How to parameterize the surface of a cylinder in the xyz-plane? "Consider the vector field $\vec{F} = \langle 5x, 6z, 4y \rangle$ and the solid $E$ bounded by the cylinder $x^2 + y^2 = 1$ and the planes $z = 0$ and $z = 2$.
Let $S$ be the boundary surface of the solid $E$.  Then $S$ consists of three surfaces, $S_1$, $S_2$, and $S_3$, where $S_1$ is the side of the cylinder, $S_2$ is the top disk and $S_3$ is the bottom disk."
We're asked to find the surface integral $\displaystyle\iint_S \vec{F} \cdot \hat{n}\, \mathrm{d}S$ by calculating
$$
\iint_{S_1} \vec{F} \cdot \hat{n}\, \mathrm{d}S 
+ \iint_{S_2} \vec{F} \cdot \hat{n}\, \mathrm{d}S 
+ \iint_{S_3} \vec{F} \cdot \hat{n}\, \mathrm{d}S. 
$$
I have no idea how to parameterize this surface. I tried using $\vec{r} = \langle \cos\theta,\, \sin\theta,\, z \rangle$, for $S_1$, but that didn't seem to get me anywhere. I don't know what to do for $S_2$ or $S_3$.
 A: This answer addresses the surfaces $S_1$ and $S_2$, but leaves $S_3$ for you to work out by analogy with $S_2$.
For $S_1$, your parametrization works just fine. We need a longitude coordinate $\theta$ (around the cylinder) and a vertical coordinate $z$, so
$$
\vec{r} = \langle \cos\theta,\, \sin\theta,\, z \rangle, 
\quad
0 \leq \theta \leq 2\pi, 
\quad 
0 \leq z \leq 2
$$
works.
For $S_2$, we need to parametrize a disk, where $z = 2$ is fixed. Polar coordinates (in plane parallel to $xy$-plane) works fine, so
$$
\vec{r} = \langle r\cos\theta,\, r\sin\theta,\, 2 \rangle, 
\quad
0 \leq r \leq 1, 
\quad 
0 \leq \theta \leq 2\pi. 
$$
A parametrization for $S_3$ is very similar. Just modify the fixed $z$-coordinate.
For the integrals, we will have to transform coordinates. For example, on $S_1$, we now have
$$
\vec{F} = \langle 5x,\, 6z,\, 4y \rangle 
= \langle 5\cos\theta,\, 6z,\, 4\sin\theta \rangle, 
$$
and on $S_2$,
$$
\vec{F} = \langle 5x,\, 6z,\, 4y \rangle 
= \langle 5r\cos\theta,\, 12,\, 4r\sin\theta \rangle. 
$$
The unit normal on $S_1$ is the outward pointing radial vector:
$$
\hat{n} = \langle x,\, y,\, 0 \rangle 
= \langle \cos\theta,\, \sin\theta,\, 0 \rangle. 
$$
On $S_2$, it's the vertical constant vector:
$$
\hat{n} = \langle 0, 0, 1 \rangle 
$$
Now, what happens to the infinitesimal surface element when we pull it back to various coordinate systems? On $S_1$,
$$
\mathrm{d}S = \mathrm{d}\theta\, \mathrm{d}z 
$$
and on $S_2$,
$$
\mathrm{d}S = \mathrm{d}x\, \mathrm{d}y 
= r\, \mathrm{d}r\, \mathrm{d}\theta. 
$$
Can you put all the pieces together?
