\magnification=\magstep1
\def\sol{\medskip\item{} {\it Solution:} }
\def\p{\partial}
\centerline{Mathematics 241 -- Multivariable Calculus}
\centerline{Solutions for Exam 3 Review Problems}
\centerline{November 28, 2007}
\bigskip
\noindent
I.  Let ${\bf F}(x,y,z) = (x + yz , y + xz , z + xy)$ be a vector field on 
${\bf R}^3$.
\medskip
\item{A)} Compute the divergence of ${\bf F}$.
\sol
The divergence of a vector field ${\bf F}$ is the scalar-valued function
${\rm div}({\bf F}) = \nabla \cdot {\bf F}$.  Here:
$$ \left({\p\over\p x},{\p\over\p y},{\p\over\p z}\right) \cdot (x + yz , y + xz , z + xy) = 3.$$
\item{B)} Compute the curl of ${\bf F}$.
\sol
The curl of a vector field ${\bf F}$ is the vector field
${\rm curl}({\bf F}) = \nabla \times {\bf F}$ (cross product).  Here:
$$ \left({\p\over\p x},{\p\over\p y},{\p\over\p z}\right) \times (x + yz , y + xz , z + xy) = 
(0,0,0).$$
\item{C)} Is there a scalar-valued function $f(x,y,z)$ such 
that ${\bf F} = \nabla f$?  Why or why not?
\sol
For a scalar-valued function, $\nabla f$ is the gradient vector field.
There are such functions: 
$$f = {1\over 2}(x^2 + y^2 + z^2) + xyz + C$$
where $C$ is an arbitrary constant.

\bigskip
\noindent
II.  All parts of this question refer to the function 
$$f(x,y) = xe^{-x^2}\cos(y).$$
\medskip
\item{A)}  Compute the second degree Taylor polynomial of $f$
at $(a,b) = (0,0)$.  Give both the form with the 
derivative and Hessian matrices and the expanded (``messy, practical'') form. 
\sol
The form with the derivative and Hessian matrices is
$$p_2(x,y) = f(0,0) + D(f)(0,0) \pmatrix{x-0\cr y-0} + {1\over 2}(x-0\ y-0) D^2(f)\pmatrix{x-0\cr y-0}.$$
Expanded out this becomes
$$\eqalign{f(0,0) + &{\p f\over \p x}(0,0)(x - 0) + {\p f\over \p x}(0,0)(x - 0)\cr
 + &{1\over 2}\left({\p^2 f\over \p x^2}(0,0)(x - 0)^2 + 
2{\p^2 f\over \p x\p y}(0,0)(x - 0)(y - 0) + {\p^2 f\over \p y^2}(0,0)(y - 0)^2\right).\cr}$$
For our function $f(x,y)$,
$$\eqalign{
{\p f\over \p x} &= \cos(y)e^{-x^2}(1 - 2x^2) \Rightarrow {\p f\over \p x}(0,0) = 1\cr
{\p f\over \p y} &= -\sin(y)xe^{-x^2} \Rightarrow {\p f\over \p x}(0,0) = 0\cr
{\p^2 f\over \p x^2} &= \cos(y)e^{-x^2}(-6x + 4x^3) \Rightarrow {\p^2 f\over \p x^2}(0,0) = 0\cr
{\p^2 f\over \p x\p y} &= -\sin(y)e^{-x^2}(1-2x^2) \Rightarrow {\p^2 f\over \p x\p y}(0,0) = 0\cr
{\p^2 f\over \p y^2} &=-\cos(y)xe^{-x^2} \Rightarrow {\p f\over \p x}(0,0) = 0\cr}$$
Hence 
$$p_2(x,y) = x.$$
\item{B)}  $f(x,y)$ has exactly three critical points with 
$0 \le y \le 2 \pi/3$.  Find them.
\sol
The critical points are the simultaneous solutions of ${\p f\over \p x} = 0$
and ${\p f\over \p y} = 0$.  From the computations of the partial derivatives in
part A, for $y$ in the range given,
$${\p f\over \p x} = \cos(y)e^{-x^2}(1 - 2x^2) = 0$$
implies $y = \pi/2$ or $x = \pm 1/\sqrt{2}$.  Then
$${\p f\over \p y} = -\sin(y)xe^{-x^2} = 0$$
implies $y = 0$ or $x = 0$.  Putting these together, we find the three critical
points:
$$(0,\pi/2),\ (1/\sqrt{2},0),\ (-1/\sqrt{2},0).$$
\item{C)} Classify the critical points from part B as 
local maxima, local minima, or saddle points using the Second Derivative
Test (Hessian Criterion).
\sol
Using the partial derivatives computed in part A,
$$D^2(f) = \pmatrix{\cos(y)e^{-x^2}(-6x + 4x^3) & -\sin(y)e^{-x^2}(1-2x^2) \cr
                    -\sin(y)e^{-x^2}(1-2x^2) & -\cos(y) x e^{-x^2} \cr}$$

\itemitem{$\bullet$}At $(0,\pi/2)$,
$$D^2(f)(0,\pi/2) = \pmatrix{0 & -1\cr -1 & 0\cr}.$$
The determinant is $\det(D^2(f)) = -1$, so this is a {\it saddle point}.
\medskip
\itemitem{$\bullet$}At $(1/\sqrt{2},0)$,
$$D^2(f)(1/\sqrt{2},0) = \pmatrix{-4e^{-1/2}/\sqrt{2} & 0\cr 0 & -e^{-1/2}/\sqrt{2}\cr}.$$
The upper left entry is negative and the 
determinant is $\det(D^2(f)) = 2e^{-1} > 0$, so this is a {\it local maximum}.
\medskip
\itemitem{$\bullet$}At $(-1/\sqrt{2},0)$,
$$D^2(f)(-1/\sqrt{2},0) = \pmatrix{4e^{-1/2}/\sqrt{2} & 0\cr 0 & e^{-1/2}/\sqrt{2}\cr}.$$
The upper left entry is positive and the 
determinant is $\det(D^2(f)) = 2e^{-1} > 0$, so this is a {\it local minimum}.
\bigskip
\noindent
III.  Let  
$f(x,y) = (x - 1)^2 + y^2$
and $g(x,y) = x^2/4 + y^2$.
\medskip
\item{A)} Find the largest and smallest values of $f$ on the constraint
curve defined by  $g(x,y) = 1$
using the method of Lagrange Multipliers.
\sol
The Lagrange multiplier equations are:
$$\eqalign{2(x-1) = \lambda x/2\cr
            2y = 2\lambda y\cr
          x^2/4 + y^2 - 1 = 0.\cr}$$
From the second equation, $y = 0$ or $\lambda = 1$.  In the first case,
from the constraint equation, $x = \pm 2$.  In the second case, from 
the first equation $2x - 2 = x/2$, so $x = 4/3$ and then $y = \pm\sqrt{5}/3$ from 
the constraint equation.  To decide which points are maxima and which
are minima of $f$ on the constraint curve we substitute:
$$\eqalign{f(2,0) &= 1\cr
           f(-2,0) &= 9\cr
           f(4/3,\pm\sqrt{5}/3) &= 2/3\cr}$$
So the maximum value is $9$ and the minimum value is $2/3$. 
\medskip 
\item{B)} Sketch the level curves of $f$ for the values $c = 2/3, 1, 9$
together with the curve  $g(x,y) = 1$ and
explain how your sketch relates to the result of part A.  
\sol
The level curves of $f$ for $c > 0$ are circles with center at $(1,0)$
and radius $\sqrt{c}$.  The constraint curve $g(x,y) = 1$ is an
ellipse with semimajor axis 2 along the $x$-axis and semiminor axis
$1$ along the $y$-axis.  The level curves of $f$ for the values $c = 2/3,1,9$
are tangent to the ellipse at the constrained critical points found in part A.
\medskip
\item{C)}  What are the absolute maximum and minimum values of 
$h(x,y)</EM> = 10 - f(x,y)$
on the region defined by $g(x,y) \le 1$? 
\sol
$h(x,y) = 10 - (x - 1)^2 - y^2$ has a critical point at $(x,y) = (1,0)$
which is a local and global maximum.  Applying Lagrange multipliers to $h$
on the constraint curve gives a system of equations with the same solutions
as those from part A.  We have $h(2,0) = 9$, $h(-2,0) = 1$, and 
$h(4/3,\pm\sqrt{5}/3) = 28/3$.  So the maximum of $h(x,y)$ is $10$ and 
the minimum is $1$.  

\bigskip
\noindent
IV.  All parts of this question refer to the region $R$ in the 
plane {\it inside} the circle $x^2 + y^2$ = 5, {\it below}
parabola $y = x^2 + 1$, and above the $x$-axis.
\medskip
\item{A)} Set up iterated integral(s) to integrate a
general function $f(x,y)$ over $R$,
integrating first with respect to $x$.
\sol
The parabola crosses the circle at $(\pm 1,2)$ and extends only down to $y = 1$.  So to 
integrate first with respect to $x$ (type 2 regions), we actually need to subdivide
$D$ into three regions:  two with $1 \le y \le 2$ and one with $0 \le y \le 1$:
$$\int_1^2\int_{-\sqrt{5 - y^2}}^{-\sqrt{y - 1}} f(x,y)\ dx\ dy + 
\int_1^2\int_{\sqrt{5 - y^2}}^{\sqrt{y - 1}} f(x,y)\ dx\ dy + 
\int_0^1\int_{-\sqrt{5 - y^2}}^{\sqrt{5 - y^2}} f(x,y)\ dx\ dy.$$
\item{B)} Same question as A, but interchanging the order of 
integration (integrating first with respect to
$y$).
\sol
The setup using type 1 regions is similar since the top edge of the 
region changes at $x = -1$ and again at $x = 1$:
$$\int_{-\sqrt{5}}^{-1}\int_0^{\sqrt{5 - x^2}} f(x,y)\ dy\ dx + 
\int_{-1}^1\int_{0}^{x^2+1} f(x,y)\ dy\ dx + 
\int_1^{\sqrt{5}}\int_{0}^{\sqrt{5 - x^2}} f(x,y)\ dy\ dx.$$ 

\bigskip
\noindent
V. 
\medskip
\item{A)} State the change of variables formula for double integrals in general.
\sol
See the text and the class notes.  The key points are that if 
$T : {\bf R}^2_{u,v} \to {\bf R}^2_{x,y}$ is a change of coordinate mapping,
and $T(D*) = D$, then
$$\int\!\int_D f(x,y)\ dx dy = \int\!\int_{D*} f(x(u,v),y(u,v)) \left|{\p (x,y)\over \p (u,v)}\right|\ du dv.$$
Here 
$${\p (x,y)\over \p (u,v)} = \det \pmatrix{ {\p x \over \p u} &  {\p x \over \p v}\cr
                                            {\p y \over \p u} &  {\p y \over \p v}\cr}$$
is the Jacobian determinant -- the determinant of the derivative matrix of the 
mapping $T$.  
\item{B)} Where does the ``extra $r$'' in the polar 
coordinate integral  $\int\!\int f(r \cos \theta, r \sin \theta) r\ dr\ d\theta$
come from?
\sol
It comes from the Jacobian determinant of the polar coordinate mapping
$T(r,\theta) = (r\cos\theta,r\sin\theta)$:
$${\p (x,y)\over \p (r,\theta)} = \det \pmatrix{ \cos\theta &  -r\sin\theta\cr
                                                 \sin\theta &  r\cos\theta\cr} = r(\cos^2\theta + \sin^2\theta) = r.$$
\item{C)} Change variables to polar coordinates to determine the integral
of $f(x,y) = xy$ 
over the quarter circle inside $x^2 + y^2 = 4$
in the {\it second} quadrant.
\sol
This is 
$$\eqalign{\int\!\int_D xy\ dx dy &= \int_{\pi/2}^\pi \int_0^2 (r\cos\theta)(r\sin\theta)\ r dr d\theta\cr
&= \int_{\pi/2}^\pi \int_0^2 r^3 \cos\theta\sin\theta\ dr d\theta\cr
&= {r^4\over 4}|_0^2 \cdot {1\over 2}{\sin^2\theta}|_{\pi/2}^\pi \cr
&= -2.}$$
\item{D)} Change variables to spherical coordinates in ${\bf R}^3$ to 
evaluate the triple integral of the function $x$ over the portion of the solid ball
of radius 1 with center at (0,0,0) in the first octant (i.e. the part of the 
region inside the sphere of radius 1, with $x, y, z$ all $\ge 0$).
\sol
This is 
$$\int_0^{\pi/2}\int_0^{\pi/2}\int_0^1 \rho\sin\phi\cos\theta\cdot \rho^2\sin\phi\ d\rho\, d\phi\, d\theta = {\pi\over 16}.$$
\bigskip
\noindent
VI.  Find the volume of the solid bounded above by
the cone $z = (x^2 + y^2)^{1/2}$, below by $z$ = 0, 
and by $0 \le x^2 + y^2 \le 4$. 
You may use any convenient coordinate system to carry out the necessary 
computations. 
\sol
Because of the circular symmetry of the region about the $z$-axis,
{\it cylindrical coordinates} are indicated for this one.  The volume is
$$V = \int_0^{2\pi} \int_0^2 \int_0^r 1\ r dz\, dr\, d\theta = {16\pi\over 3}.$$
\bye
