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\begin{document}

{\Large Key: TEST 3}


\begin{enumerate}

\item  Consider the rectangle $R = \{(x,y) \stbar 0 \leq x \leq 1, -1 \leq y \leq 1 \}$
	and compute the integral:
	\[
	\iint\limits_R xy^3 \; dA
	\]
\answer:
We just compute the interated integral:
\begin{eqnarray*}
\iint\limits_R xy^3 \; dA & = & \int_0^{1} \int_{-1}^1 xy^3 \;dy \; dx 
	= \int_0^{1} \left. \frac{xy^4}{4} \right|_{y=-1}^{y=1} \; dx \\
	& = & \int_0^{1} \left( \frac{x}{4} - \frac{x}{4} \right) \; dx 
	= \int_0^{1} 0 \; dx
	= 0
\end{eqnarray*}

\item  Find the maximum and minimum of $f(x,y,z)=x-y+z$ subject to the constraint:
\[
x^2+y^2+z^2 = 2
\]
\answer:
We have $f$ and our constraint gives: $g(x,y,z)=x^2+y^2+z^2-2$.
We then have $\nabla f=(1,-1,1)$ and $\nabla g=(2x,2y,2z)$.
We set up the three equations that come from $\nabla f= \lambda \nabla g$ (and the constraint makes four equations):
\begin{eqnarray*}
1 & = & 2\lambda x \\
-1 & = & 2 \lambda y \\
1 & = & 2 \lambda z \\
x^2+y^2+z^2 & = & 2
\end{eqnarray*}
There are lots of ways to solve this, but I solve the first equation for $\lambda$ to get $\lambda=1/(2x)$.
I then plugged this in for $\lambda$ in the other equations.
This gave me the following:
\begin{eqnarray*}
-1 =2(1/(2x))y & \ra & y=-x \\
1 =2(1/(2x))z & \ra & z=x \\
x^2+y^2+z^2=2 && 
\end{eqnarray*}
I now plut $y=-x$ and $z=x$ into the constraint equation and get:
\[
x^2+x^2+x^2=2
\]
or $x^2=2/3$.
Thus we get $x=\pm \sqrt{2/3}=\pm \sqrt{6}/3$.
Putting these back in for $y$ and $z$ give the points:
\[
\mbf{p}_1 = \left(\frac{\sqrt{6}}{3}, -\frac{\sqrt{6}}{3}, \frac{\sqrt{6}}{3} \right)
\qquad
\mbf{p}_2 = \left(-\frac{\sqrt{6}}{3}, \frac{\sqrt{6}}{3}, -\frac{\sqrt{6}}{3} \right)
\]
We now just plug $\mbf{p}_1$ and $\mbf{p}_2$ into $f$ and see which is bigger and which is smaller:
\[
f(\mbf{p}_1) = \sqrt{6} \qquad f(\mbf{p}_1) = -\sqrt{6}
\]
So, we have a maximum of $\sqrt{6}$ at $\mbf{p}_1$
and a minimum of $-\sqrt{6}$ at $\mbf{p}_2$.

\item  Let $T$ be the region in the plane bounded by the curves:
\[
y=0 \qquad x=2 \qquad y=\ln x
\]
Consider the integral
\[
\iint\limits_T x \; dA
\]
Set this integral up as an iterated integral in two ways by finding the appropriate limits of integration
\begin{enumerate}
\item  Find the limits of integration for
	\[
	\int\int x \; dx \; dy
	\]
\answer: After graphing the region, you should be able to see the limit here should be
\[
\int_0^{\ln 2} \int_{e^y}^{2} x \; dx \; dy
\]

\item  Find the limits of integration for
	\[
	\int\int x \; dy \; dx
	\]
\answer: again, you need to make sure you graphed it.  This one is probably the easier one to do first:
\[
\int_1^2\int_0^{\ln x} x \; dy \; dx
\]

\item  Pick one of your integrals above and compute it.
\answer:
The second one is tricky to integrate (you need to use integration by parts), so, we'll integrate the first one:
\begin{eqnarray*}
\int_0^{\ln 2} \int_{e^y}^{2} x \; dx \; dy 
	& = & \int_0^{\ln 2} \left. \frac{x^2}{2} \right|_{e^y}^{2} \; dy 
	= \int_0^{\ln 2} \left( 2- \frac{1}{2}e^{2y} \right) \; dy \\
	& = & \left. 2y- \frac{1}{4}e^{2y} \right|_0^{\ln 2}
	= \left( 2\ln 2 - \frac{1}{4}e^{2\ln 2} \right) - \left( - \frac{1}{4}e^{0} \right) \\
	& = & 2 \ln 2 - 1+\frac{1}{4} = 2 \ln 2 - \frac{3}{4}
\end{eqnarray*}

\end{enumerate}


\item  Let $R$ be the region in the plane defined by
	\[
	R = \{ (x,y) \stbar 1 \leq x^2+y^2 \leq 4, x \geq 0, y \geq 0 \}
	\]
 Compute the integral
\[
\iint\limits_R \frac{1}{\sqrt{x^2+y^2}} \; dA
\]
\answer:
We graph the region and convert to polar coordinate.
The region $R$ is a polar rectangle, so there isn't too much fancy going on here:
\begin{eqnarray*}
\iint\limits_R \frac{1}{\sqrt{x^2+y^2}} \; dA
	& = & \int_0^{\frac{\pi}{2}} \int_1^2 \frac{1}{\sqrt{r^2}} \cdot r \; dr \; d\theta 
	= \int_0^{\frac{\pi}{2}} \int_1^2 \; dr \; d\theta \\
	& = & \int_0^{\frac{\pi}{2}} \left. r \right|_1^2 \; d\theta
	= \int_0^{\frac{\pi}{2}} \; d\theta \\
	& = & \left. \theta \right|_0^{\frac{\pi}{2}}
	= \frac{\pi}{2}
\end{eqnarray*}










\end{enumerate}




\end{document}