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\section*{Quiz 8}
\section*{Name:\hspace{4cm}\;}
\noindent \textbf{Problem 1: Surface Area:}\\
Using any method you like, compute the surface area of the cone bounded by $0\leq z\leq 1-\sqrt{x^2+y^2}$.\\
\solution Tthis is a standard surface integral problem. The surface element is given by
\begin{align*}
dS=&\sqrt{1+(f_x)^2+(f_y)^2}\\
=&\sqrt{1+\left(\frac{x}{\sqrt{x^2+y^2}}\middle)+\middle(\frac{y}{\sqrt{x^2+y^2}}\right)}\\
=&\sqrt 2
\end{align*}
Integrating $dS=\sqrt 2$ over the unit circle gives $\pi \sqrt 2$ for the area of the ``pointy'' part of the cone.\\
The area of the base of the cone is the area of the unit circle, this is just $\pi$. \\
The total surface area of a cone is therefore
\[\pi+\pi\sqrt 2 = \pi(1+\sqrt 2)\]
\noindent \textbf{Problem 2: Cylindrical Integrals:}\\
Let $f(r, \theta)=r^2 \cos(3\theta)$. Let $D$ be the region given by $0\leq r \leq \sin(3\theta)$. Find the volume over $D$ between $f$ and the $z=0$ plane in polar coordinates. \\
\solution
This is a polar integral. The tricky thing will be making sure you have the right domain of integration. The domain on which $f(r, \theta)$ is positive is given by
\[\{[-\pi/6, \pi/6], [\pi/2, 5\pi/6], [7\pi/6, 3\pi/2]\}\]
The domain on which $\sin(3\theta)$ is positiveis given by
\[\{[0, \pi/3], [2\pi/3, \pi], [4\pi/3, 5\pi/3] \}\]
Which means that the domain of integration in $\theta$ is
\[\{[0, \pi/6], [2\pi/3, 5\pi/6], [4\pi/3, 3\pi/2]\}\]
The integral comes out to is
\begin{align*}
\int_{D_\theta}\int_{r=0}^{\sin(3\theta)} f(r, \theta)\cdot r \cdot drd\theta=& \int_{D_\theta}\int_{r=0}^{\sin(3\theta)} r^3\cos(3\theta) dr d\theta\\
=&\int_{D_\theta} \left.\frac{r^4}{4} \cos(3\theta)\right|_0^{\sin(3\theta)} d\theta\\
=&\int_{D_\theta} \frac{\cos (3\theta) \sin^4(3\theta)}{4} d\theta\\
=&\left. \frac{\sin^5(3\theta)}{4\cdot 3 \cdot 5}\right|_{D_\theta}=\frac{1}{20}\\
\end{align*}
Using the symmetry of $\sin$, you can evaluate over $0, \pi/6$ and then multiply by $3$.
\newpage
\noindent \textbf{Problem 3: Japanese Napkin Ring:}\\
A napkin ring of height one is created by taking a sphere of radius $R$ and removing from its center a cylinder from it of radius $\sqrt{R^2-(1/2)^2}$. Compute the volume of the napkin ring.\\
\solution It's best to work in polar coordinates. The function for the sphere in polar coordinates is
\[f(r, \theta)=\sqrt{R^2-r^2}\]
The region that you are integrating this over is $\sqrt{R^2-(1/2)^2}\leq r\leq R$. (Try drawing this one out!)\\
Then the integral works out to be
\begin{align*}
\int_0^{2\pi}\int_{\sqrt{R^2-(1/2)^2}}^R f(r, \theta) rdrd\theta=& \int_0^{2\pi}\int_{\sqrt{R^2-(1/2)^2}}^R r\sqrt{R^2-r^2}drd\theta\\
=&\int_0^{2\pi}\left( \left.\frac{ (R^2-r^2)^\frac{3}{2}}{3} \right| _{r=\sqrt{R^2-(1/2)^2}}^R \right)d\theta\\
=&\int_0^{2\pi}\frac{((1/2)^2)^\frac{3}{2}}{3} d\theta\\
=&\frac{\pi}{12}
\end{align*}
This only gives the top half of the napkin ring. Doubling this quantity gives the final solution of $\pi/6$. Remarkably, this volume is not dependent on $R$. \vspace{3cm}
\subsection*{Interesting Puzzle, Will not be graded}
Alice and Bob are in a long distance relationship, and are about to get married. This means that Bob needs to send Alice a ring. Alice has an envious neighbor Eve who has been trying to put an end to Alice and Bob's happy relationship for years. Every package that travels between Bob and Alice is searched by Eve, unless it is shipped inside a padlocked box. Bob has a lot of padlocks for this purpose; however, only he has the keys to his padlocks. Similarly, Alice has many padlocks so that she can ship things securely to Bob; however, only she has the keys to her own padlocks. How can Bob get a wedding ring to Alice?
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