\!y$

5. 9.

$\displaystyle S=\int_{-1}^{1}\int_{-1}^{1}\sqrt{1+4x^{2}+4y^{2}}\operatorname{% d}\!x\operatorname{d}\!y$

6. 11.

$\displaystyle S=\int_{0}^{3}\int_{-1}^{1}\sqrt{1+9+49}\operatorname{d}\!x% \operatorname{d}\!y=6\sqrt{59}\approx 46.09$

7. 13.
This is easier in polar: $\displaystyle S$ $\displaystyle=\int_{0}^{2\pi}\int_{0}^{4}r\sqrt{1+4r^{2}\cos^{2}\theta+4r^{2}% \sin^{2}\theta}\operatorname{d}\!r\operatorname{d}\!\theta$ $\displaystyle=\int_{0}^{2\pi}\int_{0}^{4}r\sqrt{1+4r^{2}}\operatorname{d}\!r% \operatorname{d}\!\theta$ $\displaystyle=\frac{\pi}{6}\big{(}65\sqrt{65}-1\big{)}\approx 273.87$
8. 15.
$\displaystyle S$ $\displaystyle=\int_{0}^{1}\int_{0}^{1}\sqrt{1+x+9y}\operatorname{d}\!x% \operatorname{d}\!y$ $\displaystyle=\int_{0}^{1}\frac{2}{3}\Bigl{(}(9y+2)^{3/2}-(9y+1)^{3/2}\Bigr{)}% \operatorname{d}\!y$ $\displaystyle=\frac{4}{135}\bigl{(}121\sqrt{11}-100\sqrt{10}-4\sqrt{2}+1\bigr{% )}\approx 2.383$
9. 17.
This is easier in polar: $\displaystyle S$ $\displaystyle=2\int_{0}^{2\pi}\int_{0}^{5}r\sqrt{1+\frac{r^{2}\cos^{2}\theta+r% ^{2}\sin^{2}\theta}{25-r^{2}\sin^{2}\theta-r^{2}\cos^{2}\theta}}\operatorname{% d}\!r\operatorname{d}\!\theta$ $\displaystyle=2\int_{0}^{2\pi}\int_{0}^{5}r\sqrt{\frac{1}{25-r^{2}}}% \operatorname{d}\!r\operatorname{d}\!\theta$ $\displaystyle=100\pi\approx 314.16$

## Exercises 14.6

1. 1.

surface to surface, curve to curve and point to point

2. 3.

Answers can vary. From this section we used triple integration to find the volume of a solid region, the mass of a solid, and the center of mass of a solid.

3. 5.

$V=\int_{-1}^{1}\int_{-1}^{1}\big{(}8-x^{2}-y^{2}-(2x+y)\big{)}\operatorname{d}% \!x\operatorname{d}\!y=88/3$

4. 7.

$V=\int_{0}^{\pi}\int_{0}^{x}\big{(}\cos x\sin y+2-\sin x\cos y\big{)}% \operatorname{d}\!y\operatorname{d}\!x=\pi^{2}-\pi\approx 6.728$

5. 9.
$\operatorname{d}\!z\operatorname{d}\!y\operatorname{d}\!x$: $\displaystyle\int_{0}^{3}\int_{0}^{1-x/3}\int_{0}^{2-2x/3-2y}\operatorname{d}% \!z\operatorname{d}\!y\operatorname{d}\!x$ $\operatorname{d}\!z\operatorname{d}\!x\operatorname{d}\!y$: $\displaystyle\int_{0}^{1}\int_{0}^{3-3y}\int_{0}^{2-2x/3-2y}\operatorname{d}\!% z\operatorname{d}\!x\operatorname{d}\!y$ $\operatorname{d}\!y\operatorname{d}\!z\operatorname{d}\!x$: $\displaystyle\int_{0}^{3}\int_{0}^{2-2x/3}\int_{0}^{1-x/3-z/2}\operatorname{d}% \!y\operatorname{d}\!z\operatorname{d}\!x$ $\operatorname{d}\!y\operatorname{d}\!x\operatorname{d}\!z$: $\displaystyle\int_{0}^{2}\int_{0}^{3-3z/2}\int_{0}^{1-x/3-z/2}\operatorname{d}% \!y\operatorname{d}\!x\operatorname{d}\!z$ $\operatorname{d}\!x\operatorname{d}\!z\operatorname{d}\!y$: $\displaystyle\int_{0}^{1}\int_{0}^{2-2y}\int_{0}^{3-3y-3z/2}\operatorname{d}\!% x\operatorname{d}\!z\operatorname{d}\!y$ $\operatorname{d}\!x\operatorname{d}\!y\operatorname{d}\!z$: $\displaystyle\int_{0}^{2}\int_{0}^{1-z/2}\int_{0}^{3-3y-3z/2}\operatorname{d}% \!x\operatorname{d}\!y\operatorname{d}\!z$ $\displaystyle V=\int_{0}^{3}\int_{0}^{1-x/3}\int_{0}^{2-2x/3-2y}\operatorname{% d}\!z\operatorname{d}\!y\operatorname{d}\!x=1.$
6. 11.
$\operatorname{d}\!z\operatorname{d}\!y\operatorname{d}\!x$: $\displaystyle\int_{0}^{2}\int_{-2}^{0}\int_{y^{2}/2}^{-y}\operatorname{d}\!z% \operatorname{d}\!y\operatorname{d}\!x$ $\operatorname{d}\!z\operatorname{d}\!x\operatorname{d}\!y$: $\displaystyle\int_{-2}^{0}\int_{0}^{2}\int_{y^{2}/2}^{-y}\operatorname{d}\!z% \operatorname{d}\!x\operatorname{d}\!y$ $\operatorname{d}\!y\operatorname{d}\!z\operatorname{d}\!x$: $\displaystyle\int_{0}^{2}\int_{0}^{2}\int_{-\sqrt{2z}}^{-z}\operatorname{d}\!y% \operatorname{d}\!z\operatorname{d}\!x$ $\operatorname{d}\!y\operatorname{d}\!x\operatorname{d}\!z$: $\displaystyle\int_{0}^{2}\int_{0}^{2}\int_{-\sqrt{2z}}^{-z}\operatorname{d}\!y% \operatorname{d}\!x\operatorname{d}\!z$ $\operatorname{d}\!x\operatorname{d}\!z\operatorname{d}\!y$: $\displaystyle\int_{-2}^{0}\int_{y^{2}/2}^{-y}\int_{0}^{2}\operatorname{d}\!x% \operatorname{d}\!z\operatorname{d}\!y$ $\operatorname{d}\!x\operatorname{d}\!y\operatorname{d}\!z$: $\displaystyle\int_{0}^{2}\int_{-\sqrt{2z}}^{-z}\int_{0}^{2}\operatorname{d}\!x% \operatorname{d}\!y\operatorname{d}\!z$ $\displaystyle V=\int_{0}^{2}\int_{0}^{2}\int_{-\sqrt{2z}}^{-z}\operatorname{d}% \!y\operatorname{d}\!z\operatorname{d}\!x=4/3.$
7. 13.
$\operatorname{d}\!z\operatorname{d}\!y\operatorname{d}\!x$: $\displaystyle\int_{0}^{2}\int_{1-x/2}^{1}\int_{0}^{2x+4y-4}\operatorname{d}\!z% \operatorname{d}\!y\operatorname{d}\!x$ $\operatorname{d}\!z\operatorname{d}\!x\operatorname{d}\!y$: $\displaystyle\int_{0}^{1}\int_{2-2y}^{2}\int_{0}^{2x+4y-4}\operatorname{d}\!z% \operatorname{d}\!x\operatorname{d}\!y$ $\operatorname{d}\!y\operatorname{d}\!z\operatorname{d}\!x$: $\displaystyle\int_{0}^{2}\int_{0}^{2x}\int_{z/4-x/2+1}^{1}\operatorname{d}\!y% \operatorname{d}\!z\operatorname{d}\!x$ $\operatorname{d}\!y\operatorname{d}\!x\operatorname{d}\!z$: $\displaystyle\int_{0}^{4}\int_{z/2}^{2}\int_{z/4-x/2+1}^{1}\operatorname{d}\!y% \operatorname{d}\!x\operatorname{d}\!z$ $\operatorname{d}\!x\operatorname{d}\!z\operatorname{d}\!y$: $\displaystyle\int_{0}^{1}\int_{0}^{4y}\int_{z/2-2y+2}^{2}\operatorname{d}\!x% \operatorname{d}\!z\operatorname{d}\!y$ $\operatorname{d}\!x\operatorname{d}\!y\operatorname{d}\!z$: $\displaystyle\int_{0}^{4}\int_{z/4}^{1}\int_{z/2-2y+2}^{2}\operatorname{d}\!x% \operatorname{d}\!y\operatorname{d}\!z$ $\displaystyle V=\int_{0}^{4}\int_{z/4}^{1}\int_{z/2-2y+2}^{2}\operatorname{d}% \!x\operatorname{d}\!y\operatorname{d}\!z=4/3.$
8. 15.
$\operatorname{d}\!z\operatorname{d}\!y\operatorname{d}\!x$: $\displaystyle\int_{0}^{1}\int_{0}^{1-x^{2}}\int_{0}^{\sqrt{1-y}}\operatorname{% d}\!z\operatorname{d}\!y\operatorname{d}\!x$ $\operatorname{d}\!z\operatorname{d}\!x\operatorname{d}\!y$: $\displaystyle\int_{0}^{1}\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}% \operatorname{d}\!z\operatorname{d}\!x\operatorname{d}\!y$ $\operatorname{d}\!y\operatorname{d}\!z\operatorname{d}\!x$: $\displaystyle\int_{0}^{1}\int_{0}^{x}\int_{0}^{1-x^{2}}\operatorname{d}\!y% \operatorname{d}\!z\operatorname{d}\!x+\int_{0}^{1}\int_{x}^{1}\int_{0}^{1-z^{% 2}}\operatorname{d}\!y\operatorname{d}\!z\operatorname{d}\!x$ $\operatorname{d}\!y\operatorname{d}\!x\operatorname{d}\!z$: $\displaystyle\int_{0}^{1}\int_{0}^{z}\int_{0}^{1-z^{2}}\operatorname{d}\!y% \operatorname{d}\!x\operatorname{d}\!z+\int_{0}^{1}\int_{z}^{1}\int_{0}^{1-x^{% 2}}\operatorname{d}\!y\operatorname{d}\!x\operatorname{d}\!z$ $\operatorname{d}\!x\operatorname{d}\!z\operatorname{d}\!y$: $\displaystyle\int_{0}^{1}\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}% \operatorname{d}\!x\operatorname{d}\!z\operatorname{d}\!y$ $\operatorname{d}\!x\operatorname{d}\!y\operatorname{d}\!z$: $\displaystyle\int_{0}^{1}\int_{0}^{1-z^{2}}\int_{0}^{\sqrt{1-y}}\operatorname{% d}\!x\operatorname{d}\!y\operatorname{d}\!z$ Answers will vary. Neither order is particularly “hard.” The order $\operatorname{d}\!z\operatorname{d}\!y\operatorname{d}\!x$ requires integrating a square root, so powers can be messy; the order $\operatorname{d}\!y\operatorname{d}\!z\operatorname{d}\!x$ requires two triple integrals, but each uses only polynomials.
9. 17.

8

10. 19.

$\pi$

11. 21.
$M=10$, $M_{yz}=15/2$, $M_{xz}=5/2$, $M_{xy}=5$; $(\overline{x},\overline{y},\overline{z})=(3/4,1/4,1/2)$
12. 23.
$M=16/5$, $M_{yz}=16/3$, $M_{xz}=104/45$, $M_{xy}=32/9$; $(\overline{x},\overline{y},\overline{z})=(5/3,13/18,10/9)\approx(1.67,0.72,1.11)$

## Exercises 14.7

1. 1.

In cylindrical, $r$ determines how far from the origin one goes in the $x$-$y$ plane before considering the $z$-component. Equivalently, if on projects a point in cylindrical coordinates onto the $x$-$y$ plane, $r$ will be the distance of this projection from the origin.

In spherical, $\rho$ is the distance from the origin to the point.

2. 3.

Cylinders (tubes) centered at the origin, parallel to the $z$-axis; planes parallel to the $z$-axis that intersect the $z$-axis; planes parallel to the $x$-$y$ plane.

3. 5.
(a) Cylindrical: $(2\sqrt{2},\pi/4,1)$ and $(2,5\pi/6,0)$ Spherical: $(3,\pi/4,\cos^{-1}(1/3))$ and $(2,5\pi/6,\pi/2)$ (b) Rectangular: $(\sqrt{2},\sqrt{2},2)$ and $(0,-3,-4)$ Spherical: $(2\sqrt{2},\pi/4,\pi/4)$ and $(5,3\pi/2,\pi-\tan^{-1}(3/4))$ (c) Rectangular: $(1,1,\sqrt{2})$ and $(0,0,1)$ Cylindrical: $(\sqrt{2},\pi/4,\sqrt{2})$ and $(0,0,1)$
4. 7.
(a) Cylindrical: $(4,\frac{\pi}{3},-1)$ and $(5\sqrt{2},\frac{3\pi}{4},6)$ Spherical: $(\sqrt{17},\frac{\pi}{3},\cos^{-1}\frac{-1}{\sqrt{17}})$ and $(\sqrt{86},\frac{3\pi}{4},\cos^{-1}\frac{6}{\sqrt{86}})$ (b) Rectangular: $(\frac{1}{2},\frac{\sqrt{3}}{2},-2)$ and $(-\sqrt{3},-1,3)$ Spherical: $(\sqrt{5},\frac{\pi}{3},\cos^{-1}(-\frac{2}{\sqrt{5}}))$ and $(\sqrt{13},\frac{5\pi}{6},\tan^{-1}\frac{2}{3})$ (c) Rectangular: $(\sqrt{2},\sqrt{6},2\sqrt{2})$ and $(1,0,0)$ Cylindrical: $(2\sqrt{2},\frac{\pi}{3},2\sqrt{2})$ and $(1,0,0)$
5. 9.
(a) A cylindrical surface or tube, centered along the $z$-axis of radius 1, extending from the $x$-$y$ plane up to the plane $z=1$ (i.e., the tube has a length of 1). (b) This is a region of space, being half of a tube with “thick” walls of inner radius 1 and outer radius 2, centered along the $z$-axis with a length of 1, where the half “below” the $x$-$z$ plane is removed. (c) This is upper half of the sphere of radius 3 centered at the origin (i.e., the upper hemisphere). (d) This is a region of space, where the ball of radius 2, centered at the origin, is removed from the ball of radius 3, centered at the origin.
6. 11.

$\displaystyle\int_{\theta_{1}}^{\theta_{2}}\int_{r_{1}}^{r_{2}}\int_{z_{1}}^{z% _{2}}h(r,\theta,z)r\operatorname{d}\!z\operatorname{d}\!r\operatorname{d}\!\theta$

7. 13.

The region in space is bounded between the planes $z=0$ and $z=2$, inside of the cylinder $x^{2}+y^{2}=4$, and the planes $\theta=0$ and $\theta=\pi/2$: describes a “wedge” of a cylinder of height 2 and radius 2; the angle of the wedge is $\pi/2$, or $90^{\circ}$.

8. 15.

Bounded between the plane $z=0$ and the cone $z=1-\sqrt{x^{2}+y^{2}}$: describes an inverted cone, with height of 1, point at $(0,0,1)$ and base radius of 1.

9. 17.

Describes a quarter of a ball of radius 3, centered at the origin; the quarter resides above the $x$-$y$ plane and above the $x$-$z$ plane.

10. 19.

Describes the portion of the unit ball that resides in the first octant.

11. 21.

Bounded above the cone $z=\sqrt{x^{2}+y^{2}}$ and below the sphere $x^{2}+y^{2}+z^{2}=4$: describes a shape that is somewhat “diamond”-like; some think of it as looking like an ice cream cone (see Figure 14.7.8). It describes a cone, where the side makes an angle of $\pi/4$ with the positive $z$-axis, topped by the portion of the ball of radius 2, centered at the origin.

12. 23.

The region in space is bounded below by the cone $z=\sqrt{3}\sqrt{x^{2}+y^{2}}$ and above by the plane $z=1$: it describes a cone, with point at the origin, centered along the positive $z$-axis, with height of 1 and base radius of $\tan(\pi/6)=1/\sqrt{3}$.

13. 25.
In cylindrical coordinates, the density is $\delta(r,\theta,z)=r+1$. Thus mass is $\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{4}(r+1)r\operatorname{d}\!z\operatorname{% d}\!r\operatorname{d}\!\theta=112\pi/3.$
14. 27.
In cylindrical coordinates, the density is $\delta(r,\theta,z)=1$. Thus mass is $\int_{0}^{\pi}\int_{0}^{1}\int_{0}^{4-r\sin\theta}r\operatorname{d}\!z% \operatorname{d}\!r\operatorname{d}\!\theta=2\pi-2/3\approx 5.617.$
15. 29.
In cylindrical coordinates, the density is $\delta(r,\theta,z)=r+1$. Thus mass is $M=\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{4}(r+1)r\operatorname{d}\!z% \operatorname{d}\!r\operatorname{d}\!\theta=112\pi/3.$

We find $M_{yz}=0$, $M_{xz}=0$, and $M_{xy}=224\pi/3$, placing the center of mass at $(0,0,2)$.

16. 31.
In cylindrical coordinates, the density is $\delta(r,\theta,z)=1$. Thus mass is $\int_{0}^{\pi}\int_{0}^{1}\int_{0}^{4-r\sin\theta}r\operatorname{d}\!z% \operatorname{d}\!r\operatorname{d}\!\theta=2\pi-2/3\approx 5.617.$

We find $M_{yz}=0$, $M_{xz}=8/3-\pi/8$, and $M_{xy}=65\pi/16-8/3$, placing the center of mass at $\approx(0,0.405,1.80)$.

17. 33.
In spherical coordinates, the density is $\delta(\rho,\theta,\varphi)=1$. Thus mass is $\int_{0}^{\pi/2}\int_{0}^{2\pi}\int_{0}^{1}\rho^{2}\sin(\varphi)\operatorname{% d}\!\rho\operatorname{d}\!\theta\operatorname{d}\!\varphi=2\pi/3.$
18. 35.
In spherical coordinates, the density is $\delta(\rho,\theta,\varphi)=\rho\cos\varphi$. Thus mass is $\int_{0}^{\pi/4}\int_{0}^{2\pi}\int_{0}^{1}\big{(}\rho\cos(\varphi)\big{)}\rho% ^{2}\sin(\varphi)\operatorname{d}\!\rho\operatorname{d}\!\theta\operatorname{d% }\!\varphi=\pi/8.$
19. 37.
In spherical coordinates, the density is $\delta(\rho,\theta,\varphi)=1$. Thus mass is $\int_{0}^{\pi/2}\int_{0}^{2\pi}\int_{0}^{1}\rho^{2}\sin(\varphi)\operatorname{% d}\!\rho\operatorname{d}\!\theta\operatorname{d}\!\varphi=2\pi/3.$

We find $M_{yz}=0$, $M_{xz}=0$, and $M_{xy}=\pi/4$, placing the center of mass at $(0,0,3/8)$.

20. 39.
In spherical coordinates, the density is $\delta(\rho,\theta,\varphi)=\rho\cos\varphi$. Thus mass is $\int_{0}^{\pi/4}\int_{0}^{2\pi}\int_{0}^{1}\big{(}\rho\cos(\varphi)\big{)}\rho% ^{2}\sin(\varphi)\operatorname{d}\!\rho\operatorname{d}\!\theta\operatorname{d% }\!\varphi=\pi/8.$

We find $M_{yz}=0$, $M_{xz}=0$, and $M_{xy}=(4-\sqrt{2})\pi/30$, placing the center of mass at $(0,0,4(4-\sqrt{2})/15)$.

21. 41.

Rectangular: $\int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{-\sqrt{1-x^{2}-y^{2}% }}^{\sqrt{1-x^{2}-y^{2}}}\operatorname{d}\!z\operatorname{d}\!y\operatorname{d% }\!x$

Cylindrical: $\int_{0}^{2\pi}\int_{0}^{1}\int_{-\sqrt{1-r^{2}}}^{\sqrt{1-r^{2}}}r% \operatorname{d}\!z\operatorname{d}\!r\operatorname{d}\!\theta$

Spherical: $\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{1}\rho^{2}\sin(\varphi)\operatorname{d}% \!\rho\operatorname{d}\!\theta\operatorname{d}\!\varphi$

Spherical appears simplest, avoiding the integration of square-roots and using techniques such as Substitution; all bounds are constants.

22. 43.

Rectangular: $\int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{\sqrt{x^{2}+y^{2}}}^% {1}\operatorname{d}\!z\operatorname{d}\!y\operatorname{d}\!x$

Cylindrical: $\int_{0}^{2\pi}\int_{0}^{1}\int_{r}^{1}r\operatorname{d}\!z\operatorname{d}\!r% \operatorname{d}\!\theta$

Spherical: $\int_{0}^{\pi/4}\int_{0}^{2\pi}\int_{0}^{\sec\varphi}\rho^{2}\sin(\varphi)% \operatorname{d}\!\rho\operatorname{d}\!\theta\operatorname{d}\!\varphi$

Cylindrical appears simplest, avoiding the integration of square-roots that rectangular uses. Spherical is not difficult, though it requires Substitution, an extra step.

23. 45.

center: $(2,3,-1)$, radius: $\sqrt{14}$

24. 47.

$(a,\theta,a\cot\phi)$

25. 49.

Hint: Use the distance formula for Cartesian coordinates.

26. 51.
27. 53.

$1-\sin 2/2$