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\centerline
{
\bf Dr. Z's Math251 Handout \#16.5 (2nd ed.) 
[Surface Integrals of Functions and Surface Integrals of Vector Fields]
}

By Doron Zeilberger

{\bf Problem Type 16.5a}: 
Evaluate the surface integral
$$
\int\int_S F(x,y,z) \, dS \quad,
$$
where $S$ is a given surface that is either given parametrically
in terms of $u$ $v$, or can be made so, or
can be written in the form $z=f(x,y)$, 
$\{ (x,y)  \vert, (x,y) \in D\}$ for some set $D$ in the $xy$-plane.

{\bf Example Problem 16.5a}: 
Evaluate the surface integral
$$
\int\int_S 2x^2z^2 \, dS \quad,
$$
where $S$ is the part of the cone $z^2=x^2+y^2$ that lies
between the planes $z=1$ and $z=3$.
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} If the surface can be described parametrically
as
$$
\{ (x(u,v),y(u,x),z(u,v)) \vert (u,v) \in D \} \quad,
$$
for some set $D$ in $uv$-plane, set-up
$r=x \,{\bf i}+y\,{\bf j}+z\,{\bf k}$,
and compute $r_u$, $r_v$, then
their cross product $r_u \times r_v$, and finally
its length  $\vert r_u \times r_v \vert$.
Put 
$$
dS=\vert r_u \times r_v \vert \, du \, dv \quad .
$$

On the other hand if the surface is given
as $z=f(x,y)$, with some description where it lives,
figure out the ``floor'' (projection on the $xy$-plane),
and put
$$
dS=\sqrt{1+ \left ( {{\partial z} \over {\partial x}} \right )^2
+ \left ( {{\partial z} \over {\partial y}} \right )^2} \, dA
$$

{\bf Note:} The first method also works in the
second case, just take $x$, $y$ as the parameters
and put $x=x,y=y,z=f(x,y)$.

&
{\bf 1.} In this problem $z^2=x^2+y^2$ means
$z=\sqrt{x^2+y^2}$, so it is possible to do it the
second way. But it is a bit easier to use cylindrical
coordinates and then the surface is $z=r$, and
the parametric representation is
$$
x=r \cos \theta \, , \, 
y=r \sin \theta \, , \, 
z=r \quad ,
$$
where the parameters are $r$ and $\theta$. The region
on the $xy$-plane below the surface is 
$$
\{ (r, \theta) \, \vert \, 1 \leq r \leq 3,
0 \leq \theta \leq 2 \pi \} \quad .
$$

So 
$$
{\bf r}(r,\theta)= r \cos \theta \, {\bf i}+
r \sin \theta \, {\bf j}+
r  \, {\bf k} \quad.
$$
$$
{\bf r}_r=  \cos \theta \, {\bf i}+
 \sin \theta \, {\bf j}+
   \, {\bf k} \quad.
$$
$$
{\bf r}_{\theta}= -r \sin \theta \, {\bf i}+
r \cos \theta \, {\bf j}+
 0 \, {\bf k} \quad.
$$
$$
{\bf r}_r \times {\bf r}_{\theta} =
- r \cos \theta \, {\bf i}-
 r \sin \theta \, {\bf j}
+ r\, {\bf k} \quad,
$$
(you do it!). Also
$$
\vert {\bf r}_r \times {\bf r}_{\theta} \vert=
\sqrt{ (- r \cos \theta)^2+(-r \sin \theta)^2 +r^2}=\sqrt{2} r \quad ,
$$
so $dS=\sqrt{2} r \, dr \, d \theta$.
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{\bf 2.} Set-up the integral
$$
\int\int_S F(x,y,z) \, dS \quad,
$$
with the $x,y,z$ replaced by their expressions in terms
of the parameters, and $S$ replaced by its description
in terms of $u,v$, and $dS$ replaced by what you found in step 1.

&
{\bf 2.} 
$$
\int\int_S 2x^2z^2 \, dS \quad
$$
$$
=\int\int_D 2(r \cos \theta)^2 r^2  \sqrt{2} r \, dr \, d\theta \quad,
$$
where $D$ is the region $1 \leq r \leq 3$, $0 \leq \theta \leq 2 \pi$.
Converting it to an iterated integral, we get
$$
=2 \sqrt{2}
\int_{0}^{2 \pi} \int_{1}^{3}  r^5 \cos^2 \theta \, dr \, d\theta \quad .
$$
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{\bf 3.} Evaluate the iterated integral.

&
{\bf 3.} 
$$
=
2\sqrt{2}
\int_{0}^{2 \pi} \int_{1}^{3}  r^5 \cos^2 \theta \, dr \, d\theta \quad .
$$
$$
=\sqrt{2}
\int_{0}^{2 \pi} 
\left [
\int_{1}^{3}  r^5 2\cos^2 \theta \, dr \right ]  \, d\theta \quad .
$$
$$
=
\sqrt{2}\, \int_{0}^{2 \pi} 2 \cos^2 \theta
\left [ {{r^6} \over {6}} \Bigl \vert_{1}^{3} \right ] \, d \theta \quad.
$$
$$
= \sqrt{2} \, \left ( {{3^6-1^6} \over {6}} \right ) \cdot
\int_{0}^{2 \pi} 2 \cos^2 \theta d \theta
$$
$$
= \sqrt{2} {{364} \over {3}} \cdot
\int_{0}^{2 \pi} (1+ \cos (2 \theta)) d \theta
$$
$$
= \sqrt{2} \, \left ( {{364} \over {3}} \right ) \cdot
(\theta + {{\sin (2 \theta)} \over {2}}) \Bigl \vert_{0}^{2 \pi}
= \sqrt{2} \, {{364} \over {3}} \cdot 2 \pi
= {{728 \sqrt{2} \pi} \over {3}}  \quad.
$$
{\bf Ans.:} ${{728 \sqrt{2} \pi} \over {3}}$.


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{\bf Problem Type 16.5b}: 
Evaluate the surface
integral $\int\int_S {\bf F} \cdot d {\bf S}$ for
the given vector field ${\bf F}$ and oriented
surface $S$.
$$
{\bf F}(x,y,z)= P(x,y,z) \, {\bf i}+Q(x,y,z) \, {\bf j}+R(x,y,z) 
\, {\bf k} \quad,
$$
where $S$ is the part of the surface $z=g(x,y)$ that lies
above some region $D$, in the $xy$-plane 
and has upward orientation.


{\bf Example Problem 16.5b}: 
Evaluate the surface
integral $\int\int_S {\bf F} \cdot d {\bf S}$ for
the given vector field ${\bf F}$ and oriented
surface $S$.
$$
{\bf F}(x,y,z)= xy \, {\bf i}+yz \, {\bf j}+zx \, {\bf k} \quad,
$$
$S$ is the part of the paraboloid $z=4-x^2-y^2$ that lies
above the square $0 \leq x \leq 1, 0 \leq y \leq 1$
and has upward orientation.


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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} 
Set-up the formula
$$
\int\int_S {\bf F} \cdot d {\bf S}=
$$
$$
\int\int_D \left (
-P {{\partial g} \over {\partial x}}
-Q {{\partial g} \over {\partial y}}
+R \right ) \, dA
$$
Be also sure to replace $z$ by $g(x,y)$.
&
{\bf 1.} Here $g=4-x^2-y^2$,
$$
P= xy \quad , \quad Q=yz \quad , \quad R=zx \quad .
$$
Plugging everything in, our surface integral
$$
=\int\int_D  -xy (-2x)-yz (-2y)+xz ) \, dA
$$
$$
=\int\int_D  (2x^2y+(2y^2+x)z ) \, dA \quad,
$$
but since $z=4-x^2-y^2$, this equals
$$
=\int\int_D  (2x^2y+(2y^2+x)(4-x^2-y^2) ) \, dA \quad,
$$
$$
=\int\int_D  (2x^2y+8y^2-2y^2x^2-2y^4+4x-x^3-xy^2) \, dA \quad,
$$
where $D$ is the square
$$
\{ \, (x,y) \, \vert \, 0 \leq x \leq 1 \, , \, 0 \leq y \leq 1 \} \quad.
$$
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{\bf 2.} Looking at $D$ convert it into an iterated integral.
&
{\bf 2.} 
$$
=\int_{0}^{1}\int_{0}^{1}  
(2x^2y+8y^2-2y^2x^2-2y^4+4x-x^3-xy^2) \, \, dx \, dy \quad.
$$
$$
=\int_{0}^{1}\int_{0}^{1}  
(-x^3+x^2(-2y^2+2y)+x(-y^2+4)-2y^4+8y^2)\, dx \, dy \quad.
$$

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{\bf 3.} Evaluate the iterated integral.
&
{\bf 3.} You do it! (it is rather tedious).
The answer turns out to be ${{713} \over {180}}$.



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{\bf A Problem from a Previous Final:}
Evaluate the surface integral
$$
\int\int_S \sqrt{3} \, x \, dS \quad ,
$$ 
where $S$ is the triangular region
with vertices $(1,0,0),(0,1,0),(0,0,1)$.

{\bf Ans.} ${{1} \over {2}}$.

\end