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\centerline
{
\bf Dr. Z's Math251 Handout \#16.3 [The Fundamental Theorem for Line Integrals]
}

By Doron Zeilberger

{\bf Problem Type 16.3a}: Determine whether or not 
$$
{\bf F}=P(x,y) {\bf i}+ Q(x,y) {\bf j} 
$$
is a conservative vector field. If it is, find a function $f$
that ${\bf F}= \g f$.

{\bf Example Problem 16.3a}: 
Determine whether or not 
$$
{\bf F}=(e^y +y \cos x) {\bf i}+ (xe^y +\sin x +2y) {\bf j}
$$
is a conservative vector field. If it is, find a function $f$
that ${\bf F}= \g f$.
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Compute 
${{\partial P} \over {\partial y}}$ and
${{\partial Q} \over {\partial x}}$. If they are {\bf not}
the same, then the vector field is {\bf not} conservative.
End of story. If they are, then it is a {\bf conservative}
vector field, and you must go on.


&
{\bf 1.} 
$$
{{\partial P} \over {\partial y}}=
{{\partial } \over {\partial y}} (e^y +y \cos x)=e^y+\cos x \quad . 
$$
$$
{{\partial Q} \over {\partial x}}=
{{\partial } \over {\partial x}} (xe^y +\sin x +2y)=
e^y+ \cos x \quad .
$$
Since they are the same, the vector field is conservative.
We must go on.
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{\bf 2.} Look for a function $f(x,y)$ such that
$$
{{\partial f} \over {\partial x}}= P \quad ,
$$
$$
{{\partial f} \over {\partial y}}= Q \quad.
$$
To this end, you integrate $P$ w.r.t. to $x$
$$
f= \int P \, dx \quad,
$$
getting an answer up to an {\bf arbitrary constant}
(from $x$'s viewpoint), that is a function of $y$, $g(y)$.
You then plug this tentative form of $f$ into the second equation
and find what $g'(y)$ is.

&
{\bf 2.}  Since
$$
{{\partial f} \over {\partial x}}= e^y +y \cos x \quad ,
$$
we get:
$$
f= \int ( e^y +y \cos x ) \, dx=
xe^y+y\sin x+ g(y) \quad .
$$
Plugging this to the second equation, we get
$$
{{\partial } \over {\partial y}}
(xe^y+y\sin x+ g(y))= xe^y +\sin x +2y \quad ,
$$
which means
$$
xe^y+\sin x+g'(y)=xe^y+\sin x +2y \quad
$$
which means $g'(y)=2y$ .

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{\bf 3.} 
Integrate the expression that you got for $g'(y)$ w.r.t. $y$,
in order to get $g$, and plug it into $f$ above.

&
{\bf 3.} 
$g(y)=\int 2y \, dy = y^2$ (now you don't have to
bother about the $C$).
Going back to $f$, we get
$$
f=xe^y+y\sin x+ y^2 \quad .
$$
{\bf Ans.}: ${\bf F}$ is conservative and the potential function
$f$ such that ${\bf F}=\g f$ is $f=xe^y+y\sin x+ y^2$ .
\cr}

{\bf Problem Type 16.3b }: 
(a) Find a function $f$ such
that ${\bf F}=\g f$ and (b) use part (a) to evaluate
$\int_C {\bf F} \cdot d {\bf r}$ along the given curve $C$.
$$
{\bf F}(x,y,z)= P(x,y,z) \, {\bf i}+ Q(x,y,z) \, {\bf j} +
R(x,y,z) \, {\bf k} \quad,
$$
$$
C: x=x(t) \, , \, y=y(t) \, , \, z=z(t) \quad ; \quad a \leq t \leq b \quad .
$$

{\bf Example Problem 16.3b}: (a) Find a function $f$ such
that ${\bf F}=\g f$ and (b) use part (a) to evaluate
$\int_C {\bf F} \cdot d {\bf r}$ along the given curve $C$.
$$
{\bf F}(x,y,z)= (2xz+y^2) \, {\bf i}+ 2xy \, {\bf j} +
(x^2+3z^2) \, {\bf k} \quad,
$$
$$
C: x=t^2 \quad , \quad y=t+1 \quad , \quad z=2t-1 \quad, 
\quad 0 \leq t \leq 1 \quad .
$$


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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} We are looking for a function $f(x,y,z)$ such
that $f_x=P$, $f_y=Q$, $f_z=R$. First integrate
$P$ w.r.t. to $x$ getting
$$
f= \int P \, dx =A(x,y,z)+g(y,z) \quad,
$$
where $A(x,y,z)$ is explicit but $g(y,z)$ is yet to be determined.
then plug it into $f_y=Q$ getting
$$
A_y+g_y=Q \quad ,
$$
so $g_y=Q-A_y$, and integrating w.r.t. $y$ gives
$$
g(y,z)= \int (Q-A_y) \, dy = B(y,z)+ h(z) \quad ,
$$
where $B(y,z)$ is explicit but $h(z)$ is yet to be determined.
Now we have
$$
f=A(x,y,z)+B(y,z)+h(z) \quad ,
$$
we plug it into $f_z=R$ and find $h'(z)$ from which we get $h$,
and plug it back into $f$.

&
{\bf 1.} 
$f_x=2xz+y^2$ means 
$$
f=x^2z+xy^2+g(y,z) \quad .
$$
$f_y=2xy$ 
means 
$$
2xy+g_y=2xy \quad ,
$$ 
so $g_y=0$ and $g(y,z)=h(z)$.
So 
$$
f=x^2z+xy^2+h(z) \quad .
$$
$f_z=x^2+3z^2$ means
$$
x^2+h'(z)=x^2+3z^2 \quad ,
$$
so $h'(z)=3z^2$ and
$$
h(z)=z^3 \quad .
$$
Going back to $f$ we have
$$
f=x^2z+xy^2+z^3 \quad.
$$
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{\bf 2.} Plug-in $t=a$, $t=b$ to get ${\bf r}(a)$ and ${\bf r}(b)$,
and compute
$$
f( {\bf r}(b))-f( {\bf r}(a)) \quad .
$$

&
{\bf 2.} 
$$
{\bf r}(t)= \langle t^2, t+1, 2t-1 \rangle \quad,
$$
So
$$
{\bf r}(0)= \langle 0, 1, -1 \rangle \quad,
$$
$$
{\bf r}(1)= \langle 1, 2, 1 \rangle \quad.
$$
Since our $f$ equals $x^2z+xy^2+z^3$,
the value of the line integral is:
$$
f(1,2,1)-f(0,1,-1)=
$$
$$
(1^2 \cdot 1+1 \cdot 2^2+1^3)-
(0^2 \cdot (-1)+0 \cdot 1^2+(-1)^3)=7 \quad.
$$
{\bf Ans.}: $7$.
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\end

