MyOpenMath/Solutions/c19SurfaceIntegralCALCULUS

	Review of dot product and a reminder that unit vectors are orthogonal and have unit magnitude. Since we are in cylindrical coordinates, we use $\rho$ for distance to the axis to avoid the use of $r$ in spherical coordinates. Define the vector area differential element $d\overrightarrow{A}=\hat{n}dA$ using the outward unit normal $\hat{n}$. From the top down, we see the top of the cylinder in what looks like polar coordinates. On the top circle have a $dA=\rho d\rho d\phi$ Also shown in is the curved side of the cylinder. Here the dimensions of the small rectangle are such that $dA=\rho d\rho dz$.
	The unit vectors $(\hat{i},\hat{j},\hat{k})$ are replaced by $(\hat{\rho },\hat{\phi },\hat{z})$.
	Evaluate the integral over the top surface.
	On the curved sides, $F\cdot d\overrightarrow{A}=F_{\rho }dA$. Note that the "odd" term vanishes: $\underset{-a}{\overset{a}{\int }}{z}^{n}dz=0\text{ if n odd}$ $\underset{-a}{\overset{a}{\int }}{z}^{n}dz=2\underset{0}{\overset{a}{\int }}{z}^{n}dz\text{ if n even}$
	When doing the integral over the entire surface, it is customary to put a little circle at the center of the integral sign.
	When adding the top and bottom integrals we have to remember that ${\hat{n}}_{top}=\hat{z}$ on the top surface but that ${\hat{n}}_{bot}=-\hat{z}$ on the bottom surface. The result depends on whether $F_z(\rho ,z)$ is an even or odd function of z.