
\section{Flux Integrals}

\begin{enumerate}[1.]
\item{ \it Let $\mathcal{S}$ be the surface ${\bf r}(u,v)=(u \cos v, u\sin v, 2v)$ with $0\leq u\leq 2$, $0\leq v\leq 2\pi$,
oriented with ${\bf N}$ pointing up.
Evaluate the integral 
$\iint_{\mathcal S} {\bf F}\cdot d{\bf S}$, where ${\bf F}=-y{\bf i}+(x+y^2){\bf j}+z{\bf k}$}


\bigskip


We have ${\bf r}_u = (\cos v, \sin v, 0),\ \ {\bf r}_v = (-u\sin  v, u\cos v, 2)$, so
$$
{\bf n}={\bf r}_u\times {\bf r}_v=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
\cos v& \sin v&  0 \\
-u\sin v & u\cos v& 2\\
            \end{matrix}\right|
=2\sin v {\bf i} -2\cos v {\bf j} + u{\bf k}.
$$
Using this ${\bf n}$, we get
\begin{align*}
\iint_{\mathcal S} {\bf F}\cdot d{\bf S}
&= \int_0^2 \int_0^{2\pi} \left[ (-u\sin v)(2\sin v) + (u\cos v + u^2\sin^2v)(-2\cos v) +2vu \right] dv du
\\
&= \int_0^2 \int_0^{2\pi} ( -2u -2u^2\sin^2 v \cos v + 2uv)  dv du
\\
&= \int_0^2 ( -2uv -\frac{2}{3} u^2\sin^3 v + uv^2) \Big|_{v=0}^{2\pi}     du
\\
&= \int_0^2 ( -4\pi u + 4\pi^2 u)    du 
\\
&= 4\pi(\pi-1)\int_0^2 u du 
= 2\pi (\pi-1) u^2\Big|_0^2 = 8\pi (\pi-1)
\end{align*}

\bigskip


\item
{\it  Evaluate the integral
$\iint_{\mathcal S} {\bf F}\cdot d{\bf S}$, where ${\bf F}= z{\bf i}+y^2{\bf j} $ and $\mathcal{S}$ is 
the part of the cylinder $x^2+y^2=4$ where $x\geq 0$, $y\geq 0$, $0\leq z\leq 1$, with the normal pointing away
from the $z$-axis.}

\medskip

We parametrize the surface using ${\bf r}(\theta,z)=\langle 2\cos\theta,2\sin\theta,z\rangle$, with
$0\leq\theta\leq\pi/2$, $0\leq z\leq 1$. Then 
${\bf r}_\theta = \langle -2\sin\theta,2\cos\theta,0\rangle ,\ \ {\bf r}_z = \langle 0,0, 1\rangle$, so
$$
{\bf n}={\bf r}_\theta \times {\bf r}_z=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
-2\sin\theta & 2\cos\theta & 0
 \\
0 & 0 & 1 \\
            \end{matrix}\right|
=2\cos\theta {\bf i} +2\sin\theta  {\bf j} .
$$
This vector points away from the $z$-axis as required. We have
$$
{\bf F}\cdot {\bf n} = 2z\cos\theta+8\sin^2\theta \sin\theta.
$$
Therefore,
\begin{align*}
\iint_{\mathcal S} {\bf F}\cdot d{\bf S}
&= \int_0^1 \int_0^{\pi/2}  (2z\cos\theta+8\sin^2\theta \sin\theta) d\theta dz \\
&= \int_0^1 \int_0^{\pi/2}  (2z\cos\theta+8(1-\cos^2\theta )\sin\theta) d\theta dz \\
&= \int_0^1   \left( 2z\sin\theta - 8\cos\theta + \frac{8}{3} \cos^3\theta \right) \Big|_{\theta=0}^{\pi/2}dz
\\
&= \int_0^1 \left( 2z+8 - \frac{8}{3} \right)dz 
= \int_0^1 \left( 2z+ \frac{16}{3} \right) \, dz = z^2+\frac{16z}{3} \Big|_{0}^1 =\frac{19}{3}
\end{align*}








\bigskip



\item
{\it  
Evaluate the surface integral $\displaystyle{\int\!\!\!\int_{S} {\bf F}\cdot d{\bf S}}$, if ${\bf F}=z\kk$ and $S$ is the rectangle with
vertices $(0,2,0)$, $(0,0,4)$, $(5,2,0)$, $(5,0,4)$, oriented so that the normal vector points upward.}

\bigskip


The plane in which the rectangle lies has the equation $2y+z=4$ (by inspection, or you can use any method you know for finding the equation of a plane passing through any three of these points). We can rewrite this as $z=4-2y$, so that 
$$d{\bf S}=(2\jj + \kk)dxdy.$$
 We need to evaluate the area of it that lies above the rectangle $0\leq x\leq 5$, $0\leq y \leq 2$. This is
$$
\int_0^5 \int_0^2 {\bf F}\cdot (2\jj+\kk )dydx = \int_0^5 \int_0^2 z dydx =\int_0^5 \int_0^2 (4-2y)  dy dx
$$
$$=\int_0^5 (4y-y^2)|_0^2 dx = \int_0^5 4 dx = 20.$$

\bigskip


\item
{\it Evaluate the surface integral
$\iint_{\mathcal S} {\bf F}\cdot d{\bf S}$, where ${\bf F}= x^2 {\bf i}$ and $\mathcal{S}$ is the part of the
plane $\displaystyle x + \frac{y}{2} + \frac{z}{3} = 1$ that lies in the first octant ($x,y,z\geq 0$),
with the normal vector pointing up.}

\bigskip

With $\mathcal{S}$ parametrized by $x,y$, and given by the equation $G(x,y,z)=0$ for
$G(x,y,z)=3x+\frac{3}{2}y+z-3$, we can use the formula
$$d{\bf S}= \frac{\nabla G}{G_z}\,dxdy = \langle 3,\frac{3}{2},1 \rangle dxdy.$$
(Note that the $\kk$-component is $1>0$, so the normal vector points up.) 
The range of $x,y$ is $0\leq x \leq 1$, $0\leq y \leq 2-2x$. Therefore
\begin{align*}
\iint_{\mathcal S} {\bf F}\cdot d{\bf S}
&= \int_0^1 \int_0^{2-2x } 3x^2   dy dx
\\
&= \int_0^1 3x^2(2-2x) dx
\\
&= \Big( 2x^3 - \frac{3x^4}{2}\Big)  \Big|_{x=0}^{1}    
= 2-\frac{3}{2} = \frac{1}{2}.
\end{align*}




\bigskip



\item
{\it Evaluate the integral
$\iint_{\mathcal S} {\bf F}\cdot d{\bf S}$, where ${\bf F}= {\bf i}+2{\bf j} + 2{\bf k}$ and $\mathcal{S}$ is 
the surface parametrized by 
${\bf r}(u,v)=\langle uv, u^2-v^2,u^2+v^2\rangle $ with $0\leq u\leq 1$, $0\leq v\leq u$, with the normal vector pointing up.}

\bigskip

We have ${\bf r}_u = (v,2u,2u),\ \ {\bf r}_v = (u,-2v,2v )$, so
$$
{\bf n}={\bf r}_u\times {\bf r}_v=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
v& 2u& 2u \\
u & -2v & 2v \\
            \end{matrix}\right|
=8uv {\bf i} - (2v^2-2u^2) {\bf j} + (-2v^2-2u^2){\bf k}.
$$
The last component of this is negative, so we need to use $-{\bf n}$ instead. We have
$$
{\bf F}\cdot {\bf n} = 8uv +2 (-2v^2+2u^2) + 2(-2v^2-2u^2) = 8uv - 8v^2.
$$
Therefore,
\begin{align*}
\iint_{\mathcal S} {\bf F}\cdot d{\bf S}
&= - \int_0^1 \int_0^u  (8uv-8v^2) \, dv du 
\\
&= - \int_0^1   \left( 4uv^2 - \frac{8}{3} v^3 \right) \Big|_{v=0}^u
\\
&= - \int_0^1 \left( 4u^3 - \frac{8}{3} u^3 \right)du 
= - \int_0^1 \frac{4}{3} u^3 \, du = \frac{u^4}{3} \Big|_{0}^1 = - \frac{1}{3}.
\end{align*}


\bigskip

\item
{\it Evaluate the integral $\iint_{\mathcal{S}}{\bf F}\cdot d{\bf S}$, where ${\bf F}=x^2\ii+y^4\jj$ and
$\mathcal{S}$ is the part of the surface $xy^2z=3$ that lies above the rectangle $1\leq x\leq 2$, $1\leq y\leq 3$,
with the normal vector pointing up.}

\bigskip

Our surface $\mathcal{S}$ has the form $G(x,y,z)=3$ with $G=xy^2z$, so that
$$
d{\bf S}=\frac{\nabla{G}}{G_z}=\frac{y^2z\ii +2xyz\jj + xy^2 \kk}{xy^2}
=\frac{z}{x}\ii+\frac{2z}{y}\jj+\kk$$
$$=\frac{3/xy^2}{x}\ii + \frac{6/xy^2}{y}\jj + \kk
 =\frac{3}{x^2y^2}\ii + \frac{6}{xy^3}\jj + \kk
$$
$$
{\bf F}\cdot d{\bf S}=\frac{3}{y^2}+\frac{6y}{x}$$
\begin{align*}
\iint_{\mathcal{S}}{\bf F}\cdot d{\bf S}
&=\int_1^2 \int_1^3 \Big(\frac{3}{y^2}+\frac{6y}{x}\Big) dydx
=\int_1^2  \Big(-\frac{3}{y}+\frac{3y^2}{x}\Big)\Big|_1^3  dx 
\\
&=\int_1^2  \Big(-1+3+\frac{27}{x}-\frac{3}{x}\Big) dx 
=\int_1^2  \Big(2+\frac{24}{x}\Big) dx \\
&= 2x+24\ln x \Big|_1^2 = 2 + 24 \ln 2.
\end{align*}
\end{enumerate}

