Problem 1. Compute each of the following volumes enclosed by two surfaces.

(a) The sphere $x^2+y^2+z^2=a^2$ and cylinder $x^2+y^2=ax$;

(b) The cone $z=\sqrt{x^2+y^2}$ and the cylinder $z^2=2x$;

Problem 2. Find the center of mass of the following plane regions.

(a) $D$ enclosed by $y=\sqrt{2px}, \,\, x=x_0>0, \,\, y=0$;

(b) $D$ is the semi-ellipse $\left\{\left(x,y\right)\mid \frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1, y\geq 0 \right\}$;

(b) $D$ enclosed by two disks $\rho=a\cos\theta$ and $\rho=b\cos\theta$ ($0<a<b$, $-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2})$.

Problem 3. Find the center of mass of the plane region $D$ enclosed by $y=x^2$ and $y=x$, with density function

$$\mu \left(x,y\right) = x^2y.$$

Problem 4. Compute the center of mass of the following solid bodies (using triple integrals, of course). Assume the density is identically equal to $1$.

(a) $\Omega$ enclosed by $z^2=x^2+y^2$ and $z=1$;

(b) $\Omega$ enclosed by $z=\sqrt{A^2-x^2-y^2}$, $z=\sqrt{a^2-x^2-y^2}$, and $z=0$ (assuming $A>a>0$);

Problem 5. Compute the moments of inertia $I_x$, $I_y$, $I_0$ for the following plane regions.

(a) $D=\left\{\left(x,y\right)\mid \frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1\right\}$;

(b) $D$ enclosed by $y^2=\frac{9}{2}x$ and $x=2$;

Problem 6. Consider the solid body $\Omega$ (with constant density $\rho_0$) enclosed by the surface $z=x^2+y^2$, $z=0$, $\left|x\right|=a, \left|y\right|=a$. Compute:

(1) The volume of $\Omega$;

(2) The center of mass of $\Omega$;

(3) The moment of inertia of $\Omega$ with respect to the $z$ axis.

Problem 7. Find the moment of inertial of a cylinder with respect to its axis of symmetry, assuming the height of the cylinder is $h$, the radius of its base is $a$, and the density is identically equal to $1$.

Problem 8. Use triple integral to compute the volume of the unit ball $\left\{\left(x,y,z\right)\mid x^2+y^2+z^2\leq 1\right\}$.

Answers

Problem 1.

(a) $\frac{2}{3}a^3\left(\pi-\frac{4}{3}\right)$ $\qquad$ (b) $2\pi-\frac{32}{9}$ $\qquad$ (c) $\frac{16}{3}R^3$

Problem 2.

(a) $\bar{x} = \frac{3}{5}x_0$, $\bar{y}=\frac{3}{8}\sqrt{2px_0}$. $\quad$ (b) $\bar{x}=0$, $\bar{y}=\frac{4b}{3\pi}$. $\quad$ (c) $\bar{x}=\frac{a^2+ab+b^2}{2\left(a+b\right)}$, $\bar{y}=0$.

Problem 3.

$\bar{x} = \frac{35}{48}$, $\bar{y} = \frac{35}{54}$.

Problem 4.

(a) $\left(0,0,\frac{3}{4}\right)$ $\quad$ (b) $\left(0,0,\frac{3\left(A^4-^4\right)}{8\left(A^3-a^3\right)}\right)$ $\quad$ (c) $\left(\frac{2}{5}a, \frac{2}{5}a, \frac{7}{30}a^2\right)$

Problem 5.

(a) $I_x = \frac{1}{4}\pi ab^3$, $I_y = \frac{1}{4}\pi a^3b$ $\quad$ (b) $I_x = \frac{72}{5}$, $I_y=\frac{96}{7}$ $\quad$ (c) $I_x=\frac{1}{3}ab^3$, $I_y=\frac{1}{3}a^3b$

Problem 6.

(1) $\frac{8}{3}a^4$ $\quad$ (2) $\bar{x}=\bar{y}=0,\,\, \bar{z}=\frac{7}{15}a^2$ $\quad$ (3) $\frac{112}{45}a^6\rho_0$

Problem 7.

$\frac{\pi}{2}a^4h$

Problem 8.

$\frac{4}{3}\pi$

Solutions to Selected Problems

Problem 1 (a). The volume is bounded by two graphs $z_\mathrm{top}=\sqrt{a^2-x^2-y^2}$ and $z_\mathrm{bottom}=-\sqrt{a^2-x^2-y^2}$ over the region $x^2+y^2=ax$. The volume is thus equal to the double integral

\begin{align*} &\iint_{\left\{\left(x,y\right)\mid x^2+y^2-ax\leq 0\right\}}\left[\sqrt{a^2-x^2-y^2}-\left(-\sqrt{a^2-x^2-y^2}\right)\right]dA\\ &=\iint_{\left\{\left(x,y\right)\mid x^2+y^2-ax\leq 0\right\}}2\sqrt{a^2-x^2-y^2}\,dA \end{align*}

Note that the region defined by $x^2+y^2-ax\leq 0$ can be written in polar coordinates as $r\leq a\cos\theta$, with $\theta$ ranging from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. Thus the above integral equals

\begin{align*} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^{a\cos\theta}2\sqrt{a^2-r^2}\,rdr\,d\theta. \end{align*}

The inner integral can be computed via a substitution $u=r^2$:

\begin{align*} &\int_0^{a\cos\theta}2\sqrt{a^2-r^2}\,rdr=\int_0^{a^2\cos^2\theta}\sqrt{a^2-u}\,du\\ &=-\frac{2}{3}\left(a^2-u\right)^{\frac{3}{2}}\Bigg|_{u=0}^{u=a^2\cos^2\theta}=-\frac{2}{3}a^3\sin^3\theta+\frac{2}{3}a^3=\frac{2}{3}a^3\left(1-\sin^3\theta\right). \end{align*}

Plug this back into the outer integral, one has

\begin{align*} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{2}{3}a^3\left(1-\sin^3\theta\right)\,d\theta&=\frac{2}{3}a^3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(1-\sin^3\theta\right)\,d\theta\\ &=\frac{2}{3}a^3\left(\pi-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^3\theta\,d\theta\right)=\frac{2}{3}a^3\left(\pi-\frac{4}{3}\right). \end{align*}

The computation of the last integral involving $\sin^3\theta$ is standard (see page TA-3 in the textbook). It can be done as follows:

\begin{align*} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^3\theta\,d\theta &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2\theta\cdot\sin\theta\,d\theta\\ &=-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(1-\cos^2\theta\right)d\cos\theta=-2\int_{0}^{\frac{\pi}{2}}\left(1-\cos^2\theta\right)d\cos\theta\\ &\stackrel{t=\cos\theta}{=\!=\!=\!=\!=\!=}-2\int_1^0\left(1-t^2\right)dt=2\int_0^1\left(1-t^2\right)dt=2\left(1-\frac{2}{3}\right)=\frac{4}{3}. \end{align*}

Problem 1 (b). The volume is bounded by two graphs $z_\mathrm{top}=2x$ and $z_\mathrm{bottom}=\sqrt{x^2+y^2}$ over the region $x^2+y^2-2x\leq 0$. (The boundary of the region, which is $x^2+y^2-2x=0$, can be determined by computing the intersection of $z=\sqrt{x^2+y^2}$ and $z^2=2x$.) Similar to Problem 1 (a), this region can be represented in polar coordinates as $r\leq 2\cos\theta$ with $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. The volume can thus be computed as

\begin{align*} &\iint_{\left\{\left(x,y\right)\mid x^2+y^2-2x\leq 0\right\}}\left[2x-\sqrt{x^2+y^2}\right]dA=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^{2\cos\theta}\left(2r\cos\theta-r\right)\,rdr\,d\theta\\ =&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(2\cos\theta-1\right)\int_0^{2\cos\theta}r^2dr\,d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(2\cos\theta-1\right)\cdot\frac{8}{3}\cos^3\theta\,d\theta\\ =&\frac{16}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^4\theta\,d\theta-\frac{8}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^3\theta\,d\theta=\frac{16}{3}\times \frac{3\pi}{8}-\frac{8}{3}\times \frac{4}{3}=2\pi-\frac{32}{9}. \end{align*}

The integrals involving $\cos^3\theta$ and $\cos^4\theta$ can be found on page TA-3 in the textbook, or computed as follows:

\begin{align*} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^3\theta\,d\theta&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2\theta\,d\sin\theta\stackrel{u=\cos\theta}{=\!=\!=\!=\!=\!=}\int_{-1}^1\left(1-u^2\right)du=2-\frac{2}{3}=\frac{4}{3},\\ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^4\theta\,d\theta&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^3\theta\,d\sin\theta=\sin\theta\cos^3\theta\Bigg|_{\theta=-\frac{\pi}{2}}^{\theta=\frac{\pi}{2}}-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin\theta\,d\cos^3\theta\\ &=0-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin\theta\cdot 3\cos^2\theta \left(-\sin\theta\right)\,d\theta\\ &=3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2\theta \cos^2\theta \,d\theta=\frac{3}{4}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(2\sin\theta\cos\theta\right)^2d\theta\\ &=\frac{3}{4}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2 2\theta\,d\theta=\frac{3}{4}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1-\cos 4\theta}{2}d\theta\\ &=\frac{3}{4}\left(\frac{\pi}{2}-\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos 4\theta\,d\theta\right)\\ &=\frac{3}{4}\left(\frac{\pi}{2}-\frac{1}{2}\times\frac{1}{4}\sin 4\theta\Bigg|_{\theta = -\frac{\pi}{2}}^{\theta=\frac{\pi}{2}}\right)=\frac{3}{4}\times\frac{\pi}{2}=\frac{3\pi}{8}. \end{align*}

Problem 1 (c). The volume is two graphs $z_\mathrm{top}=\sqrt{R^2-x^2}$ and $z_\mathrm{bottom}=-\sqrt{R^2-x^2}$ over the (disk of radius $R$) region $x^2+y^2\leq R^2$. Thus the desired volume is

two_cylinders

\begin{align*} &\iint_{\left\{\left(x,y\right)\mid x^2+y^2\leq R^2\right\}}\left[\sqrt{R^2-x^2}-\left(-\sqrt{R^2-x^2}\right)\right]dA=\iint_{\left\{\left(x,y\right)\mid x^2+y^2\leq R^2\right\}}2\sqrt{R^2-x^2}\,dA\\ =&\int_0^{2\pi}\!\!\!\int_0^R 2\sqrt{R^2-r^2\cos^2\theta}\,\,rdr\,d\theta=\int_0^{2\pi}\!\!\!\int_0^R \sqrt{R^2-r^2\cos^2\theta}\,\,d\left(r^2\right)\,d\theta\\ =&4\int_0^{\frac{\pi}{2}}\!\!\!\int_0^R \sqrt{R^2-r^2\cos^2\theta}\,\,d\left(r^2\right)\,d\theta\\ &\!\!\!\!\stackrel{u=r^2}{=\!=\!=\!=\!=\!=}4\int_0^{\frac{\pi}{2}}\!\!\!\int_0^{R^2}\sqrt{R^2-u\cos^2\theta}\,\,du\,d\theta=4\int_0^{\frac{\pi}{2}}-\frac{2}{3\cos^2\theta}\left(R^2-u\cos^2\theta\right)^{\frac{3}{2}}\Bigg|_{u=0}^{u=R^2}\,d\theta\\ =&4\int_0^{\frac{\pi}{2}}\frac{2}{3\cos^2\theta}R^3-\frac{2}{3\cos^2\theta}\left(R^2-R^2\cos^2\theta\right)^{\frac{3}{2}}d\theta=4\int_0^{\frac{\pi}{2}}\frac{2R^3}{3\cos^2\theta}\left(1-\sin^3\theta\right)\,d\theta\\ =&\frac{8R^3}{3}\int_0^{\frac{\pi}{2}}\frac{1-\sin^3\theta}{\cos^2\theta}\,d\theta=\frac{16R^3}{3}. \end{align*}

Note that the last step follows from the following computation

\begin{align*} &\int_0^{\frac{\pi}{2}}\frac{1-\sin^3\theta}{\cos^2\theta}\,d\theta=\int_0^{\frac{\pi}{2}}\frac{\left(1-\sin\theta\right)\left(1+\sin\theta+\sin^2\theta\right)}{1-\sin^2\theta}\,d\theta\\ =&\int_0^{\frac{\pi}{2}}\frac{\left(1-\sin\theta\right)\left(1+\sin\theta+\sin^2\theta\right)}{\left(1-\sin\theta\right)\left(1+\sin\theta\right)}\,d\theta\\ =&\int_0^{\frac{\pi}{2}}\frac{1+\sin\theta+\sin^2\theta}{1+\sin\theta}\,d\theta=\int_0^{\frac{\pi}{2}}\frac{1+\sin\theta\left(1+\sin\theta\right)}{1+\sin\theta}\,d\theta\\ =&\int_0^{\frac{\pi}{2}}\frac{d\theta}{1+\sin\theta}+\int_0^{\frac{\pi}{2}}\sin\theta\,d\theta=\int_0^{\frac{\pi}{2}}\frac{d\theta}{1+\sin\theta}+1\\ =&\int_0^{\frac{\pi}{2}}\frac{d\theta}{1+2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}+1\\ =&\int_0^{\frac{\pi}{2}}\frac{d\theta}{\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2}+2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}+1\\ =&\int_0^{\frac{\pi}{2}}\frac{d\theta}{\left(\sin\frac{\theta}{2}+\cos\frac{\theta}{2}\right)^2}+1\\ =&\int_0^{\frac{\pi}{2}}\frac{d\theta}{\cos^2\frac{\theta}{2}\left(\tan\frac{\theta}{2}+1\right)^2}+1\\ =&\int_0^{\frac{\pi}{2}}\frac{d\left(2\tan\frac{\theta}{2}\right)}{\left(\tan\frac{\theta}{2}+1\right)^2}+1\qquad\textrm{becase $\frac{d}{d\theta}\tan\frac{\theta}{2}=\frac{1}{2\cos^2\frac{\theta}{2}}$}\\ &\!\!\!\!\stackrel{u=tan\frac{\theta}{2}}{=\!=\!=\!=\!=\!=}\int_0^{1}\frac{2du}{\left(u+1\right)^2}\,d\theta+1=-\frac{2}{u+1}\Bigg|_{u=0}^{u=1}+1=-1+2+1=2. \end{align*}

The region enclosd by two cylinders of equal radii is a highly interesting object in geometry and goes a long way back in the pre-history of calculus. Check this article from Wolfram MathWorld and this page from TAMU for more details.

Problem 2 (a). By definition,

\begin{align*} m&=\iint_D\,dx\,dy=\int_0^{x_0}\!\!\!\int_0^{\sqrt{2px}}dy\,dx=\int_0^{x_0}\sqrt{2px}\,dx=\frac{2}{3}\sqrt{2px_0^3},\\ \bar{x}&=\frac{1}{m}\iint_Dx\,dx\,dy=\frac{1}{m}\int_0^{x_0}\!\!\!\int_0^{\sqrt{2px}}x\,dy\,dx=\frac{1}{m}\int_0^{x_0}\sqrt{2px^3}\,dx\\ &=\frac{3}{2\sqrt{2px_0^3}}\cdot\frac{2}{5}\sqrt{2px_0^5}=\frac{3}{5}x_0,\\ \bar{y}&=\frac{1}{m}\iint_Dy\,dx\,dy=\frac{1}{m}\int_0^{x_0}\!\!\!\int_0^{\sqrt{2px}}y\,dy\,dx=\frac{1}{m}\int_0^{x_0}\frac{1}{2}\cdot 2px\,dx\\ &=\frac{3}{2\sqrt{2px_0^3}}\cdot\frac{1}{2}px_0^2=\frac{3}{8}\sqrt{2px_0}. \end{align*}

Problem 3. By definition,

\begin{align*} m&=\iint_Dx^2y\,dA=\int_0^1\!\!\!\int_{x^2}^xx^2y\,dy\,dx=\int_0^1\frac{x^2}{2}\left(x^2-x^4\right)\,dx\\ &=\left(\frac{x^5}{10}-\frac{x^7}{14}\right)\Bigg|_{x=0}^{x=1}=\frac{1}{10}-\frac{1}{14}=\frac{1}{35},\\ \bar{x}&=\frac{1}{m}\iint_Dx\cdot x^2y\,dA=35\int_0^1\!\!\!\int_{x^2}^xx^3y\,dy\,dx=35\int_0^1\frac{x^3}{2}\left(x^2-x^4\right)\,dx\\ &=\frac{35}{2}\left(\frac{x^6}{6}-\frac{x^8}{8}\right)\Bigg|_{x=0}^{x=1}=\frac{35}{2}\cdot\frac{1}{24}=\frac{35}{48},\\ \bar{y}&=\frac{1}{m}\iint_Dy\cdot x^2y\,dA=35\int_0^1\!\!\!\int_{x^2}^xx^2y^2\,dy\,dx=35\int_0^1\frac{x^2}{3}\left(x^3-x^6\right)\,dx\\ &=\frac{35}{3}\left(\frac{x^6}{6}-\frac{x^9}{9}\right)\Bigg|_{x=0}^{x=1}=\frac{35}{3}\cdot\frac{3}{54}=\frac{35}{54}. \end{align*}

Problem 4 (a). By symmetry, $\bar{x}=0$, $\bar{y}=0$. The total mass is (note that the density is identically $1$)

\begin{align*} m&=\iiint_\Omega dV=\int_0^1\iint_{\left\{\left(x,y\right)\mid x^2+y^2\leq z^2\right\}}\,dA\,dz\\ &=\int_0^1\pi z^2\,dz=\frac{\pi}{3}z^3\Bigg|_{z=0}^{z=1}=\frac{\pi}{3}, \end{align*}

thus the $z$-coordinate of the center of mass is

\begin{align*} \bar{z}&=\frac{1}{m}\iiint_\Omega z\,dV=\frac{3}{\pi}\int_0^1\iint_{\left\{\left(x,y\right)\mid x^2+y^2\leq z^2\right\}}z\,dA\,dz\\ &=\frac{3}{\pi}\int_0^1\pi z^2\cdot z\,dz=\frac{3}{\pi}\cdot\frac{\pi}{4}z^4\Bigg|_{z=0}^{z=1}=\frac{3}{4}. \end{align*}

Problem 5 (a). Direct computation gives

\begin{align*} I_y&=\iint_D x^2\,dA=\int_{-a}^a\!\!\!\int_{-\frac{b}{a}\sqrt{a^2-x^2}}^{\frac{b}{a}\sqrt{a^2-x^2}}x^2\,dy\,dx\\ &=\frac{2b}{a}\int_{-a}^ax^2\sqrt{a^2-x^2}\,dx=\frac{4b}{a}\int_{0}^ax^2\sqrt{a^2-x^2}\,dx\\ &\stackrel{x=\sin\theta}{=\!=\!=\!=\!=\!=}\frac{4b}{a}\int_{0}^{\frac{\pi}{2}}a^3\sin^2\theta\cos\theta\cdot a\cos\theta\,d\theta\\ &=4a^3b\int_{0}^{\frac{\pi}{2}}\left(\sin^2\theta\left(1-\sin^2\theta\right)\right)\,d\theta\\ &=4a^3b\left[\int_{0}^{\frac{\pi}{2}}\sin^2\theta\,d\theta-\int_{0}^{2\pi}\sin^4\theta\,d\theta\right]\\ &=4a^3b\left(\frac{\pi}{4}-\frac{3\pi}{8}\right)=\frac{1}{4}\pi a^3b. \end{align*}

(The integral involving $\sin^4\theta$ is computed using the same idea as that used for integrating $\cos^4\theta$ in the solution to Problem 1 (b).)

By symmetry, $I_x=\frac{1}{4}\pi ab^3$. Thus

$$I_0=I_x+I_y=\frac{\pi}{4}ab\left(a^2+b^2\right).$$

Problem 6. Take a look at the graph of the function $z=x^2+y^2$, as follows. (Rotate/Zoom with mouse.)

(1) The volume of $\Omega$ equals to the volume under the graph $z=x^2+y^2$ over the rectangle bounded by $\left|x\right|\leq a$, $\left|y\right|\leq a$. Thus

\begin{align*} \mathrm{Vol}\left(\Omega\right)&=\iint_{\left\{\left(x,y\right)\mid \left|x\right|\leq a, \left|y\right|\leq a\right\}}\left(x^2+y^2\right)\,dA\\ &=\int_{-a}^a\int_{-a}^a\left(x^2+y^2\right)\,dx\,dy\\ &=\int_{-a}^a \left(\frac{1}{3}x^3+xy^2\right)\Bigg|_{x=-a}^{x=a} \,dy\\ &=\int_{-a}^a\left(\frac{2}{3}a^3+2ay^2\right)\,dy\\ &=\frac{2}{3}a^3\cdot 2a+2a\cdot \frac{2}{3}a^3=\frac{8}{3}a^4. \end{align*}

(2) Since the density of $\Omega$ is constant $\rho_0$, the total mass is

$$m=\rho_0\cdot\mathrm{Vol}\left(\Omega\right)=\frac{8}{3}a^4\rho_0$$

The $z$-coordinate of the center of mass is thus

\begin{align*} \bar{z}&=\frac{1}{m}\iiint_{\Omega}\rho_0 z\,dV=\frac{1}{m}\int_{-a}^a\!\!\int_{-a}^a\!\!\int_0^{x^2+y^2}\rho_0 z\,dz\,dy\,dx\\ &=\frac{\rho_0}{m}\int_{-a}^a\!\!\int_{-a}^a\frac{1}{2}\left(x^2+y^2\right)^2\,dy\,dx\\ &=\frac{\rho_0}{2m}\int_{-a}^a\!\!\int_{-a}^a\left(x^4+2x^2y^2+y^4\right)\,dy\,dx\\ &=\frac{\rho_0}{2m}\int_{-a}^a\left(x^4y+\frac{2}{3}x^2y^3+\frac{1}{5}y^5\right)\Bigg|_{y=-a}^{y=a}dx\\ &=\frac{\rho_0}{2m}\int_{-a}^a\left(2ax^4+\frac{4a^3}{3}x^2+\frac{2a^5}{5}\right)\,dx\\ &=\frac{\rho_0}{2m}\left(2a\cdot\frac{1}{5}x^5+\frac{4a^3}{3}\cdot\frac{1}{3}x^3+\frac{2a^5}{5}x\right)\Bigg|_{x=-a}^{x=a}\\ &=\frac{3\rho_0}{16a^4\rho_0}\left(2a\cdot\frac{2}{5}a^5+\frac{4a^3}{3}\cdot\frac{2a^3}{3}+\frac{2a^5}{5}\cdot 2a\right)\\ &=\frac{3}{16 a^4}\cdot \frac{112}{45}a^6=\frac{7}{15}a^2. \end{align*}

By symmetry, $\bar{x}=0$, $\bar{y}=0$.

(3) The moment of inertia is

\begin{align*} I_0&=\iiint_\Omega \rho_0\left(x^2+y^2\right)dV=\rho_0\iiint_\Omega\left(x^2+y^2\right)dV\\ &=\rho_0\int_{-a}^a\!\!\int_{-a}^a\!\!\int_0^{x^2+y^2}\left(x^2+y^2\right)\,dz\,dy\,dx\\ &=\rho_0\int_{-a}^a\!\!\int_{-a}^a\left(x^2+y^2\right)^2\,dy\,dx\\ &=\rho_0\cdot \frac{112}{45}a^6=\frac{112}{45}a^6\rho_0. \end{align*}

Note that in the computation above we reused the intermediate result in (2) that

$$\int_{-a}^a\!\!\int_{-a}^a\left(x^2+y^2\right)^2\,dy\,dx=\frac{112}{45}a^6.$$

Problem 7. Consider a Cartesian coordinate system with origin at the geometric center of the cylinder, with $z$-axis aligned with the axis of symmetry. Then the cylinder occupies the following region in $\mathbb{R}^3$:

$$\Omega = \left\{\left(x,y,z\right) \mid x^2+y^2\leq a^2, -\frac{h}2\leq z\leq \frac{h}{2}\right\}.$$

The desired moment of inertia is with respect to the $z$-axis. Therefore,

\begin{align*} I_0&=\iiint_{\Omega}\left(x^2+y^2\right)dV=\int_{-\frac{h}{2}}^{\frac{h}{2}}\iint_{\left\{\left(x,y\right)\mid x^2+y^2\leq a^2\right\}}\left(x^2+y^2\right)dA\,dz\\ &=\int_{-\frac{h}{2}}^{\frac{h}{2}}\int_0^{2\pi}\int_{0}^{a}r^2\cdot rdr\,d\theta\,dz=\int_{-\frac{h}{2}}^{\frac{h}{2}}\int_0^{2\pi}\frac{a^4}{4}\,d\theta\,dz\\ &=\int_{-\frac{h}{2}}^{\frac{h}{2}}2\pi\cdot \frac{a^4}{4}\,dz=\frac{\pi a^4}{2}h. \end{align*}

Problem 8. Note that the ball is symmetric so it doesn't matter which direction you pick to slice. Let's try slicing along the $z$-axis. The bounds in $z$ is from $-1$ to $1$. For each fixed $z$, the slice of the ball has equation $x^2+y^2\leq 1-z^2$, i.e. it is a disk in the $xy$-plane of radius $\sqrt{1-z^2}$. Thus

\begin{align*} \mathrm{Vol}\left(B\right)&=\iiint_BdV=\int_{-1}^1\iint_{\left\{\left(x,y\right)\mid x^2+y^2\leq 1-z^2\right\}}\,dA \,dz\\ &=\int_{-1}^1\mathrm{Area}\left(\left\{\left(x,y\right)\mid x^2+y^2\leq 1-z^2\right\}\right)\,dz\\ &=\int_{-1}^1\pi \left(1-z^2\right)\,dz=\pi\left(z-\frac{1}{3}z^3\right)\Bigg|_{z=-1}^{z=1}\\ &=\pi\left(\frac{2}{3}-\left(-\frac{2}{3}\right)\right)=\frac{4}{3}\pi. \end{align*}

In principio erat Verbum

Quiz 2 Practice Problems

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Solutions to Selected Problems

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Solutions to Selected Problems

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