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\noindent
\parbox{2in}{\bf Math 2130}
\hfill {\Large \bf Test \#3} \hfill
\parbox{2in}{\bf \hfill July $27^\mathrm{th}$\!,\  2016}

\vspace{0.2in}

\noindent {\large\bf Name:} \underline{\hskip 3.0 truein}
\hfill {\bf Be sure to show your work!}

\vspace{0.05in}

\noindent
\begin{equation*}
\begin{array}{ccc}
 x & = & \rho \cos(\theta) \sin(\varphi) \\
 y & = & \rho \sin(\theta) \sin(\varphi) \\
 z & = & \rho \cos(\varphi) 
\end{array} 
\hspace*{2.5in}
\begin{array}{c}
J=\rho^2\sin(\varphi) \\
{}\\
\cos^2(\theta) = \frac{1}{2}\left(1+\cos(2\theta)\right)
\end{array}
\end{equation*}

\vspace{0.05in}

\noindent {\bf\large 1. (12 points)} Let $R$ be the region bounded by $y=5-x^2$ and $y=-x+3$. 

\force \hfill [Warning: One of the integrals below will have to be split into 2 pieces.]

\vspace{-0.15in}

\begin{enumerate}[(a)]
   \item Sketch the region $R$.
   \item Set up the integral $\iint_R f(x,y)\,dA$ using the order of integration ``$dy\,dx$''. \hfill [{\it Note:} You cannot evaluate this integral.]
   \item Set up the integral $\iint_R f(x,y)\,dA$ using the order of integration ``$dx\,dy$''. \hfill [{\it Note:} You cannot evaluate this integral.]
\end{enumerate}   

\vspace{0.5in}

\vfill

\noindent {\bf\large 2. (12 points)} Evaluate $\displaystyle \int_0^4 \int_{\sqrt{x}}^2 \sin(y^3)\,dy\,dx$.    % Reusing Spring 2012 #1

\vspace{0.05in}

{\it Note/Hint:} You cannot integrate $\int \sin(y^3)\,dy$ in terms of elementary functions.

\vfill

\newpage

\noindent {\bf\large 3. (12 points)} Let $R$ be the left half (i.e. $x \leq 0$) of the annulus lying between $x^2+y^2=1$ and $x^2+y^2=4$. 

\vspace{0.05in}

{\bf Sketch a picture} of $R$ then compute its centroid. You should be able to use symmetry and geometry to cut down the amount of integration you need to do.

\vspace{-0.1in}

$$ m = \iint_R 1\,dA \qquad M_{y} = \iint_R x\,dA \qquad M_{x} = \iint_R y\,dA \qquad (\bar{x},\bar{y}) = \left(\dfrac{M_y}{m}, \dfrac{M_x}{m} \right)$$

\vfill

\noindent {\bf\large 4. (13 points)} Compute $\displaystyle \iint_R (x+y)e^{x-y}\,dA$ where $R$ is the region bounded by $x+y=1$, $x-y=0$ and $y=0$. 

\vspace{0.05in}

Use a (natural) change of coordinates which simplifies the region $R$ and\dots{}don't forget the Jacobian!

\vfill

\newpage


\noindent {\bf\large 5. (15 points)} Let $E$ be the region bounded by $z=5\sqrt{x^2+y^2}$ and $z=10$. Set up integrals which computes the volume of $E$ using the following orders of integration: \quad [Do {\bf not} evaluate these integrals.]


\begin{enumerate}[(a)]
\item $\int_?^? \  \int_?^? \  \int_?^? \  ??? \,dz\,dy\,dx$
\item $\int_?^? \  \int_?^? \  \int_?^? \  ??? \,dx\,dz\,dy$
\item Set up this integral in cylindrical coordinates.
\item Set up this integral in spherical coordinates.
\end{enumerate}

\vfill

\vfill

\vfill

\noindent {\bf\large 6. (12 points)} Compute $\iiint_E \dfrac{1}{\sqrt{x^2+y^2+z^2}} \,dV$ where $E$ is the solid region outside $x^2+y^2+z^2=1$, inside $x^2+y^2+z^2=4$, and in the first octant.

\vfill

\newpage 

\noindent {\bf\large 7. (12 points)} Consider the region $E$ bounded below by $z=x^2+y^2$ and above by $z=4$. Compute the centroid of $E$.

\vspace{0.05in}

{\bf Free information:} The volume of $E$ is $8\pi$.

\vspace{-0.2in}

$$ m = \iiint_E 1\,dV \qquad M_{yz} = \iiint_E x\,dV \qquad M_{xz} = \iiint_E y\,dV \qquad  M_{xy} = \iiint_E z\,dV \qquad (\bar{x},\bar{y},\bar{z}) = \left(\dfrac{M_{yz}}{m}, \dfrac{M_{xz}}{m},\dfrac{M_{xy}}{m} \right)$$

\vfill


\noindent {\bf\large 8. (12 points)} Consider the integral: $\displaystyle I = \int_0^7 \int_{-\sqrt{49-x^2}}^{0} \int_{-\sqrt{49-x^2-y^2}}^0 \ z\cos(x^2+y^2+z^2) \,dz\,dy\,dx$.   

\begin{enumerate}[(a)]
   \item Rewrite $I$ in the following order of integration: $\iiint \quad \,dy\,dz\,dx$.
   
   Do {\bf not} evaluate the integral.

\vspace{1in}
   
   \item Rewrite $I$ in terms of cylindrical coordinates.
   
   Do {\bf not} evaluate the integral.
   
\vspace{1in}

   \item Rewrite $I$ in terms of spherical coordinates.
   
   Do {\bf not} evaluate the integral.

\vspace{1in}

\end{enumerate}

\end{document}

