\documentclass[12pt]{article}
%\pagestyle{empty}
\setlength{\topmargin}{-1in}
\setlength{\oddsidemargin}{-.3in}
\setlength{\textheight}{9in}
\setlength{\textwidth}{7.2in}

\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{enumerate}

\newcommand{\adj}{\mathrm{Adj}}
\newcommand{\mbf}{\mathbf}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ra}{\rightarrow}
\newcommand{\answer}{\textbf{Answer}}
\newcommand{\ol}{\overline}

\begin{document}


{\Large Key: Take home 3}

\begin{enumerate}[1.]





\item  Recall that the average value of a function of one variable on the interval $[a,b]$ is defined by
	\[
	f_{ave} = \frac{\int_a^b f(x) \; dx}{b-a}
	\]
	Similarly, we can define average value for functions of two variables on a bounded set $D$.
	If $f(x,y)$ is a function of two variables, we define
	\[
	f_{ave} = \frac{\iint_D f(x,y) \; dA}{\iint_D dA}
	\]
\begin{enumerate}[(a)]
\item  Find the average value of $f(x,y)=e^{x+y}$ over the triangle with vertices $(0,0), (0,1), (1,0)$.

\answer: Notice that the integral in the denominator is the area of the region, which for us is $\frac{1}{2}$,
so there is no need to compute this integral.
Next the only issue is really to find the limit of the integral.
\begin{eqnarray*}
f_{ave} & = & \frac{\iint_D f(x,y) \; dA}{1/2} 
	= 2 \iint_D e^{x+y} \; dA \\
	& = & 2\int_0^1 \int_0^{-x+1} e^{x+y} \; dy \; dx \\
	& = & 2\int_0^1 e-e^x \; dx \\
	& = & 2(1) = 2
\end{eqnarray*}

\item  How would you define average value for a function of three variables?

\answer:
	\[
	f_{ave} = \frac{\iiint_D f(x,y,z) \; dV}{\iiint_D dV}
	\]


\item  Find the average value of $f(x,y,z)=2x+3y+z$ over the domain $D$ defined by
	\[
	1 \leq x \leq 2 \qquad
	-2 \leq y \leq 1 \qquad
	0 \leq z \leq 1
	\]

\answer:
Again, the denominator is the volume of the domain, which for us is $3$.
Thus,
\begin{eqnarray*}
f_{ave} & = & \frac{\iiint_D f(x,y,z) \; dV}{3} \\
	& = & \frac{1}{3} \int_1^2 \int_{-2}^1 \int_0^1 (2x+3y+z) \; dz \; dy \; dx \\
	& = & \frac{1}{3}(6) = 2
\end{eqnarray*}


\end{enumerate}


















\item  Consider the solid cylinder $x^2+y^2 \leq 1$, $1 \leq z \leq 2$ with density $\delta = (x^2+y^2)z^2$.
\begin{enumerate}[(a)]
\item  Find the mass of the solid.

\answer:  Let $C$ be the solid cylinder.  We need to compute
\[
M = \iiint_C \delta \; dV
\]
We convert to cylindrical coordinate and we have $\delta = r^2z^2$.
Our limits of integration will be:
\[
0 \leq \theta \leq 2\pi \qquad
0 \leq r \leq 1 \qquad
1 \leq z \leq 2
\]
So we get:
\begin{eqnarray*}
M & = & \iiint_C \delta \; dV \\
	& = & \int_1^2 \int_0^{2\pi} \int_0^1 r^2 z^2 r \; dr \; d\theta \; dz \\
	& = & \frac{7\pi}{6}
\end{eqnarray*}

\item  Find the center of mass of the solid.

\answer: we need to find the integrals:
\[
\ol{x}=\frac{\iiint_C x \delta \; dV}{M} \qquad
\ol{y}=\frac{\iiint_C y \delta \; dV}{M} \qquad
\ol{z}=\frac{\iiint_C z \delta \; dV}{M} \qquad
\]
We compute these using cylindrical coordinates.
We first do $\ol{x}$:
\begin{eqnarray*}
\ol{x} & = & \frac{\iiint_C x \delta \; dV}{M} \\
	& = & \frac{6}{7\pi} \int_1^2 \int_0^{2\pi} \int_0^1 (r \cos \theta) (r^2 z^2) r \; dr \; d\theta \; dz \\
	& = & \frac{6}{7\pi} \int_1^2 \int_0^{2\pi} \int_0^1 r^4 z^2 \cos \theta \; dr \; d\theta \; dz \\
	& = & 0
\end{eqnarray*}
(We could have seen this by using symmetry).
Next for $\ol{y}$:
\begin{eqnarray*}
\ol{y} & = & \frac{\iiint_C y \delta \; dV}{M} \\
	& = & \frac{6}{7\pi} \int_1^2 \int_0^{2\pi} \int_0^1 (r \sin \theta) (r^2 z^2) r \; dr \; d\theta \; dz \\
	& = & \frac{6}{7\pi} \int_1^2 \int_0^{2\pi} \int_0^1 r^4 z^2 \sin \theta \; dr \; d\theta \; dz \\
	& = & 0
\end{eqnarray*}
(We could have also seen this by using symmetry).
Now, for $\ol{z}$:
\begin{eqnarray*}
\ol{z} & = & \frac{\iiint_C z \delta \; dV}{M} \\
	& = & \frac{6}{7\pi} \int_1^2 \int_0^{2\pi} \int_0^1 (z) (r^2 z^2) r \; dr \; d\theta \; dz \\
	& = & \frac{6}{7\pi} \int_1^2 \int_0^{2\pi} \int_0^1 r^3 z^3 \; dr \; d\theta \; dz \\
	& = & \frac{6}{7\pi} \left( \frac{15\pi}{8} \right) \\
	& = & \frac{45}{28}
\end{eqnarray*}
So the center of mass is
\[
\left( 0, 0, \frac{45}{28} \right)
\]
\end{enumerate}












\item  Let $E$ be the ellipsoid
\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1
\]
where $a,b,c$ are all positive.
\begin{enumerate}[(a)]
\item  Find the volume of $E$.

\answer: We need to compute:
	\[
	\iiint_E dV
	\]
To do this we first change variables:
	\[
	x=au \qquad
	y=bv \qquad
	z=cw 
	\]
The Jacobian of this transformation is
	\begin{eqnarray*}
	J & = & \det \left[
		\begin{array}{ccc}
		a & 0 & 0 \\
		0 & b & 0 \\
		0 & 0 & c 
		\end{array} \right]
	= abc
	\end{eqnarray*}
Our integral now becomes:
	\[
	\iiint_E dV = \iiint abc \; du \; dv \; dw
	\]
Notice that the transformation changes the ellipsoid into a sphere of radius 1.
So, our integral (in $u,v,w$) is now over the unit ball $B$: $u^2+v^2+w^2 \leq 1$.
So we now have
	\[
	\iiint_E dV = \iiint_B abc \; du \; dv \; dw
	\]
Now, this is perfect to convert it into spherical coordinates:
	\[
	u= \rho \sin \phi \cos \theta \qquad
	v = \rho \sin \phi \sin \theta \qquad
	w = \rho \cos \phi
	\]
And, of course our limits of integration will be
	\[
	0 \leq \rho \leq 1 \qquad
	0 \leq \phi \leq \pi \qquad
	0 \leq \theta \leq 2\pi
	\]
So, finally we have:
	\begin{eqnarray*}
	\iiint_E dV & = & \int_0^\pi \int_0^1 \int_0^{2\pi} \rho^2 \sin \phi (abc) \; d\theta \; d\rho \; d\phi \\
		& = & \frac{4\pi abc}{3}
	\end{eqnarray*}
A short cut would have been to not convert into spherical coordinates, but to notice that 
	$\iiint_B dV$ is the volume of the sphere which we know to be $4\pi/3$.

\item  Find the integral
	\[
	\iiint_E \left( \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \right) \; dV
	\]
\answer:
We make the same change of coordinates above (using the same Jacobian) and get:
\begin{eqnarray*}
\iint_B (u^2+v^2+w^2) (abc) \; du \; dv \; dw
\end{eqnarray*}
We now convert to spherical coordinates, as above and get
\begin{eqnarray*}
\int_0^\pi \int_0^1 \int_0^{2\pi} (\rho^2) \rho^2 \sin \phi (abc) \; d\theta \; d\rho \; d\phi
	& = & abc \int_0^\pi \int_0^1 \int_0^{2\pi} \rho^4 \sin \phi \; d\theta \; d\rho \; d\phi \\
	& = & \frac{4 \pi a b c}{5}
\end{eqnarray*}
\end{enumerate}














\item  The cylinder $x^2+y^2=x$ divides the unit sphere ($x^2+y^2+z^2=1$) into two regions $S_1$ and $S_2$.
	$S_1$ is the region of the sphere inside the cylinder and 
	$S_2$ is the region of the sphere outside the cylinder.
\begin{enumerate}[(a)]
\item  Find the area of $S_1$.

\answer: The upper surface of the sphere is given by the function 
	\[
	f(x,y) = \sqrt{1-x^2-y^2}
	\]
We compute the partials
	\[
	f_x = \frac{-x}{\sqrt{1-x^2-y^2}} \qquad
	f_y = \frac{-y}{\sqrt{1-x^2-y^2}}
	\]
Then, the area of $S_1$ is given by
	\[
	\frac{1}{2}A(S_1) = \iint_D \sqrt{1+f_x^2+f_y^2} \; dA
	\]
Where $D$ is the disc in the $x-y$ plane bounded by the circle $x^2+y^2=x$.
The formula is $\frac{1}{2}A(S_1)$ because $S_1$ has a top bottom halves and the integral is only the top half.
The area of the bottom half is the same as the area of the top half.
So, we substitute $f_x$ and $f_y$ into the integral and simplify:
	\[
	\frac{1}{2}A(S_1) = \iint_{D} \frac{1}{\sqrt{1-x^2-y^2}} \; dA
	\]
We now convert to polar coordinate.  Notice that the equation $x^2+y^2=x$ transforms into the equation
$r=\cos \theta$.
Making $\theta$ go from $0$ to $\pi$ gives the whole circle and making $\theta$ go from $0$ to $\pi/2$ 
gives half the circle (graph it!).

Next, using symmetry and geometry, notice that we can actually compute the integral only over half of
the disc, the half with $y>0$ (and now we will have $\frac{1}{4}A(S_1)$ because we cut the area in half again).
The reason I do this is because it made it easier for me to see the limits for integration in polar coordinate.
	\begin{eqnarray*}
	\frac{1}{4}A(S_1) & = & \iint_{D} \frac{1}{\sqrt{1-x^2-y^2}} \; dA \\
		& = & \iint_{D} \frac{1}{\sqrt{1-r^2}} r \; dr \; d\theta \\
		& = & \int_0^{\pi/2} \int_0^{\cos \theta} \frac{1}{\sqrt{1-r^2}} r \; dr \; d\theta \\
		& = & \frac{\pi}{2} - 1
	\end{eqnarray*}
This gives
	\[
	A(S_1) = 2\pi - 4
	\]

\item  Find the area of $S_2$.

\answer: we know the surface area of a sphere is $4\pi r^2$.
So the surface area of the unit sphere is $4\pi$.
Therefore the area of $S_2$ is
	\begin{eqnarray*}
	A(S_2) & = & A(\mbox{sphere}) - A(S_1) \\
		& = & 4\pi - (2\pi-4) \\
		& = & 2\pi +4
	\end{eqnarray*}

\end{enumerate}




\end{enumerate}

\end{document}







