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\begin{center}{\bf Math 214 Advanced Calculus II \\
Midterm Exam II\\
SOLUTIONS }\end{center}
April 21, 2001 \hfill Ali Sinan Sert\"{o}z \atla
{\bf 1) }
\parbox[t]{\olcu}{
Suppose $V$ is open in $\Rr^n$ and $f:V\rightarrow \Rr$ is continuously
differentiable with $\Delta_f\neq 0$ on $V$. Prove that
\[ \lim_{r\rightarrow 0}\frac{{\rm Vol}(f(B_r(\mathbf{x}_0)))}{{\rm
Vol}(B_r(\mathbf{x}_0))} = |\Delta_f(\mathbf{x}_0)| \]
for every $\mathbf{x}_0\in V$.
}
\atla
{\bf Solution:}
\newcommand{\Bx}{B_r(\mathbf{x}_0)}
\newcommand{\VBx}{{\rm Vol}(\Bx )}
\newcommand{\VfBx}{{\rm Vol}(f(\Bx ))}
Fix an arbitrary $\mathbf{x}_0\in V$. Note that $\overline{\Bx}$ is
compact, $\Delta_f$ attains its $inf$ and $sup$ there, and
\begin{eqnarray}
\VfBx &=& \int_{f(\Bx )} d\mathbf{u} \\
&=&\int_{\Bx} |\Delta_f (\mathbf{x})|\; d\mathbf{x} \\
&=&c\; \VBx \\
&=&|\Delta_f (\mathbf{x}_1)|\; \VBx
\end{eqnarray}
where \\
(1) follows from Cor. 7.5 (p349), \\
(2) follows from Thm. 7.14 (p375), \\
(3) holds for some $c$ between the $inf$ and $sup$ of $\Delta_f$ on $\Bx$,
Thm.~7.9~(p354), \\
(4) holds for some $\mathbf{x}_1\in\overline{\Bx}$ since
$|\Delta_f(\mathbf{x})|$ is continuous there.\\
Finally taking the limit as $r\rightarrow 0$ gives the required
result.\\
Note that this is Exercise 7 on page 382. (The problem would be more
meaningful if the range of $f$ is $\Rr^n$)...!
\\
\mbox{}\hrulefill\mbox{}\atla
\mbox{}
\newpage
{\bf 2) } \parbox[t]{\olcu}{
Suppose $V$ is open in $\Rr^n$ and $f:V\rightarrow \Rr$ is
continuous. Prove that if
\[ \int_E f(\mathbf{x})d\mathbf{x}=0 \]
for all Jordan regions $E\subset V$, then $f=0$ on $V$.
}
\atla
{\bf Solution:}
\newcommand{\bfx}{\mathbf{x}}
Assume that there is an $\bfx_0\in V$ where $f(\bfx_0 )=2\epsilon >0$. \\
Let $U=\{ \bfx\in V \; | \; f(\bfx )> \epsilon >0\; \}$. We know that
$U$ is not empty since $\bfx_0\in U$. Since $f$ is continuous, $U$ is
open. ($U$ is the preimage of the open interval $(\epsilon,\infty
)$). By continuity of $f$ there is an open ball $B_{2r}$ of radius
$2r$ around $\bfx_0$,
where $f$ is strictly positive. The
integral of $f$ on $B_r$ is zero by the assumption of the theorem. (Note
that $B_r$ is a Jordan region). By
the mean value theorem this is equal to $c\; Vol(B_r)$ for some $c$
between the inf and sup of $f$ on $B_r$. Since $f$ is continuous on
$B_{2r}$, there is a point $\bfx_1\in \overline{B_r}$ where
$f(\bfx_1)=c$, so $c\neq 0$. But this forces the volume of $B_r$ to be
zero, a contradiction. So there is no point where $f$ is strictly
positive. Similarly $f$ cannot take a strictly negative value. So $f$
is identically zero on $V$. \\
Note that this is Exercise 9 on page 357. \\
\mbox{}\hrulefill\mbox{}\atla
{\bf 3) } \parbox[t]{\olcu}{
Let $S$ be the surface defined by $z=7\sqrt{x^2+y^2}$, $7\leq z\leq
28$. Evaluate the integral
\[ \iint_{\substack{\mbox{~} \\\! \! \! S}} F\cdot \vec{\mathbf{n}}\; d\sigma \]
where $F(x,y,z)=(x^2,y(1-2x),-z)$ and $\vec{\mathbf{n}}$ is the unit
normal vector pointing outward.
}
\atla
{\bf Solution:}
\newcommand{\del}{\partial}
Evaluating this integral directly requires a lot of tedious work, so
we try using Stokes' theorem which reduces this integral to an
integral of the boundary. Since the boundary consists of two circles,
we expect that the resulting integral will be easy. But to apply
Stokes' theorem we must express $F$ as curl$G$. Since div$F=0$, such a
$G$ exists, see Theorem 8.6 on page 450. Set $G=(P,Q,R)$. Trial and
error, with some luck, will give you a $G$ with curl$G=F$. For example
I set $Q=0$ and tried to solve the remaining equalities: \\
$R_y=x^2$, $P_z-R_x=y-2xy$, $-P_y=-z$. This easily yields a solution,
for example $G=(yz,0,yx^2)$. Now Stokes' theorem says that the
required integral is equal to $\int_{\del S}G\cdot T\; ds$. \\
We now parametrize the boundary $\del S=C_1-C_2$. \\
$C_1$: $\phi (t)=(\cos t, \sin t, 7)$, $t\in [0,2\pi ]$, \\
$C_2$: $\psi (t)=(4\cos t, 4\sin t, 28)$, $t\in [0, 2\pi ]$. \\
$G(\phi (t))\cdot \phi'(t)dt=-7\sin^2 t\; dt$. \\
$G(\psi (t))\cdot \psi'(t)dt=-448\sin^2 t\; dt$. \\
So we integrate $441 \sin^2 t dt$ from $0$ to $2\pi$ to obtain
$441\pi$ as the answer. \\
\mbox{}\hrulefill\mbox{}\atla
{\bf 4) } \parbox[t]{\olcu}{
Find the surface area of the cap $x^2+y^2+z^2=R^2$,
$z\geq \sqrt{R^2-A^2}$, where
\linebreak
$R\geq A \geq 0$.
}
\atla
{\bf Solution:}
We first parametrize the surface: \\
$\phi (r,\theta )=( r\cos \theta,r\sin\theta,\sqrt{R^2-r^2})$,
$(r,\theta )\in [0,A]\times [0,2\pi ]=E$. \\
To find the surface area we need to evaluate $\int_E
\parallel N_{\phi_r \times \phi_\theta} \parallel
d(r,\theta )$, see
Definition~8.12 on page 424. A straightforward calculation gives
$\parallel N_{\phi_r \times \phi_\theta} \parallel =r\left(
R^2/(R^2-r^2)\right)^{1/2}$. Integrating this over $E$ gives $2\pi
R^2-2\pi R\sqrt{R^2-A^2}$. \\
\mbox{}\hrulefill\mbox{}\atla
{\bf 5) } \parbox[t]{\olcu}{
{\bf (i) }
Construct a $C^\infty$ vector field $F:\Rr^3\backslash \{ (0,0,0)\} \rightarrow
\Rr^3$ such that $F$ is not defined at the origin and ${\rm div}
F=0$. \\
{\bf (ii) }
Let $E_1$ be the surface given by $x^2+y^2+z^2=1$ and $E_2$ the
surface given by $x^2/25+y^2/36+z^2/49=1$. Show that
\[ \iint_{\substack{\mbox{~}\\ \! \! \! E_1}} F\cdot
\vec{\mathbf{n}_1}\; d\sigma
=
\iint_{\substack{\mbox{~}\\ \! \! \! E_2}} F\cdot
\vec{\mathbf{n}_2}\; d\sigma \]
where $F$ is the vector field you found in the first part, and
$\vec{\mathbf{n}_i}$ is the unit normal pointing outward of the
surface $E_i$, $i=1,2$.
}
\atla
{\bf Solution:}
The easiest way to construct a vector field $F$ with zero divergence
is to start with a vector field $G$ and set $F$ to be curl$G$. Since
we want $F$ to be defined everywhere except at the origin we can start
with an arbitrary $C^\infty$ vector field $G$ which is defined
everywhere except at the origin. One such easy example is
$G=(\log (x^2+y^2+z^2), \log (x^2+y^2+z^2), \log (x^2+y^2+z^2))$. \\
To prove the second part let the region between these surfaces be
denoted by $E$. Then use Gauss' theorem, Theorem~8.4 on page 441, and
observe that the right hand side is zero since div$F=0$. This gives,
after observing correct orientations of the surfaces $E_1$ and $E_2$,
that the required equality holds. \\
Note that this is Exercise 10.c on page 454. I solved parts (a) and
(b) in class and promised to ask part (c) in the exam! \\
\mbox{}\hrulefill\mbox{}\atla
\end{document}
\atla
{\bf Notes to the novice:} You must write your answers in a readable
and self explanatory manner. I do not try to read what you mean. I
simply read what you write. If what you have in mind and what you have
on paper are different, beware that I mark only what is on the
paper. Intentions don't count! Moreover, if your method is correct,
whatever that means, you do not automatically get partial
credits. Your partials are determined on how close you get to the
correct result, and let me be the judge of it!! \\
Good luck...
{\bf 6) } \parbox[t]{\olcu}{
Find $G(\mathbb{C},\mathbb{R})$, the group of automorphisms of the
field $\mathbb{C}$ keeping the elements of the field $\mathbb{R}$
pointwise fixed. Is $\mathbb{C}/\mathbb{R}$ a normal extension? \\
{\tiny (See page 239, and also 245.)}}
\atla
{\bf 7) } \parbox[t]{\olcu}{
Let $K/F$ be some field extension and $a\in K$.
Let $f(x)\in F[x]$ be a monic polynomial
of smallest positive degree which has $a$ as a root. Let $g(x)\in
F[x]$ be another monic polynomial which has $a$ as a root. Show that
if $\deg f=\deg g$ then $f=g$. \\
{\tiny (See page 211.)}
}
\atla
{\bf 8) } \parbox[t]{\olcu}{
If $G$ is a group, $Z$ its center, and if $G/Z$ is cyclic,
prove that $G$ must be abelian. {\tiny (Problem 16 on page 91.)}
} \atla
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\settowidth{\eksi}{\bf Bonus: }
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{\bf Bonus: } \parbox[t]{\olcu}{
If $G$ is a group and $H$ is a subgroup of index $p$ in
$G$, where $p$ is the smallest prime dividing the order of $G$, then
prove that $H$ is a normal subgroup of $G$
} \atla
{\bf All questions are 7.5 points except the Bonus which is 15 points.}
\end{document}
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