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\centerline{Mathematics 241, section 1 -- Multivariable Calculus}
\centerline{Solutions for Exam III -- November 26, 2013}
\vskip 10pt
\noindent
I.  Given that
\begin{eqnarray*}
f(x,y) &=& \frac{y}{x^2 + y^2 + 4}\\
f_x(x,y) &=& \frac{-2xy}{x^2 + y^2 + 4)^2}\\
f_y(x,y) &=& \frac{x^2 - y^2 + 4}{(x^2 + y^2 + 4)^2}\\
f_{xx}(x,y) &=& \frac{2y(3x^2-y^2-4)}{(x^2 + y^2 + 4)^3}\\
f_{xy}(x,y) &=& \frac{-2x(x^2-3y^2+4)}{(x^2 + y^2 + 4)^3}\\
f_{yy}(x,y) &=& \frac{-2y(3x^2-y^2+12)}{(x^2 + y^2 + 4)^3},
\end{eqnarray*}
find all critical points of $f$ and apply the Second
Derivative Test to each to determine the types of the critical points.
\vskip 10pt
\noindent
{\it Solution:}  The critical points are the solutions of $f_x = f_y = 0$.  From $f_x = 0$,
we get either $x = 0$ or $y = 0$.  There are no solutions of the equation $f_y = 0$ with $y = 0$,
but if $x = 0$, we get $y = \pm 2$.  So the critical points are $(0,2)$ and $(0,-2)$.
At $(0,2)$, $f_{xx}(0,2)f_{yy}(0,2) - (f_{xy}(0,2))^2 = (-1/16)(-1/16) - 0^2 = 1/256 > 0$
and $f_{xx}(0,2) = -1/16 < 0$.
Therefore, $f$ has a {\it local maximum} at $(0,2)$.  At $(0,-2)$,  
$f_{xx}(0,-2)f_{yy}(0,-2) - (f_{xy}(0,-2))^2 = (1/16)(1/16) - 0^2 = 1/256 > 0$.
and $f_{xx}(0,-2) = 1/16 > 0$.
Therefore $f$ has a {\it local minimum} at $(0,-2)$.  
\vskip 10pt
\noindent
II.  
\begin{enumerate}
\item[A)]  (20) Find the maximum and minimum values of 
$$f(x,y) = xy + 2y$$
on the closed and bounded set 
$$S = \{(x,y) \in \R^2 \mid x^2 + y^2 \le 9\}.$$
\vskip 10pt
\item[] {\it Solution:}  We have 
\begin{eqnarray*}
f_x &=& y\\
f_y &=& x + 2
\end{eqnarray*}
So $f$ has one critical point at $(-2,0) \in S$ with $f(-2,0) = 0$.  (This is a saddle point for $f$).  
We use Lagrange Multipliers next.  With the constraint equation $x^2 + y^2 = 9$,
the Lagrange equations are 
\begin{eqnarray*}
y &=& 2\lambda x\\
x + 2 &=& 2\lambda y\\
x^2 + y^2 - 9 &=&0. 
\end{eqnarray*}
If we multiply the first equation by $y$ and the second by $x$, then
we can equate the right sides to obtain an equation in which $\lambda$
has been eliminated:  $y^2 = x(x + 2) = x^2 + 2x$.  We can then use this
equation to substitute for the $y^2$ and eliminate $y$ in the constraint equation:
$$2x^2 + 2x - 9 = 0.$$
This equation can be solved by the quadratic formula:
$$x = \frac{-2 \pm \sqrt{4 + 72}}{4} \doteq -2.68, 1.68.$$
Then from the constraint equation we get two $y$-values for each of these
$x$-values, so there are four solutions, approximately
$$(-2.68,\pm 1.35), (1.68,\pm 2.49)$$
We have 
\begin{eqnarray*}
f(-2,0) &= & 0\\ 
f(-2.68,1.35) &\doteq & -.92\\
f(-2.68,-1.35) &\doteq & .92\\
f(1.68,2.49) &\doteq & -9.2\\
f(1.68,-2.49) &\doteq & 9.2\\
\end{eqnarray*}
So putting everything together, we see that the maximum 
value of $f$ on $S$ is about $9.2$ and the minimum value
of $f$ on $S$ is about $-9.2$. 
\item[B)] (5)  Give a sketch of the contours of $f$ for the 
maximum and minimum values you found in part A.  
\item[] {\it Solution:}  The function is $f(x,y) = (x+2)y$.   Hence
the contours of $f$ for values not equal to zero are {\it rectangular hyperbolas}
with asymptotes equal to the lines $x = -2$ and $y = 0$.  
\begin{center}
\includegraphics[width=3.0truein,height=3.0truein]{II.png}
\end{center}
The graphs shows two hyperbolas, {\it each tangent to the boundary circle of $S$
at the points where the maximum and minimum values are 
attained}.  The relevant portions of the branches of the hyperbolas are in 
the first and fourth quadrants of the plane.  (Recall the geometry of Lagrange Multipliers we saw in the Lab
portion of Problem Set 7.)
\end{enumerate}
\vskip 10pt
\noindent
III.  All parts of this problem refer to the region
$$R = \{(x,y) \in \R^2 \mid x^2 + y^2 \le 4, y \ge 0, y \le 2x\}.$$
and the solid $S \subset \R^3$ with upper boundary $z = x^2$ and 
lower boundary the $xy$-plane \emph{and the ``shadow'' of $S$ in the $xy$-plane is $R$}. 
\vskip 10pt
\begin{enumerate}
\item[A)] (10) Sketch the region $R$ in the plane.
\item[] {\it Solution:} This is the region inside a circle of radius 2, above the 
$x$-axis, and below the line $y = 2x$:
Note: the region lies entirely in the first quadrant.
\item[B)] (10)  Set up iterated integral(s) of the form $\int\int f(x,y)\ dy\ dx$
to compute the volume of $S$.  Do not compute the value.
\vskip 10pt
\item[] {\it Solution:}  Since the upper boundary of $R$ changes when $y = 2x$
crosses the circle, we
need the sum of two interated integrals.
The line $y = 2x$ crosses $x^2 + y^2 = 4$ when $5x^2 = 4$ so $x = \pm 2/\sqrt{5}$.
Since our region is entirely in the first quadrant, it's
$$\int_0^{2/\sqrt{5}} \int_{0}^{2x} x^2\ dy\ dx + 
\int_{2/\sqrt{5}}^2 \int_0^{\sqrt{4-x^2}} x^2\ dy\ dx.$$
\item[C)]  (10)  Now set up iterated integral(s) of the form $\int\int f(x,y)\ dx\ dy$
to compute the volume of $S$.  Do not compute the value.
\vskip 10pt
\item[] {\it Solution:}  This one is slightly tricky because you 
have to notice that, even though the integral does not 
need to be split, the region $R$ does not extend all the way up to 
$y = 2$.  The range of $y$-values is determined like this:  
The line $y = 2x$ crosses $x^2 + y^2 = 4$ when $5x^2 = 4$ so $x = 2/\sqrt{5}$
and $y = 4/\sqrt{5}$.  The integral is:
$$\int_0^{4/\sqrt{5}} \int_{y/2}^{\sqrt{4-y^2}} x^2\ dx\ dy.$$
\item[D)]  (10)  Finally, set up a polar coordinate integral that 
will compute the volume of $S$.  Do not compute the value.
\vskip 10pt
\item[] {\it Solution:} With polar coordinates, you do not need to 
split into two integrals, and the limits of integration are all constant.
The part of the line $y = 2x$ in the first quadrant is the ray $\theta = \arctan(2)$.   
With the ``extra $r$'' from the $dA$ in polar coordinates: 
$$\int_0^{\arctan(2)} \int_0^2 r^3\cos^2\theta\ dr\ d\theta.$$
\item[E)]  (10)  Finally, compute the volume of $S$ using any one
of your answers from parts B,C,D (your choice). 
\vskip 10pt
\item[] {\it Solution:}  Don't know about you, but the first two
look pretty bad to me!  Let's evaluate the last one:
\begin{eqnarray*}
\int_0^{\arctan(2)} \int_0^2 r^3\cos^2\theta\ dr\ d\theta 
&=& \int_0^2 r^3\ dr \cdot \int_0^{\arctan(2)} \cos^2\theta\ d\theta\\
&=& \left. \frac{r^4}{4}\right|_0^2 \cdot \left(\left. \frac{\theta}{2} + \frac{\sin(2\theta)}{4}\right)\right|_0^{\arctan(2)}\\
&=& 4 \cdot \left(\frac{1}{2}\arctan(2) + \frac{1}{5}\right).
\end{eqnarray*}
\end{enumerate}

\vskip 10pt     
\noindent
{\it Extra Credit} (10)  
Evaluate the improper double integral
$$\int\int_{{\R}^2} e^{1-x^2-y^2}\ dA = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{1-x^2 -y^2}\ dy\ dx.$$
\vskip 10pt
\noindent
We can evaluate this by switching to polar coordinates!  The integral over
all of $\R^2$ is 
$$\int_0^{2\pi} \int_0^\infty e^{1-r^2} r\ dr\ d\theta$$
This equals 
$$
2\pi \cdot \lim_{B\to\infty} \left. -\frac{1}{2} e^{1-r^2}\right|_0^B
= 2\pi \cdot \lim_{B\to\infty} \left(\frac{e}{2} - \frac{1}{2} e^{1-B^2}\right)
= \pi e.
$$
Wow! 

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