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\begin{document}
\title{Solutions, 311-XXII}
\maketitle

{\bf 21( 4/ 8) Introduction to Divergence and Stokes theorems\\
          Sec. 4.(9)  p.262(7,8*,12*,13,14*,16*)}\\
\section{Problem 4.9.8}
Use Stokes' theorem to solve 4.1.6:\\
Find the integral $\oint {\bf F}\cdot {\bf R}$ around the circumference of the
circle $x^2 - 2x + y^2 = 2$, $z=1$, where ${\bf F} = y {\bf i} + x{\bf j}
+xyz^2 {\bf k}$.
\\ 
{\bf Solution:}\\
By Stokes theorem we have
\[ 
\oint_C {\bf F} \cdot d {\bf R} 
   = 
\int \int_\Sigma \left(\nabla \times {\bf F} \right) \cdot {\bf n} dS
\] 
where $\Sigma$ is the interior of the circle $(x-1)^2+y^2 = 2$, $z=1$
with $C$ its boundary, oriented clockwise as seen from the positive $z$-axis.


The curl of ${\bf F}$ is needed for the surface integral:
\begin{eqnarray*}
 \nabla \times {\bf F} & = & \left|
\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k} \\
\partial_x &\partial_y & \partial_z  \\
y   & x & xyz^2 \end{array} \right| =  
z^2\left(x{\bf i}- y{\bf j}\right)
\end{eqnarray*}
while
\[ {\bf n} dS = {\bf k}dx dy \ , \]
since the circle lies on the plane $z=1$, parallel to the $xy$ plane.
Then
\begin{eqnarray*}
\int \int_\Sigma \left(\nabla \times {\bf F} \right) \cdot {\bf n} dS
& = & 0 \ . 
\end{eqnarray*}

To get the same result without using Stokes' them, we evaluate the line
integral directly.  On the circle
$z=1$ and
$d{\bf R} = 2(x-1) {\bf i} dx + 2y {\bf j} dy $, while
${\bf F} = y {\bf i} + x{\bf j}
+xy {\bf k}$. 
Converting to polars, 
$ x = 1 + \sqrt{2} \cos{\theta} \ ,\  y = \sqrt{2} \sin{\theta}$,
with  $ dx = -\sqrt{2}\sin{\theta} d\theta$ 
and $dy  = \sqrt{2}\cos{\theta} d\theta$.
\begin{eqnarray*} 
\oint_C {\bf F} \cdot d {\bf R} & = &
  \oint_C \left( 
y {\bf i} + x{\bf j} +xy {\bf k} \right) \cdot
\left( 2(x-1) {\bf i} dx + 2y {\bf j} dy \right) \\
& = & \oint_C \left(2y(x-1) dx + 2xy dy \right) \\
& = & \int_{\theta = 0}^{2\pi}
\left(-4\sqrt{2} \sin^2{\theta} \cos{\theta} + 4 \sin{\theta}\cos{\theta}
\left(1 + \sqrt{2} \cos{\theta}\right)\right) d\theta \\
& = & \int_{\theta = 0}^{2\pi} 
\left(-4\sqrt{2}\cos{\theta}\sin^2{\theta} 
+ 4\sqrt{2} \cos^2{\theta}\sin{\theta}
+ 4 \cos{\theta} \sin{\theta} \right)d\theta \\
& = &   0 
\end{eqnarray*}

\begin{center}
\psfig{figure=49_8.eps,width=3.5in}
\end{center}


\section{Problem 4.9.12}
Let $S$ be the portion of the paraboloid $z=9-x^2-y^2$ that lies above the plane
$z=0$, and let ${\bf F} = (y-z) {\bf i} - (x+z){\bf j} +(x+y) {\bf k}$.
Find $\int \int_S \left(\nabla \times {\bf F} \right) \cdot {\bf n} dS$.\\
{\bf Solution:}\\
By Stokes theorem we have
\[ \int \int_S \left(\nabla \times {\bf F} \right) \cdot {\bf n} dS
   = \oint_C {\bf F} \cdot d {\bf R} 
= \int \int_{\Sigma} \left(\nabla \times {\bf F} \right) \cdot {\bf n} dS  \]
where $\Sigma$ is the interior of the circle $x^2+y^2 = 9$, $z=1$ 
on the $x$-$y$ plane and $C$ is the common boundary of $S$ and $\Sigma$
oriented clockwise as seen from the positive $z$-axis.

The curl of ${\bf F}$ is needed:
\[ \nabla \times {\bf F} = \left|
\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k} \\
\partial_x &\partial_y & \partial_z  \\
(y-z)   & - (x+z) & (x+y) \end{array} \right| =  
2\left({\bf i}- {\bf j}\right)\]
Any one of these integrals will give the desired answer. For practice I show
all three:
\begin{enumerate}
\item The original integral 
$\int \int_S \left(\nabla \times {\bf F} \right) \cdot {\bf n} dS$.\\
On $S$, we have that $z = f(x,y) = 9-x^2-y^2$. Then
\begin{eqnarray*}
 {\bf n} dS & = & \partial_x{\bf R}\times \partial_y{\bf R} dx dy 
              =   \left({\bf i} - 2x {\bf k}\right)
            \times\left({\bf j} - 2y {\bf k}\right) dx dy \\
            & = & \left|
\begin{array}{rrr}
{\bf i} & {\bf j} & {\bf k} \\
1       &      0  &   -2x   \\
0       &      1  &   -2y \end{array} \right|   
              =   2x {\bf i} + 2y {\bf j} + {\bf k}
\end{eqnarray*}
\begin{eqnarray*}
\int \int_S \left(\nabla \times {\bf F} \right) \cdot {\bf n} dS & = &
\int \int_{\Sigma} 2\left({\bf i}- {\bf j}\right)\cdot
            \left( 2x {\bf i} + 2y {\bf j} + {\bf k}\right) dx dy \\
& = &4\int \int_{\Sigma} (x-y) dx dy = 0
\end{eqnarray*}
by the symmetry of the domain about $(x,y)=(0,0)$ and the oddness of the
integrand.

\item The integral over the circle 
$\int \int_{\Sigma} \left(\nabla \times {\bf F} \right) \cdot {\bf n} dS$.\\
On the circle ${\bf n} dS = {\bf k} dx dw $, so that
$ \left( \nabla \times {\bf F} \right) \cdot {\bf n} =
2\left({\bf i}- {\bf j}\right)\cdot {\bf k} = 0$
so that
\[
\int \int_{\Sigma} \left(\nabla \times {\bf F} \right) \cdot {\bf n} dS = 0 
\ .\] 
\item The line integral $\oint_C {\bf F} \cdot d {\bf R}$.\\
Now, since again $z=0$ and 
$d{\bf R} = 2x {\bf i} dx + 2y {\bf j} dy$,
we have (we will use polar coordinates, $x=3\cos{\theta}$ 
and $y=3\sin{\theta}$):
\begin{eqnarray*} 
\oint_C {\bf F} \cdot d {\bf R} & = &
  \oint_C \left( 
(y-z) {\bf i} - (x+z){\bf j} +(x+y) {\bf k} \right) \cdot
\left( 2x {\bf i} dx + 2y {\bf j} dy \right) \\
& = & \oint_C 2xy \left( dx - dy \right)
 =  \int_{\theta = 0}^{2\pi} 18\cos{\theta} \sin{\theta}
\left( -3 \sin{\theta} + 3\cos{\theta} \right) d\theta \\
& = & 54 \int_{\theta = 0}^{2\pi} 
\left(\cos^2{\theta}\sin{\theta} - \cos{\theta}\sin^2{\theta}\right)d\theta \\
& = &   -18 \left. \left(  \cos^3{\theta} + \sin^3{\theta}\right) 
\right|_{\theta = 0}^{2\pi} = 0 
\end{eqnarray*}
\end{enumerate}
Thus, all versions produce the same answer, $0$, and in this case the easiest
computation was the surface integral over the planar surface (circle $\Sigma$).
\begin{center}
\psfig{figure=49_12.eps,width=3.5in}
\end{center}

\newpage

\section{Problem 4.9.14}
Use Stokes' theorem to evaluate 
\[ \int_C \left[ x\sin{y} {\bf i} - y\sin{x}{\bf j}+(x+y)z^2{\bf k} \right]
\cdot d{\bf R} \]
along the path consisting of straight-line segments successively joining
the points $P_0 = (0,0,0)$ to $P_1 = (\pi/2,0,0)$ to $P_2 = (\pi/2,0,1)$
to $P_3 = (0,0,1)$ to $P_4 = (0,\pi/2,1)$ to $P_5 = (0,\pi/2,0)$ and back to
$P_0 = (0,0,0)$.\\
{\bf Solution:}\\
We will use Stokes' theorem so we can convert to a surface integral over
the two planar rectangles one with vertices $P_0,P_1,P_2,P_3$ on the 
$y=0$ plane with unit normal ${\bf j}$, 
the other with vertices $P_3,P_4,P_5,P_0$ on the $x=0$ plane
with unit normal ${\bf i}$.
We need the curl of 
$ {\bf F} := x\sin{y} {\bf i} - y\sin{x}{\bf j}+(x+y)z^2{\bf k}$:
\begin{eqnarray*}
 \nabla \times {\bf F} & = & \left|
\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k} \\
\partial_x &\partial_y & \partial_z  \\
x\sin{y}   & - y\sin{x} & (x+y)z^2 \end{array} \right| =  
z^2\left({\bf i}- {\bf j}\right) - (y\cos{x}+x\cos{y}){\bf k} \ .
\end{eqnarray*}
Then 
\[ \left. \nabla \times {\bf F} \cdot {\bf j} \right|_{y=0}
= - z^2 \ ,\]
and
\[ \left. \nabla \times {\bf F} \cdot {\bf i} \right|_{x=0}
= z^2 \ .\]
so that
\begin{eqnarray*}
\oint_C {\bf F} \cdot d {\bf R} & = & -\int\int_{xz} z^2 dx dz 
                                      +\int\int_{yz} z^2 dy dz \\
& = & -\int_{x=0}^{2} dx \int_{z=0}^1 z^2 dz 
      +\int_{y=0}^{\pi/2}\int_{z=0}^1 z^2 dz \\
& = & - \left. x \right|_0^{\pi/2} \left. \frac{z^3}{3} \right|_0^1
      + \left. y \right|_0^{\pi/2} \left. \frac{z^3}{3} \right|_0^1 \\
& = & -\frac{\pi}{6} + \frac{\pi}{6} = 0 \ . 
\end{eqnarray*}

\begin{center}
\psfig{figure=49_14.eps,width=5.in}
\end{center}

\newpage

\section{Problem 4.9.16}
If ${\bf F} = xz {\bf i} - y{\bf j}+x^2y{\bf k}$, use Stokes' theorem
to evaluate $\int_C {\bf F}\cdot {\bf R}$, where $C$ is the closed path
consisting of the edges of the triangle with vertices at the points 
$P_1 = (1,0,0)$, $P_2 = (0,0,1)$, $P_3 = (0,0,0)$ transversed from $P_1$ to
$P_2$ to $P_3$, and back to $P_1$.\\
{\bf Solution:}\\
We will use Stokes' theorem so we can convert to a surface integral over
the surface of the triangle
with vertices $P_0,P_1,P_2$ on the
$y=0$ plane with unit normal ${\bf j}$.
We need the curl of ${\bf F}$:
\begin{eqnarray*}
 \nabla \times {\bf F} & = & \left|
\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k} \\
\partial_x &\partial_y & \partial_z  \\
xz   & - y & x^2y \end{array} \right| =  
x^2{\bf i}+x(1-2y) {\bf j}\ .
\end{eqnarray*}
Then
\[ \left. \nabla \times {\bf F} \cdot {\bf j} \right|_{y=0}
= x \ ,\]
and we have
\begin{eqnarray*}
\oint_C {\bf F} \cdot d {\bf R} & = & \int\int_{xz} x dx dz  \\
& = & \int_{x=0}^{1} dx x \int_{z=0}^{1-x} dz \\
& = & \int_{x=0}^{1} dx x(1-x) \\ 
& = & \left. \frac{x^2}{2} - \frac{x^3}{3}\right|_0^1\\
& = & \frac{1}{6}
\end{eqnarray*}
\begin{center}
\psfig{figure=49_16.eps,width=5.in}
\end{center}


\end{document}

