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\line{Math 174\hfill Extra Credit\hfill D.~Gottlieb}
\bigskip\noindent
Peter Schlamiel tried to use Stokes' theorem
$$
\oint\limits_C \vec F\cdot d\vec r=\iint_S (\vec \nabla \times \vec F)\cdot \vec n\ d\sigma
$$
where $S$ is the surface $\{(x,y,z)$ so that $x^2+y^2+4z^2=4$ and $y\leq 1\}$ and $C$ is the boundary closed curve on the surface given by $y=1$.
Here $\vec F=\ds{{x\over x^2+y^2}\ \vec i + {y\over x^2+y^2}\ \vec j+x\vec k}$.

\noindent
Find Schlamiel's mistakes.
One point for each error you find, minus one point for each correct thing you say is an error.
Give reasons for your statements.
Unclear or incorrect reasons will receive 0 or $-1$ points.
Correct reasons will get $+1$ points each:
$$
\align
\vec\nabla\times\vec F&=-\vec j\\
\vec n&=2x\vec i+2 y\vec j+8z\vec k\\
(\vec\nabla\times\vec F)\cdot \vec n&=-2y.
\endalign
$$
Parameterize $S$ by $\vec r(u,v)=v\cos u\vec i+v\sin u\vec j+{1\over 2}\sqrt{4-v^2}\ \vec k$ where $0\leq u\leq 2\pi$ and $-2\leq v\leq 2$.
$$
\align
\vec r_u&=-v\sin u\vec i+v\cos u\vec j\\
\vec r_v&=\cos u\vec i+\sin u\vec j-{v\over 2\sqrt{4-v^2}}\ \vec k\\
\vec r_u\times\vec r_v&={-v^2\cos u\over 2\sqrt{4-v^2}}\ \vec i+
\bigg( {-v^2\sin u\over 2\sqrt{4-v^2}}\bigg)\ \vec j-v\vec k\\
d\sigma&=\sqrt{{v^4\over 4(4-v^2)}+v^2}\ du dv\\
\iint_S (\vec\nabla\times\vec F)\cdot \vec n d\sigma&=
\int^2_{-2} \int_0^{2\pi} (-2v\cos u)
\sqrt{{v^4\over 16-4v^2}+v^2}\ du\ dv
\endalign
$$
Now $C=\{(x,y,z)|x^2+1+4z^2=4,\ y=1\}$.
Let
$\vec r(x)$ parametrize $C$ where
$$
\vec r(x)=x\vec i+\vec j+{1\over 2}\sqrt{3-x^2}\ \vec k,\qquad\qquad
\text{ where }-\sqrt{3}\leq x\leq \sqrt{3}
$$
$$
\align
{d\vec r\over dx}&=(\vec i-{x\over 2\sqrt{3-x^2}}\ \vec k)\\
\vec F\cdot {d\vec r\over dx}&=\bigg({x\over x^2+1}\ i+{1\over x^2+1}\ j+x\vec k\bigg)\cdot \bigg(\vec i-{x\over 2\sqrt{3-x^2}}\ k\bigg)\\
\vec F\cdot {dr\over dx}&={x\over x^2+1}-{x^2\over 2\sqrt{3-x^2}}\\
\oint_C\vec F\cdot d\vec r&=\int^{\sqrt{3}}_{-\sqrt{3}}\bigg({x\over x^2+1}-{x^2\over 2\sqrt{3-x^2}}\bigg)\ dx\\
\endalign
$$
So
$$
\align
& \int^2_{-2}\int_0^{2\pi}  (-2v\cos u)\sqrt{{v^4\over 16-4v^2}+v^2}\ dv\ du\\
&\quad =\int^{\sqrt{3}}_{-\sqrt{3}}\ \bigg({x\over x^2+1}-{x^2\over 2\sqrt{3}-x^2}\bigg)\ dx
\endalign
$$
\bye
