Outward Flux Through Cylinder

1. Feb 28, 2015

BOAS

1. The problem statement, all variables and given/known data
Calculate the outward flux of $F = 3z\mathbf{e_{\rho}} + cos(\phi)\mathbf{e_{\phi}} + 2z\mathbf{e_{z}}$ through the base of a cylinder centered at the origin with radius $\mathrm{R}$ and height $\mathrm{2H}$.
2. Relevant equations

3. The attempt at a solution

I am unsure of how to tell if I have calculated the outward flux, and not entirely confident that it makes sense to have 'H' in my final answer. I'd really appreciate it if you could take a look.

$\Phi = \int_{s} \mathbf{F}. \mathbf{n} dA$

$\mathbf{F}. \mathbf{n} = (3z\mathbf{e_{\rho}} + cos(\phi)\mathbf{e_{\phi}} + 2z\mathbf{e_{z}}).(\mathbf{e_{z}}) = 2z$

$\Phi = \int_{s} 2z dA$

$\Phi = \int_{s} 2z \rho d\rho d\phi$

$\Phi = \int_{0}^{2\pi} \int_{0}^{R} 2z \rho d\rho d\phi$

$\Phi = \int_{0}^{2\pi} z \rho^{2} d\phi = z \rho^{2} [\phi]^{2\pi}_{0}$

$\Phi = 2 \pi z \rho^{2}$

$\Phi = -2 \pi H R^{2}$

Thanks!

Last edited: Feb 28, 2015
2. Feb 28, 2015

Dick

Seems ok. Except I would say the OUTWARD normal on the bottom of the cylinder is $-\mathbf{e_z}$. You could actually have done this calculation in your head, since you are integrating over a disk and the integrand is constant.

3. Feb 28, 2015

BOAS

Ah, of course! I should have picked up on that.

Thanks!