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\centerline
{
\bf Solutions to the ``QUIZ'' for Lecture 23
}
\bigskip
{\bf 1.} 
Determine whether or not the vector field is conservative.
If it is, find a function $f$ such that ${\bf F}=\g f$.
$$
{\bf F}(x,y,z)= 
(3x^2y^3z^3+yz)\, {\bf i}+ 
(3x^3y^2z^3+xz) \, {\bf j}+ (3x^3y^3z^2+xy) \, {\bf k}
$$

{\bf Sol.}: Here:

$P=3x^2y^3z^3+yz$, $Q=3x^3y^2z^3+xz$, $R=3x^3y^3z^2+xy$.

We have:

$\frac{\partial P}{\partial y}=9x^2y^2z^3+z$ \quad , \quad 
$\frac{\partial Q}{\partial x}=9x^2y^2z^3+z$ \quad ,

these are equal.

$\frac{\partial P}{\partial z}=9x^2y^3z^2+y$ \quad , \quad 
$\frac{\partial R}{\partial x}=9x^2y^3z^2+y$ \quad ,

these are equal.

$\frac{\partial Q}{\partial z}=9x^3y^2z^2+x$ \quad , \quad 
$\frac{\partial R}{\partial y}=9x^3y^2z^2+x$ \quad ,

these are equal.

So {\bf F} is indeed conservative.

We now need  to find a function $f(x,y,z)$ such that
$$
\langle \frac{\partial f}{\partial x},
\frac{\partial f}{\partial y},
\frac{\partial f}{\partial z} \rangle =
\langle P,Q,R \rangle \quad .
$$
Since $\frac{\partial f}{\partial x}=P$,
$$
f(x,y,z)= \int (3x^2y^3z^3+yz) \, dx = x^3y^3z^3+xyz+ g(y,z) \quad,
$$
where $g(y,z)$ is yet-to-be-determined.
Using $\frac{\partial f}{\partial y}=Q$,
$$
3x^3y^2z^3+xz+ \frac{\partial g}{\partial y} =3x^3y^2z^3+xz \quad,
$$
giving 
$$
\frac{\partial g}{\partial y}=0 \quad,
$$
So
$$
g(y,z)= \int \, 0 \, dy = 0+ h(z) \quad ,
$$
where $h(z)$ is yet-to-be-determined.
Going back above:
$$
f(x,y,z)= x^3y^3z^3+xyz+ h(z) \quad,
$$
Using $\frac{\partial f}{\partial z}=R$, we get:
$$
3x^3y^3z^2+xy+ h'(z) =3x^3y^3z^2+xy \quad,
$$
giving $h'(z)=0$ so $h(z)=C$ and we have
$$
f(x,y,z)=  x^3y^3z^3+xyz+ C \quad,
$$
but we can forget about the $C$ (we were asked to find {\it a} potential
function, not all of them), so 

{\bf Ans. to second part:} $f(x,y,z)=x^3y^3z^3+xyz$.




{\bf 2.} 
Evalute
$$
\int_C \,\, 5y \, dx + 10x \, dy \quad ,
$$
where $C$ is the closed curve consisting of the boundary of the rectangle
$$
\{ \, (x,y) \, | \, 0 \leq x \leq 1 \quad , \quad 0 \leq y \leq 1 \, \} .
$$

{\bf Sol.}: We are supposed to use {\bf Green's Theorem}.
$$
\int_C {\bf F} \cdot d{\bf r} = \int \int_D 
\left ( \frac{\partial Q}{\partial x} -\frac{\partial P}{\partial y} \right )
\, dA \quad .
$$
Since no orientation is mentioned we take the {\bf default} one of
{\it positive} (counter-clockwise).

In this problem $P=5y$ and $Q=10x$, so we have
$$
\frac{\partial Q}{\partial x} -\frac{\partial P}{\partial y}=
10-5=5 \quad,
$$
and the vector-field line-integral that we have to compute equals:
$$
\int\int_D \, 5 \, dA =
5 \int\int_D \, 1 \, \, dA = 5 Area(D) \quad ,
$$
(since an {\bf area integral} with the integrand being $1$ equals the 
{\bf area}). By elementary-school geometry, the area of $D$ is $1 \cdot 1=1$,
so we have:

{\bf Ans.}: $5$ (type number).

{\bf Comments:} Everyone got it right!

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