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\centerline
{
\bf Solutions to the ``QUIZ'' for Lecture 25
}

{\bf 1.}
Let
$$
F(x,y,z) =
$$
$$
\langle \, \cos(\sqrt{1+x^7} +zy^9) \quad , \quad \tan(x^7+y^2+1/z) 
\quad , \quad
\tan^{-1} (e^{xyz}+\cos^6(x^8-y+3z) \, \rangle \quad ,
$$
and let $\langle P,Q,R \rangle = curl \,\, {\bf F}$. Compute
$$
{{\partial P} \over {\partial x}}+
{{\partial Q} \over {\partial y}}+
{{\partial R} \over {\partial z}} \quad .
$$
Be sure to explain everything.
(Note: there was a typo in the problem that has now been corrected).
\bigskip
{\bf Sol.} We have to compute $div \, ( \langle P,Q,R \rangle )$ but $\langle P,Q,R \rangle = curl ({\bf F})$ so
we have to compute $div (curl ({\bf F}) )$. But a famous theorem says that this is always the $zero$ function.

{\bf Ans.}: $0$.

{\bf Comment:} Almost everyone got it right.

{\bf 2.} Calculate the surface integral \hfill\break
$\int\int_{S} {\bf F} \cdot d \, {\bf S}$, where
$$
{\bf F}(x,y,z)= \langle \, 2x+y+z \, , \, x+2y+z \, , \, x+y+2z \, \rangle
$$
where $S$ is the surface of the box bounded by
the planes 
$x=0$, $x=1$, $y=0$, $y=4$, $z=0$, $z=5$.
\bigskip
{\bf Sol.}:
We are supposed to use the {\bf Divergence Theorem}:
$$
\int\int_{S} {\bf F} \cdot d \, {\bf S} \, = \,
\int\int\int_E div ({\bf F}) dV \quad,
$$
where $E$ is the {\bf inside} of the box.
We have
$$
div ({\bf F})=\frac{\partial (2x+y+z)}{\partial x}+\frac{\partial (x+2y+z)}{\partial y}+\frac{\partial (x+y+2z)}{\partial z}
=2+2+2=6 \quad .
$$
So
$$
\int\int_{S} {\bf F} \cdot d \, {\bf S} \, = \,
\int\int\int_E 6 dV \quad,
$$
Taking $6$ out, we get:
$$
6 \int\int\int_E 1 dV \quad .
$$
But a {\bf volume-integral} whose integrand is $1$ equals the volume, so this equals
$$
6 \cdot Volume(Box)
$$
The volume of the box is $1 \cdot 4 \cdot 5=20$, so the answer is $6 \cdot 20=120$.

{\bf Ans.}: $120$.

{\bf Comment:} About \%95 of the students got it right.

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