

\section{Stokes's Theorem}

\begin{enumerate}[1.]


\item
{\it Evaluate $\iint_{\mathcal S} {\bf curl\, F}\cdot d{\bf S}$, if ${\bf F}=(x^4-2y){\bf i}+(\cos y){\bf j}
+z\sin(xy){\bf k}$ and $\mathcal{S}$ is the part of the paraboloid $z=9-x^2-y^2$ that lies above the $xy$-plane.}



\medskip

We apply Stokes's Theorem twice, first to $\mathcal S$ and its boundary $C=\{(x,y,z):\ x^2+y^2=9,\ z=0\}$
oriented counterclockwise, then to $C$ and $D==\{(x,y,z):\ x^2+y^2\leq 9,\ z=0\}$:
$$
\iint_{\mathcal S} {\bf curl\, F}\cdot d{\bf S}
= \int_C {\bf F}\cdot d{\bf r}
=
\iint_{D} {\bf curl\, F}\cdot {\bf k} dA.
$$
On $D$, we have ${\bf curl\, F}\cdot {\bf k} 
= \frac{\partial (\cos y )}{\partial x} - \frac{\partial (x^4-2y)}{\partial y}=2$, so that the last integral is
$$
\iint_D 2dA = 2\cdot (\hbox{area of }D) = 2\cdot 9\pi = 18\pi.$$

\bigskip


\item {\it 
Use Stokes's Theorem to evaluate $\int_C(x-z)dx+(x+y)dy+(y+z)dz$,
if $C$ is the ellipse where the plane $z=y$ intersects the cylinder
$x^2+y^2=1$, oriented counterclockwise as viewed from above.}

\medskip

Let ${\bf F}=(x-z)\ii+(x+y)\jj+(y+z)\kk$.
By Stokes's Theorem, we have 
$$\displaystyle{\int_C {\bf F}\cdot d{\bf s}=
\iint_\mathcal{S} (\nabla\times {\bf F})\cdot d{\bf S}},$$
where $\mathcal{S}$ is the region in the plane $z=y$ bounded by $\mathcal{C}$.
We parametrize $\mathcal{S}$
as $\{(x,y,y):(x,y)\in D\}$, where $D$ is the disc $x^2+y^2=1$ in 
the $xy$-plane.  Then ${\bf n}= -\jj+\kk$ and 
$\nabla\times {\bf F}=\ii-\jj+\kk$ (you should verify this!), so that
$$\iint_\mathcal{S} (\nabla\times {\bf F})\cdot d{\bf S}
=\iint_D 2 dxdy =2(\hbox{area of }D)=2\pi.$$




\bigskip
















\item {\it Evaluate the integral $\int_C {\bf F}\cdot d{\bf r}$, where $C$ is the curve of intersection of the cylinder $x^2+y^2=1$
and the paraboloid $z=100-(x-2)^2 - (y+4)^2$, oriented counterclockwise when viewed
from above, and ${\bf F}=(3x^2y+y^3)  {\bf i}+z^4 {\bf k}$.}

\bigskip



We apply Stokes's Theorem to $C$ and to the surface $\mathcal{S}$ consisting of:

\smallskip

\begin{itemize}
\item the part 
$\mathcal{S}_1$ of the cylinder $x^2+y^2=1$ between $C$ and the $xy$-plane, with normal pointing inward, 
\item and
the disc $D=\{(x,y,z): x^2+y^2\leq 1,\ z=0\}$, with normal $\kk$.
\end{itemize}

\smallskip

By Stokes's Theorem, $\int_C {\bf F}\cdot d{\bf r}=
\iint_{\mathcal S} {\bf curl\, F}\cdot d{\bf S}$.
The curl of ${\bf F}$ is
$$
{\bf curl\, F}=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
\partial_x & \partial_y & \partial_z  \\
3x^2 y+y^3 & 0 & z^4 \\
            \end{matrix}\right|
= - (3x^2 +3y^2)\kk.
$$
On the cylindrical surface, ${\bf curl\, F}\cdot {\bf N}=0$. Therefore 
\begin{align*}
\int_C {\bf F}\cdot d{\bf r}&= - \int_D (3x^2+3y^2) dA
\\
&= - \int_0^1\int_0^{2\pi} 3r^2\cdot r \, d\theta dr
\\
&= - 2\pi \cdot \frac{3r^4}{4}\Big|_0^1 = - \frac{3\pi}{2}.
\end{align*}

\bigskip



\item {\it 
Let $R$ be the cube whose one face is the square in the $xy$-plane with vertices $(\pm 1,\pm 1,0)$, and the opposite face lies in the plane $z=2$ and has vertices $(\pm 1, \pm 1, 2)$. Let
$${\bf F}=xyz^2  {\bf i}+xy^2\jj +x^3yz{\bf k}.$$
Evaluate
$\iint_\mathcal{S} (\nabla\times {\bf F})\cdot d{\bf S}$, if $\mathcal{S}$ is the surface consisting of the 4 vertical faces and the top of $R$ (without the bottom), oriented so that the normal vector points away from the origin.}


\bigskip

We apply Stokes's Theorem twice:
\begin{align*}
\iint_\mathcal{S} (\nabla\times {\bf F})\cdot d{\bf S}
&= \int_C {\bf F}\cdot d{\bf r}
\\
&= \iint_D (\nabla\times {\bf F})\cdot d{\bf S},
\end{align*}
where $D$ is the square with vertices $(\pm 1,\pm 1,0)$ in the $xy$-plane, oriented with ${\bf N}=\kk$, and $C$ is the common oriented boundary of both $\mathcal{S}$ and $D$. 
The curl of ${\bf F}$ is
$$
{\bf curl\, F}=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
\partial_x & \partial_y & \partial_z  \\
xyz^2 & xy^2 & x^3yz \\
            \end{matrix}\right|
= x^3z\ii - (3x^2yz-2xyz)\jj + (y^2-xz^2)\kk.
$$
In the $xy$-plane, we have $z=0$, so that ${\bf curl\, F}=y^2\kk$.
Therefore 
\begin{align*}
\iint_D (\nabla\times {\bf F})\cdot {\bf k}dA&= 
\int_D y^2 dA
\\
&=  \int_{-1}^1\int_{-1}^{1} y^2\,dxdy =  \int_{-1}^1 2y^2 dy = \frac{2y^3}{3}\Big|_{-1}^1 = \frac{4}{3}.
\end{align*}


\end{enumerate}

