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\centerline
{
\bf Solutions to the ``QUIZ'' for Nov. 9, 2009
}
\bigskip


{\bf 1.} 
Determine whether or not the vector field 
$$
F(x,y,z)=y^2z^3 \, {\bf i} + 2xyz^3 \, {\bf j} + 3xy^2z^2\, {\bf k} 
$$ 
is conservative. If it is conservative, find a function $f$ such that
${\bf F}=\nabla f$.

{\bf Sol.} Here 
${\bf F}= \langle y^2z^3 \, , \, 2xyz^3 \, , \, 3xy^2z^2 \rangle$,

so $F_1=y^2z^3$, $F_2=2xyz^3$, $F_3=3xy^2z^2$.

${{\partial F_1} \over {\partial y}}= 2yz^3$,
${{\partial F_2} \over {\partial x}}= 2yz^3$,

so ${{\partial F_1} \over {\partial y}}={{\partial F_2} \over {\partial x}}$
and the first condition is OK. Also

${{\partial F_1} \over {\partial z}}= 3y^2z^2$,
${{\partial F_3} \over {\partial x}}= 2y^2z^2$,

so ${{\partial F_1} \over {\partial z}}={{\partial F_3} \over {\partial x}}$
and the second condition is OK. Finally

${{\partial F_2} \over {\partial z}}= 6xyz^2$,
${{\partial F_3} \over {\partial y}}=6xyz^2$,

so ${{\partial F_2} \over {\partial z}}={{\partial F_3} \over {\partial y}}$
and the third condition is OK. 

Now it is time to find the potential function $f$ .

From 
$$
{{\partial f} \over {\partial x}}=F_1 \quad,
$$
We get
$$
{{\partial f} \over {\partial x}}=y^2z^3 \quad .
$$
Integrating with respect to $x$, we get
$$
f(x,y,z)= \int y^2 z^3 \, dx = xy^2z^3+ g(y,z) \quad.
$$
Using
$$
{{\partial f} \over {\partial y}}=F_2 \quad,
$$
in other word
$$
{{\partial f} \over {\partial y}}=2xyz^3 \quad,
$$
We get:
$$
2xyz^3+g_y(y,z)= 2xyz^3 \quad .
$$
This meant
$$
g_y(y,z)=0 \quad .
$$
Integrating with respect to $y$, we get:
$$
g(y,z)= \int 0 \, dy = 0+ h(z) \quad.
$$
Going back above, we have:
$$
f(x,y,z)= xy^2z^3+ h(z) \quad.
$$
Using
$$
{{\partial f} \over { \partial z}}=F_3 \quad,
$$
$$
{{\partial f} \over { \partial z}}=3xy^2z^2 \quad,
$$
we get
$$
3xy^2z^2+h'(z)=3xy^2z^2 \quad
$$
Doing the algebra, we get
$$
h'(z)=0 \quad.
$$
Integrating, we get
$$
h(z) = \int 0 \, dz = 0 + C
$$
So 
$$
f(x,y,z)=xy^2z^3 + C \quad.
$$
But you don't have to write the $+C$.

{\bf Ans.} $f(x,y,z)=xy^2 z^3$ (type: multivariable function).

{\bf Comments}: 1. About \%50 got it perfectly. Most people knew
how to do it.
2. This particular problem could be done ``by inspection'', but you
are still expected to follow the general method.

{\bf 2.} 
Show that the line integral
$$
\int_C 2xy^2 \, dx \, + \, 2x^2y \, dy \quad ,
$$
is independent of the path $C$, and evaluate it if
$C$ is {\it any} path from $(1,0)$ to $(0,1)$.

{\bf Sol.} Here ${\bf F}= \langle \, 2xy^2 \, , 2x^2y \, \rangle$.
So $F_1=2xy^2, F_2=2x^2y$. Since $(F_1)_y=4xy,(F_2)_x=4xy$,
$(F_1)_y=(F_2)_x$ so this is a {\bf conservative} vector field.

Let's find a potential function.

$$
f(x,y) = \int 2xy^2 \, dx = x^2y^2 + g(y) \quad ,
$$
So
$$
{{\partial f} \over {\partial y}}= 2x^2y+ g'(y) \quad .
$$
Using
$$
{{\partial f} \over {\partial y}}= 2x^2y \quad .
$$
We get
$$
2x^2y+g'(y)=2x^2y \quad
$$
So $g'(y)=0$, and so $g(y)=C$, and
$$
f(x,y)= x^2 y^2 \quad .
$$


The line-integral (regadless of the path) is always
$$
f(END) - f(START) \quad.
$$
The question tells you that the path starts at $(1,0)$ and
ends at $(0,1)$.
So the answer is:
$$
f(0,1) - f(1,0)= 0^2 \cdot 1 - 1^2 \cdot 0 = 0 \quad .
$$

{\bf Ans.} $0$ (type number).

{\bf Comments}: About \%40 got it, but many people ran out of time.






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