Surface Element Conversion for Flux Through Uncapped Cylinder

[RoyalFlush100]

Homework Statement

In the attached image.

Homework Equations

Gradient(x, y, z) * <f, g, h> = <fx, gy, hz>

The Attempt at a Solution

Because the cylinder's not capped, I know that all the flux will be in the radial direction. So, I can find a normal vector by finding the gradient of the cylinder: n = <2x, 0, 2z>/(2sqrt(x^2+z^2)) = <x, 0, z>/sqrt(x^2+z^2)

Now, I want to put this in terms of t (the angle) and h (y):
r(t, y) = <acos(t), h, asin(t)>
Where: y: (-2, 2) and t: [0, 2pi)

Now we can rewrite the integrand:
<acos(t)/sqrt(a), 0, asin(t)/sqrt(a)> * <acos(t)/sqrt(a), 0, asin(t)/sqrt(a)> dS
=(a^2cos^2(t) + a^2sin^2(t))/a dS
=a dS

Now, the only thing I'm confused by (assuming everything else is right), is what to do with dS. I know it needs to be put in terms of dt and dh (where I already have the limits of integration), but I am unsure of how to perform this conversion.

 

[Charles Link]

Isn't the vector $\vec{E }$ of the field used in this flux equation simply a unit vector in the radially outward direction? If it is, (and I don't know that I should be giving you this much of the answer, but the problem is almost trivial), isn't the flux equal to the surface area?

 

[RoyalFlush100]

Isn't the vector $\vec{E }$ of the field used in this flux equation simply a unit vector in the radially outward direction? If it is, (and I don't know that I should be giving you this much of the answer, but the problem is almost trivial), isn't the flux equal to the surface area?

I believe the answer should be a multiplied by the surface area, but I a unsure how to properly evaluate the integral, converting dS to be in terms of dt and dh.

 

[Charles Link]

I believe the answer should be a multiplied by the surface area, but I a unsure how to properly evaluate the integral, converting dS to be in terms of dt and dh.

You don't need to do any conversion. Surface area $S=\pi a^2 h$.

 

[RoyalFlush100]

You don't need to do any conversion. Surface area $S=\pi a^2 h$.

Okay. Just curious though, how would I perform that conversion in a less trivial case?

 

[Charles Link]

Surface integrals in the general case can take much effort to evaluate. The problem, I think, was designed to keep it simple.

 

[Orodruin]

In the general case, where you have a surface parametrised by the two variables $t$ and $s$, the directed surface element is given by
$$
\newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}}
d\vec S = \vec n \, dS = \dd{\vec x}{t} \times \dd{\vec x}{s} dt\, ds,
$$
where $\vec x$ is the position vector (which on the surface is parametrised by $t$ and $s$.