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\centerline
{
\bf Dr. Z's Math251 Handout \#16.8 [Stokes' Theorem]
}

By Doron Zeilberger

{\bf Problem Type 16.8a}: 
Use Stokes' Theorem to evaluate
$\int\int_S \, curl \, {\bf F} \cdot d {\bf S}$, where
$$
{\bf F}(x,y,z)= P(x,y,z) \, {\bf i}+
Q(x,y,z) \, {\bf j}+
R(x,y,z) \, {\bf k} \quad,
$$
and $S$ is some surface with a given orientation
(that should boil down to either outwards or inwards).


{\bf Example Problem 16.8a}:  Use Stokes' Theorem to evaluate
$\int\int_S \, curl \, {\bf F} \cdot d {\bf S}$, where
$$
{\bf F}(x,y,z)= x^2 e^{2yz} \, {\bf i}+
y^2 e^{3xz} \, {\bf j}+
z^2 e^{4xy} \, {\bf k} \quad,
$$
and $S$ is the hemisphere $x^2+y^2+z^2=9$,
$z \geq 0$, oriented upwards.


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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} 
You are going to use Stokes' Theorem
$$
\int\int \, curl \, {\bf F} \cdot d {\bf S} 
= \int_C {\bf F} \cdot d\,{\bf r}
\quad .
$$
The challenge is to find the bounding curve $C$.

For hemispheres $x^2+y^2+z^2=R^2$, $z \geq 0$,
like in this problem, the bounding curve
is simply the circle $x^2+y^2=R^2$, on the $xy$-plane
$z=0$ and the parametric representation is
$$
x=R \cos t \, , \, y= R \sin t \, , \, z=0 \quad.
$$
If it is the hemisphere $z \leq 0$ then it is the same
but in the negative direction .
If it is the hemisphere $x^2+y^2+z^2=R^2$, $y \leq 0$,
then  the parametric representation is
$$
x=R \cos t \, , \, z= R \sin t \, , \, y=0 \quad,
$$
etc.

If it is the part of a surface $z=g(x,y)$ that lies
above a plane $z=a$, oriented outwards, then
$C$ is obtained by solving $g(x,y)=a$ and
representing $g(x,y)=a$ in parametric notation and adding to it
$z=a$. The orientation of $C$ is such as to obey the
right-hand rule.

At the end you need to represent $C$ in parametric form
$$
{\bf r}(t)= \langle x(t) , y(t), z(t) \rangle \, , \,
a \leq t  \leq b \quad,
$$
for some expressions $x(t),y(t),z(t)$ of $t$ and some
numbers $a$ and $b$.

&
{\bf 1.} Here $C$ is simply the circle
$x^2+y^2=3^2$ that  lives in the $xy$-plane, and its parametric
representation is
$$
x=3 \cos t \, , \, y= 3 \sin t \, , \, z=0 \quad.
$$
So
$$
{\bf r}(t)=3 \cos t \,{\bf i}+ 3 \sin t \,{\bf j} + 0 \, {\bf k} \quad,
$$
$$
0 \leq t \leq 2 \pi \quad .
$$

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{\bf 2.} Plug-in the expressions for $x,y,z$ in terms of $t$
in ${\bf F}$, in order to express it in terms of the parameter $t$.
Also figure out $d{\bf r}={\bf r}'(t) \, dt$.

&
{\bf 2.} 
$$
{\bf F}(x,y,z)= 
(3 \cos t)^2 e^{0} \, {\bf i}+
(3 \sin t)^2 e^{0} \, {\bf j}+
0 \, {\bf k} 
$$
$$
=9 \cos^2 t  \, {\bf i}+
9 \sin^2 t  \, {\bf j}+
0 \, {\bf k} \quad,
$$
$$
d{\bf r}(t)=(-3 \sin t \,{\bf i}+ 3 \cos t \,{\bf j} + 0\, {\bf k}) 
\, dt \quad .
$$
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{\bf 3.} Find the dot product ${\bf F} \cdot d {\bf r}$ and
integrate from $t=a$ to $t=b$.

&
{\bf 3.} 
$$
{\bf F} \cdot d {\bf r} =
((9 \cos^2 t) \cdot (-3 \sin t)+
(9 \sin^2 t) \cdot (3 \cos t)+ 0 \cdot 0)\, dt=
$$
$$
(-27 \cos^2 t \sin t+27 \sin^2 t \cos t) \, dt \quad .
$$
Finally, integrating from $t=0$ to $t=2 \pi$, we get
$$
\int_C {\bf F} \cdot d {\bf r} =
\int_0^{2 \pi}
(-27 \cos^2 t \sin t+27 \sin^2 t \cos t) \, dt
$$
$$
=9\cos^3 t+9 \sin^3 t \Bigl \vert_{0}^{2 \pi}
$$
$$
=
(9 \cos^3 (2\pi)+9 \sin^3 (2\pi))
-
(9 \cos^3 (0)+9 \sin^3 (0))
$$
$$
=9-9=0 \quad.
$$
{\bf Ans.}: $0$.

\cr}

{\bf Problem Type 16.8b}: 
Use Stokes' Theorem to evaluate
$\int\int_S \, curl \, {\bf F} \cdot d\, {\bf S}$, if
$$
{\bf F}=P(x,y,z) \, {\bf i}+
Q(x,y,z) \, {\bf j}+
R(x,y,z) \, {\bf k} \quad ,
$$
and $S$ consists of the top and four sides (but not the bottom)
of the cube with vertices $(\pm A, \pm A, \pm A)$.

{\bf Example Problem 16.8b}:  
Use Stokes' Theorem to evaluate
$\int\int_S \, curl \, {\bf F} \cdot d\, {\bf S}$, if
$$
{\bf F}= x^2 y^2\, {\bf i}+
y^2z \, {\bf j}+
yz^2 \, {\bf k} \quad ,
$$
and $S$ consists of the top and four sides (but not the bottom)
of the cube with vertices $(\pm 2, \pm 2, \pm 2)$.


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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} In this problem, it is possible to find
the bounding curve $C$, and use Stokes's Theorem directly,
{\bf but}, in this case $C$ is a square with four sides
and we would have to do four integrals, and it is a pain.
We will use Stokes's theorem {\bf indirectly} by
finding {\bf another} surface with the same bounding curve.
Naturally for a box in which the given surface consists
of the top and the four walls, the bottom is such a surface.
&
{\bf 1.} The bottom face is 
$$
-2 \leq x, y \leq 2 \quad , \quad z=-2 \quad.
$$
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{\bf 2.} Find $curl \, {\bf F}$.
&
{\bf 2.} 
$$
curl \, {\bf F}=
(z^2-y^2) \, {\bf i} +0 \, {\bf j} -2x^2y \, {\bf k} 
$$
(You do it!)
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{\bf 3.} Plug-in $z=-A$ and note that $dS=d \,x \, d \, y \, {\bf k}$,
and the region of inegration is
$$
\{ (x,y) \, \vert \, -A \leq x \leq A, -A \leq y \leq A \, \} \quad .
$$
Do the integration
&
{\bf 3.} When $z=-2$,
$$
curl \, {\bf F}=
(4-y^2) \, {\bf i} +0 \, {\bf j} -2x^2y \, {\bf k}  \quad .
$$
So 
$$
curl \, {\bf F} \cdot d {\bf S}=
((4-y^2) \, {\bf i} +0 \, {\bf j} -2x^2y \, {\bf k}) \cdot
{\bf k}=-2x^2y \quad.
$$
Finally,
$$
\int\int_S curl \, {\bf F} \cdot d {\bf S}=
\int_{-2}^{2}\int_{-2}^{2} -2x^2y \, dx \, dy
$$
$$
=\int_{-2}^{2} \left [ \int_{-2}^{2} -2x^2y \, dx \right ] \, dy
=
\int_{-2}^{2}
(-2y) \left [ 
{{x^3} \over {3}} \Bigl \vert_{-2}^{2}  \right ] \, dy
$$
$$
=\int_{-2}^{2} (-2y)  {{16} \over {3}} \, dy
$$
$$
={{-32} \over {3}} \int_{-2}^{2} y   \, dy
={{-32} \over {3}} \cdot {{y^2} \over {2}} \Bigl \vert_{-2}^{2} 
={{-32} \over {3}} \cdot 0=
0 \quad .
$$
{\bf Ans.}: $0$ .

\cr}

{\bf Problem Type 16.8c}: 
Use Stokes' Theorem to
evaluate $\int_C {\bf F} \cdot d \, {\bf r}$,  where
$$
{\bf F}(x,y,z)=
P(x,y,z) \, {\bf i}+
Q(x,y,z) \, {\bf j}+
R(x,y,z) \, {\bf k} \quad ,
$$
and $C$  is a curve that bounds some surface
(that you have to figure out!), $z=g(x,y)$,
above the region $\{ (x,y) \vert (x,y) \in D \, \}$
(that you have to find!).
$C$ is oriented counterclockwise as viewed from above.

{\bf Example Problem 16.8c}:  
Use Stokes' Theorem to
evaluate $\int_C {\bf F} \cdot d \, {\bf r}$,  where
$$
{\bf F}(x,y,z)=
(2x+y^2) \, {\bf i}+
(2y+z^2) \, {\bf j}+
(3z+x^2) \, {\bf k} \quad ,
$$
and $C$  is the triangle with vertices
$(2,0,0)$,$(0,2,0)$, $(0,0,2)$, and
is oriented counterclockwise as viewed from above.

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Find a convenient surface that our curve bounds,
and express it in terms of $z=g(x,y)$. Also figure
out its projection on the $xy$-plane.

&
{\bf 1.} The three vertices of our triangle lie on the plane
$x+y+z=2$ ({\bf you do it!}), so $z=2-x-y$, and $g(x,y)=2-x-y$.
Also the projection of the triangle on the $xy$ plane
is bounded by the line $x+y=2$ and the axes, so
it is the type I region
$$
D=\{ (x,y) \, \vert \, 0 \leq x \leq 2 \, , 0 \leq y \leq 2-x \, \} \quad .
$$

\cr

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{\bf 2.} Find $curl \, {\bf F}$.

&
{\bf 2.} 
$$
curl \, {\bf F} =
-2z \, {\bf i}-2x \, {\bf j}-2y \, {\bf k} \quad .
$$
(You do it!)
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{\bf 3.} You have to use Stokes' Theorem
$$
\int_C {\bf F} \cdot d \, {\bf r} =
\int_S  curl \, {\bf F} \cdot d \, {\bf S}  \quad .
$$
Convert the surface integral into an area integral using the
formula from 16.7
$$
\int\int_S {\bf F} \cdot d {\bf S}=
$$
$$
\int\int_D \left (
-P {{\partial g} \over {\partial x}}
-Q {{\partial g} \over {\partial y}}
+R \right ) \, dA
$$
Be also sure to replace $z$ by $g(x,y)$.
{\bf Note:} The ${\bf F}$ from this formula is
{\bf not} the same as our ${\bf F}$, it is
rather our $curl \, {\bf F}$, so use ${\bf F}$
as a {\it local variable}.
&
{\bf 3.} 
$$
\int_S  curl {\bf F} \cdot d \, {\bf S}=
\int_S   (-2z \, {\bf i}-2x \, {\bf j}-2y \, {\bf k} )
\cdot d \, {\bf S} \quad
$$
where the surface is the one from step 1, i.e.
$z=2-x-y$ over $D$ given above.
Here we have
$P=-2z$, $Q=-2x$, $R=-2y$, $g=2-x-y$, so
$$
\int\int_D (-(-2z) (-1)-(-2x) (-1)+(-2y) ) \, dA \quad .
$$
Now we have to replace $z$ by $2-x-y$, so
this equals
$$
\int\int_D (-2(2-x-y)-2x-2y ) \, dA 
=\int\int_D -4 \, dA \quad 
$$
$$
=-4 \int\int_D \, dA \quad ,
$$
where $D$ is the triangle 
$$
D=\{ (x,y) \, \vert \, 0 \leq x \leq 2 \, , 0 \leq y \leq 2-x \, \} \quad .
$$
In this case the area integral is simply the area of the triangle
($2\cdot 2/2=2$) times $-4$, so the answer is $-8$,
but of course you are welcome to to do it without
the shortcut:
$$
\int\int_D (-4) \, dA =
\int_{0}^{2}\int_0^{2-x} (-4) \, dx \, dy =
-8 \quad.
$$
{\bf Ans.}: $-8$.


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