# Manually integrating to find flux through hemisphere

1. Sep 15, 2015

### yango_17

1. The problem statement, all variables and given/known data
Basically, I am being asked to calculate the electric flux through the top of a hemisphere centered on the z-axis using "brute force integration" of the surface area.

2. Relevant equations
Gauss' Law

3. The attempt at a solution
Using intuition and Gauss' law, I know that the flux is Φ=Eπr^2, but I'm not sure how to prove this using "brute force" integration as my teacher requires. If anyone could help me at least set up the integral, it would be much appreciated.

2. Sep 15, 2015

### BiGyElLoWhAt

I'm not sure what you mean by "brute force". It can be readily shown that the projection of a sphere onto theal plane of said sphere is pi*r^2, and since phi=E(dot)dA or just E(dot)A for approximation, then you end up with that expression. Perhaps show that E(dot)A is E*pir*r^2

3. Sep 15, 2015

### yango_17

I'm not sure exactly what he means either haha. Anyway, he said it'd be easier to do if we use the spherical coordinate system to represent the hemisphere, which sounds like a hint, but I'm still not too sure how to proceed.

4. Sep 15, 2015

### BiGyElLoWhAt

Hmmm... I'm not sure I agree with that, but it can definitely be done in spherical coordinates. You are familiar with the equation of a sphere, no?

Edit* Wait, I lied. It makes the limits way nicer.

5. Sep 15, 2015

### yango_17

No, I'm afraid I don't know how to represent a hemisphere in spherical coordinates

6. Sep 15, 2015

### BiGyElLoWhAt

I see. Well a sphere takes the form $<r,\phi,\theta>$. That is an arbitrary point on the surface of a sphere, and you need to find the normal to the tangent plane at each point, then dot it with the electric field, and integrate over the surface.
$\Phi = \int_S \vec{E} \cdot d\vec{A}$

7. Sep 15, 2015

### BiGyElLoWhAt

By brute force, he might have just meant make the long calculation.

8. Sep 15, 2015

### yango_17

Alright, I'll see if I can work it out from here. Thanks!

9. Sep 15, 2015

No problemo!