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\centerline
{
\bf Dr. Z's Math251 Handout \#16.9 [The Divergence Theorem]
}

By Doron Zeilberger

{\bf Problem Type 16.9a}: 
Use the Divergence  Theorem
to calculate the surface integral
$\int\int_{S} {\bf F} \cdot d \, {\bf S}$, where
$$
{\bf F}(x,y,z)= P(x,y,z) \, {\bf i}+
Q(x,y,z) \, {\bf j}
+ R(x,y,z) {\bf k} \quad,
$$
where $S$ is a surface 
bounding some solid region that can be expressed
in some coordinate system (Cartesian, cylindrical, or spherical).

{\bf Example Problem 16.9a}: 
Use the Divergence  Theorem
to calculate the surface integral \hfill\break
$\int\int_{S} {\bf F} \cdot d \, {\bf S}$, where
$$
{\bf F}(x,y,z)= 
e^{2x} \sin 2y \, {\bf i}+
e^{2x} \cos 2y \, {\bf j}
+ y^2z^2 {\bf k} \quad,
$$
where $S$ is the surface of the box bounded by
the planes 
$x=0$, $x=2$, $y=0$, $y=1$, $z=0$, $z=3$.


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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Compute the divergence of ${\bf F}$:
$$
div \, {\bf F}=
{{\partial P} \over {\partial x}}+
{{\partial Q} \over {\partial y}}+
{{\partial R} \over {\partial z}} \quad .
$$

&
{\bf 1.} 
$$
div \, {\bf F}=
{{\partial } \over {\partial x}} (e^{2x} \sin 2y)+
{{\partial } \over {\partial y}} (e^{2x} \cos 2y ) +
{{\partial } \over {\partial z}} ( y^2z^2 )
$$
$$
=2e^{2x} \sin 2y
-2 e^{2x} \sin 2y +
2y^2z=2y^2z  \quad .
$$
\cr

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{\bf 2.} Write the solid region $E$ in `iterated form'
in either Cartesian, Cylindrical, or Spherical coordinates.

&
{\bf 2.} The solid region $E$ bounding our surface is clearly
the box
$$
\{ \, (x,y,z) \, \vert
\,  0 \leq x \leq 2 \, , \,
0 \leq y \leq 1 \, , \,  0 \leq z \leq 3 \, \} \quad.
$$

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{\bf 3.} Set-up the Divergence Theorem, and evaluate the
triple integral.
$$
\int\int_{S} {\bf F} \cdot d \, {\bf S}=
\int\int\int_E div \, {\bf F} \, dV \quad .
$$

&
{\bf 3.} 
$$
\int\int_{S} {\bf F} \cdot d \, {\bf S}=
\int\int\int_E div \, {\bf F} \, dV \quad .
$$
$$
=\int_0^2\int_0^1\int_0^3 2y^2z\, dz \, dy \, dx \quad .
$$
$$
=2\left ( \int_0^2 \, dx \right )
\left ( \int_0^1 y^2 \, dy \right )
\left ( \int_0^3 z \, dz \right )
$$
$$
=2(2-0) \cdot{{1^3-0^3} \over {3}} \cdot{{3^2-0^2} \over {2}}
=2 \cdot 3 =6 \quad .
$$
{\bf Ans.}: $6$.
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