Solutions To Selected Problems Chapter 13 Chapter 15

Chapter 14

Exercises 14.1

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 $x$ .
3.

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

area of $R=\displaystyle\int_{0}^{4}\int_{-\sqrt{4-y}}^{\sqrt{4-y}}\ dx\ dy$
19.

area of $R=\displaystyle\int_{0}^{4}\int_{-\sqrt{4-x^{2}/4}}^{\sqrt{4-x^{2}/4}}\ dy\ dx$
21.

area of $R=\displaystyle\int_{-1}^{2}\int_{x^{2}}^{x+2}\ dy\ dx$

Exercises 14.2

1.

volume
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.
5.

6; $\displaystyle\int_{-1}^{1}\int_{1}^{2}\left(\frac{x}{y}+3\right)\ dy\ dx$
7.

112/3; $\displaystyle\int_{0}^{2}\int_{0}^{4-2y}\left(3x^{2}-y+2\right)\ dx\ dy$
9.

16/5; $\displaystyle\int_{-1}^{1}\int_{0}^{1-x^{2}}\left(x+y+2\right)\ dy\ dx$
11.
(a) (b) $\displaystyle\int_{0}^{1}\int_{x^{2}}^{\sqrt{x}}x^{2}y\ dy\ dx=\int_{0}^{1}% \int_{y^{2}}^{\sqrt{y}}x^{2}y\ dx\ dy$ . (c) $\frac{3}{56}$
13.
(a) (b) $\displaystyle\int_{-1}^{1}\int_{-1}^{1}x^{2}-y^{2}\ dy\ dx=\int_{-1}^{1}\int_{% -1}^{1}x^{2}-y^{2}\ dx\ dy$ . (c) 0
15.
(a) (b) (c) $\displaystyle\int_{0}^{2}\int_{0}^{3-3/2x}\big{(}6-3x-2y\big{)}\ dy\ dx=\int_{% 0}^{3}\int_{0}^{2-2/3y}\big{(}6-3x-2y\big{)}\ dx\ dy$ . (d) 6
17.
(a) (b) $\displaystyle\int_{-3}^{3}\int_{0}^{\sqrt{9-x^{2}}}\big{(}x^{3}y-x\big{)}\ dy% \ dx=\int_{0}^{3}\int_{-\sqrt{9-y^{2}}}^{\sqrt{9-y^{2}}}\big{(}x^{3}y-x\big{)}% \ dx\ dy$ . (c) 0
19.

Integrating $e^{x^{2}}$ with respect to $x$ is not possible in terms of elementary functions. $\displaystyle\int_{0}^{2}\int_{0}^{2x}e^{x^{2}}\ dy\ dx=e^{4}-1$ .
21.

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

$\displaystyle\int_{0}^{1}\int_{0}^{x}\frac{2y}{x^{2}+y^{2}}\ dy\ dx=\ln 2$ .
23.

average value of $f=6/2=3$
25.

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

Exercises 14.3

1.

$f\big{(}r\cos\theta,r\sin\theta\big{)}$ , $r\ dr\ d\theta$
3.

$\displaystyle\int_{0}^{2\pi}\int_{0}^{1}\big{(}3r\cos\theta-r\sin\theta+4\big{% )}r\ dr\ d\theta=4\pi$
5.

$\displaystyle\int_{0}^{\pi}\int_{\cos\theta}^{3\cos\theta}\big{(}8-r\sin\theta% \big{)}r\ dr\ d\theta=16\pi$
7.

$\displaystyle\int_{0}^{2\pi}\int_{1}^{2}\big{(}\ln(r^{2})\big{)}r\ dr\ d\theta% =2\pi\big{(}\ln 16-3/2\big{)}$
9.

$\displaystyle\int_{-\pi/2}^{\pi/2}\int_{0}^{6}\big{(}r^{2}\cos^{2}\theta-r^{2}% \sin^{2}\theta\big{)}r\ dr\ d\theta\lx@parboxnewline=\int_{-\pi/2}^{\pi/2}\int% _{0}^{6}\big{(}r^{2}\cos(2\theta)\big{)}r\ dr\ d\theta=0$
11.

$\displaystyle\int_{-\pi/2}^{\pi/2}\int_{0}^{5}\big{(}r^{2}\big{)}\ dr\ d\theta% =125\pi/3$
13.

$\displaystyle\int_{0}^{\pi/4}\int_{0}^{\sqrt{8}}\big{(}r\cos\theta+r\sin\theta% \big{)}r\ dr\ d\theta=16\sqrt{2}/3$
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) $\displaystyle\int_{0}^{2\pi}\int_{0}^{a}re^{r^{2}}\ dr\ d\theta=\pi(1-e^{-a^{2% }})$ . (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 $\displaystyle\left(\int_{-\infty}^{\infty}e^{-t^{2}}\ dt\right)^{2}=\pi$ .
17.

$3\pi/4-9\sqrt{3}/16$
19.

$2$
21.

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

Exercises 14.4

1.

Because they are scalar multiples of each other.
3.

“little masses”
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.
7.

$\overline{x}=5.25$
9.

$(\overline{x},\overline{y})=(0,3)$
11.

$M=150$ g;
13.

$M=2$ lb
15.

$M=16\pi\approx 50.27$ kg
17.

$M=54\pi\approx 169.65$ lb
19.

$M=150$ g; $M_{y}=600$ ; $M_{x}=-75$ ; $(\overline{x},\overline{y})=(4,-0.5)$
21.

$M=2$ lb; $M_{y}=0$ ; $M_{x}=2/3$ ; $(\overline{x},\overline{y})=(0,1/3)$
23.

$M=16\pi\approx 50.27$ kg; $M_{y}=4\pi$ ; $M_{x}=4\pi$ ; $(\overline{x},\overline{y})=(1/4,1/4)$
25.

$M=54\pi\approx 169.65$ lb; $M_{y}=0$ ; $M_{x}=504$ ; $(\overline{x},\overline{y})=(0,2.97)$
27.

$I_{x}=64/3$ ; $I_{y}=64/3$ ; $I_{O}=128/3$
29.

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

Exercises 14.5

1.

arc length
3.

surface areas
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.
7.

$\displaystyle S=\int_{0}^{2\pi}\int_{0}^{2\pi}\sqrt{1+\cos^{2}x\cos^{2}y+\sin^% {2}x\sin^{2}y}\ dx\ dy$
9.

$\displaystyle S=\int_{-1}^{1}\int_{-1}^{1}\sqrt{1+4x^{2}+4y^{2}}\ dx\ dy$
11.

$\displaystyle S=\int_{0}^{3}\int_{-1}^{1}\sqrt{1+9+49}\ dx\ dy=6\sqrt{59}% \approx 46.09$
13.
This is easier in polar: $\displaystyle S$ $\displaystyle=\int_{0}^{2\pi}\int_{0}^{4}r\sqrt{1+4r^{2}\cos^{2}t+4r^{2}\sin^{% 2}t}\ dr\ d\theta$ $\displaystyle=\int_{0}^{2\pi}\int_{0}^{4}r\sqrt{1+4r^{2}}\ dr\ d\theta$ $\displaystyle=\frac{\pi}{6}\big{(}65\sqrt{65}-1\big{)}\approx 273.87$
15.
$\displaystyle S$ $\displaystyle=\int_{0}^{1}\int_{0}^{1}\sqrt{1+x+9y}\ dx\ dy$ $\displaystyle=\int_{0}^{1}\frac{2}{3}\Big{(}(9y+2)^{3/2}-(9y+1)^{3/2}\Big{)}\ dy$ $\displaystyle=\frac{4}{135}\big{(}121\sqrt{11}-100\sqrt{10}-4\sqrt{2}+1\big{)}% \approx 2.383$
17.
This is easier in polar: $\displaystyle S$ $\displaystyle=2\int_{0}^{2\pi}\int_{0}^{5}r\sqrt{1+\frac{r^{2}\cos^{2}t+r^{2}% \sin^{2}t}{25-r^{2}\sin^{2}t-r^{2}\cos^{2}t}}\ dr\ d\theta$ $\displaystyle=2\int_{0}^{2\pi}\int_{0}^{5}r\sqrt{\frac{1}{25-r^{2}}}\ dr\ d\theta$ $\displaystyle=100\pi\approx 314.16$

Exercises 14.6

1.

surface to surface, curve to curve and point to point
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.
5.

$V=\int_{-1}^{1}\int_{-1}^{1}\big{(}8-x^{2}-y^{2}-(2x+y)\big{)}\ dx\ dy=88/3$
7.

$V=\int_{0}^{\pi}\int_{0}^{x}\big{(}\cos x\sin y+2-\sin x\cos y\big{)}\ dy\ dx=% \pi^{2}-\pi\approx 6.728$
9.
$dz\ dy\ dx$ : $\displaystyle\int_{0}^{3}\int_{0}^{1-x/3}\int_{0}^{2-2x/3-2y}\ dz\ dy\ dx$ $dz\ dx\ dy$ : $\displaystyle\int_{0}^{1}\int_{0}^{3-3y}\int_{0}^{2-2x/3-2y}\ dz\ dx\ dy$ $dy\ dz\ dx$ : $\displaystyle\int_{0}^{3}\int_{0}^{2-2x/3}\int_{0}^{1-x/3-z/2}\ dy\ dz\ dx$ $dy\ dx\ dz$ : $\displaystyle\int_{0}^{2}\int_{0}^{3-3z/2}\int_{0}^{1-x/3-z/2}\ dy\ dx\ dz$ $dx\ dz\ dy$ : $\displaystyle\int_{0}^{1}\int_{0}^{2-2y}\int_{0}^{3-3y-3z/2}\ dx\ dz\ dy$ $dx\ dy\ dz$ : $\displaystyle\int_{0}^{2}\int_{0}^{1-z/2}\int_{0}^{3-3y-3z/2}\ dx\ dy\ dz$ $\displaystyle V=\int_{0}^{3}\int_{0}^{1-x/3}\int_{0}^{2-2x/3-2y}\ dz\ dy\ dx=1.$
11.
$dz\ dy\ dx$ : $\displaystyle\int_{0}^{2}\int_{-2}^{0}\int_{y^{2}/2}^{-y}\ dz\ dy\ dx$ $dz\ dx\ dy$ : $\displaystyle\int_{-2}^{0}\int_{0}^{2}\int_{y^{2}/2}^{-y}\ dz\ dx\ dy$ $dy\ dz\ dx$ : $\displaystyle\int_{0}^{2}\int_{0}^{2}\int_{-\sqrt{2z}}^{-z}\ dy\ dz\ dx$ $dy\ dx\ dz$ : $\displaystyle\int_{0}^{2}\int_{0}^{2}\int_{-\sqrt{2z}}^{-z}\ dy\ dx\ dz$ $dx\ dz\ dy$ : $\displaystyle\int_{-2}^{0}\int_{y^{2}/2}^{-y}\int_{0}^{2}\ dx\ dz\ dy$ $dx\ dy\ dz$ : $\displaystyle\int_{0}^{2}\int_{-\sqrt{2z}}^{-z}\int_{0}^{2}\ dx\ dy\ dz$ $\displaystyle V=\int_{0}^{2}\int_{0}^{2}\int_{-\sqrt{2z}}^{-z}\ dy\ dz\ dx=4/3.$
13.
$dz\ dy\ dx$ : $\displaystyle\int_{0}^{2}\int_{1-x/2}^{1}\int_{0}^{2x+4y-4}\ dz\ dy\ dx$ $dz\ dx\ dy$ : $\displaystyle\int_{0}^{1}\int_{2-2y}^{2}\int_{0}^{2x+4y-4}\ dz\ dx\ dy$ $dy\ dz\ dx$ : $\displaystyle\int_{0}^{2}\int_{0}^{2x}\int_{z/4-x/2+1}^{1}\ dy\ dz\ dx$ $dy\ dx\ dz$ : $\displaystyle\int_{0}^{4}\int_{z/2}^{2}\int_{z/4-x/2+1}^{1}\ dy\ dx\ dz$ $dx\ dz\ dy$ : $\displaystyle\int_{0}^{1}\int_{0}^{4y}\int_{z/2-2y+2}^{2}\ dx\ dz\ dy$ $dx\ dy\ dz$ : $\displaystyle\int_{0}^{4}\int_{z/4}^{1}\int_{z/2-2y+2}^{2}\ dx\ dy\ dz$ $\displaystyle V=\int_{0}^{4}\int_{z/4}^{1}\int_{z/2-2y+2}^{2}\ dx\ dy\ dz=4/3.$
15.
$dz\ dy\ dx$ : $\displaystyle\int_{0}^{1}\int_{0}^{1-x^{2}}\int_{0}^{\sqrt{1-y}}\ dz\ dy\ dx$ $dz\ dx\ dy$ : $\displaystyle\int_{0}^{1}\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}\ dz\ dx\ dy$ $dy\ dz\ dx$ : $\displaystyle\int_{0}^{1}\int_{0}^{x}\int_{0}^{1-x^{2}}\ dy\ dz\ dx+\int_{0}^{% 1}\int_{x}^{1}\int_{0}^{1-z^{2}}\ dy\ dz\ dx$ $dy\ dx\ dz$ : $\displaystyle\int_{0}^{1}\int_{0}^{z}\int_{0}^{1-z^{2}}\ dy\ dx\ dz+\int_{0}^{% 1}\int_{z}^{1}\int_{0}^{1-x^{2}}\ dy\ dx\ dz$ $dx\ dz\ dy$ : $\displaystyle\int_{0}^{1}\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}\ dx\ dz\ dy$ $dx\ dy\ dz$ : $\displaystyle\int_{0}^{1}\int_{0}^{1-z^{2}}\int_{0}^{\sqrt{1-y}}\ dx\ dy\ dz$ Answers will vary. Neither order is particularly “hard.” The order $dz\ dy\ dx$ requires integrating a square root, so powers can be messy; the order $dy\ dz\ dx$ requires two triple integrals, but each uses only polynomials.
17.

8
19.

$\pi$
21.
$M=10$ , $M_{yz}=15/2$ , $M_{xz}=5/2$ , $M_{xy}=5$ ; $(\overline{x},\overline{y},\overline{z})=(3/4,1/4,1/2)$
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.

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.
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.
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 $(0,0,1)$
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 $(1,0,0)$
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.
11.

$\displaystyle\int_{\theta_{1}}^{\theta_{2}}\int_{r_{1}}^{r_{2}}\int_{z_{1}}^{z% _{2}}h(r,\theta,z)r\ dz\ dr\ d\theta$
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 $90^{\circ}$ .
15.

Bounded between the plane $z=1$ 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.
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.
19.

Describes the portion of the unit ball that resides in the first octant.
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.
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 $\tan(\pi/6)=1/\sqrt{3}$ .
25.
In cylindrical coordinates, the density is $\delta(r,\theta,z)=r+1$ . Thus mass is $\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{4}(r+1)r\ dz\ dr\ d\theta=112\pi/3.$
27.
In cylindrical coordinates, the density is $\delta(r,\theta,z)=1$ . Thus mass is $\int_{0}^{\pi}\int_{0}^{1}\int_{0}^{4-r\sin\theta}r\ dz\ dr\ d\theta=2\pi-2/3% \approx 5.617.$
29.
In cylindrical coordinates, the density is $\delta(r,\theta,z)=r+1$ . Thus mass is $M=\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{4}(r+1)r\ dz\ dr\ d\theta=112\pi/3.$

We find $M_{yz}=0$ , $M_{xz}=0$ , and $M_{xy}=224\pi/3$ , placing the center of mass at $(0,0,2)$ .
31.
In cylindrical coordinates, the density is $\delta(r,\theta,z)=1$ . Thus mass is $\int_{0}^{\pi}\int_{0}^{1}\int_{0}^{4-r\sin\theta}r\ dz\ dr\ d\theta=2\pi-2/3% \approx 5.617.$

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 $\approx(0,0.405,1.80)$ .
33.
In spherical coordinates, the density is $\delta(\rho,\theta,\varphi)=1$ . Thus mass is $\int_{0}^{\pi/2}\int_{0}^{2\pi}\int_{0}^{1}\rho^{2}\sin(\varphi)\ d\rho\ d% \theta\ d\varphi=2\pi/3.$
35.
In spherical coordinates, the density is $\delta(\rho,\theta,\varphi)=\rho\cos\varphi$ . Thus mass is $\int_{0}^{\pi/4}\int_{0}^{2\pi}\int_{0}^{1}\big{(}\rho\cos(\varphi)\big{)}\rho% ^{2}\sin(\varphi)\ d\rho\ d\theta\ d\varphi=\pi/8.$
37.
In spherical coordinates, the density is $\delta(\rho,\theta,\varphi)=1$ . Thus mass is $\int_{0}^{\pi/2}\int_{0}^{2\pi}\int_{0}^{1}\rho^{2}\sin(\varphi)\ d\rho\ d% \theta\ d\varphi=2\pi/3.$

We find $M_{yz}=0$ , $M_{xz}=0$ , and $M_{xy}=\pi/4$ , placing the center of mass at $(0,0,3/8)$ .
39.
In spherical coordinates, the density is $\delta(\rho,\theta,\varphi)=\rho\cos\varphi$ . Thus mass is $\int_{0}^{\pi/4}\int_{0}^{2\pi}\int_{0}^{1}\big{(}\rho\cos(\varphi)\big{)}\rho% ^{2}\sin(\varphi)\ d\rho\ d\theta\ d\varphi=\pi/8.$

We find $M_{yz}=0$ , $M_{xz}=0$ , and $M_{xy}=(4-\sqrt{2})\pi/30$ , placing the center of mass at $(0,0,4(4-\sqrt{2})/15)$ .
41.

Rectangular: $\int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{-\sqrt{1-x^{2}-y^{2}% }}^{\sqrt{1-x^{2}-y^{2}}}\ dz\ dy\ dx$

Cylindrical: $\int_{0}^{2\pi}\int_{0}^{1}\int_{-\sqrt{1-r^{2}}}^{\sqrt{1-r^{2}}}r\ dz\ dr\ d\theta$

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

Spherical appears simplest, avoiding the integration of square-roots and using techniques such as Substitution; all bounds are constants.
43.

Rectangular: $\int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{\sqrt{x^{2}+y^{2}}}^% {1}\ dz\ dy\ dx$

Cylindrical: $\int_{0}^{2\pi}\int_{0}^{1}\int_{r}^{1}r\ dz\ dr\ d\theta$

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

Cylindrical appears simplest, avoiding the integration of square-roots that rectangular uses. Spherical is not difficult, though it requires Substitution, an extra step.
45.

center: $(2,3,-1)$ , radius: $\sqrt{14}$
47.

$(a,\theta,a\cot\phi)$
49.

Hint: Use the distance formula for Cartesian coordinates.
51.
53.

$1-\sin 2/2$