# Flux over sphere between 2 planes

1. Jan 29, 2013

### sandy.bridge

1. The problem statement, all variables and given/known data
$\vec{F}=(4x, 3y, 0)$
Find the flux over the part of the sphere centered at (0, 0, 0) with radius 4 between the planes z=-1.19 and z=0.87.

3. The attempt at a solution
$\hat{N}=(x, y, z)/4$
$\int\int_{S}(4x, 3y, 0)\bullet([(x, y, z)/4])dxdy$
$0.25\int\int_{S}(4x^2+3y^2)dxdy$

The issue I am having is the limits of integration. I converted to polar but got a bit confused because there was three values for r due to the locations of the planes that intersect the sphere. Any recommendations?

Last edited: Jan 29, 2013
2. Jan 29, 2013

### Dick

Split the sphere into two parts, one with z>=0 and another z<0. Integrate over the two and add the result. Now you only have two limits for r in each half sphere.

3. Jan 29, 2013

### sandy.bridge

Another quick question: does the lower limit coincide with r at the origin, and the upper limit would be r at the intersection of the plane?

4. Jan 29, 2013

### Dick

The flux should be positive in each half sphere, right? I would say in each case the lower limit should be intersection with the plane and the upper limit should be 'at the origin' i.e. r=4.

5. Jan 30, 2013

### sandy.bridge

Alright, yeah that was what I was doing. I must be messing up somewhere because I am not getting the right result.

After converting to polar I get:
$$0.25\int\int_{D}(4r^3cos^2\theta+3r^3sin^2\theta)drd\theta$$
which after evaluating theta from 0 -> 2pi I get
$$0.25\pi\int(4r^3+3r^3)dr$$
and the limits for r are determined via
$$r=\sqrt(4-z^2)$$

Is this alright so far?

6. Jan 30, 2013

### Dick

Seem ok so far.

7. Jan 31, 2013

### sandy.bridge

Alright, when I finish the rest of the problem I don't get the right answer though.
$$7/8\pi(r^4)$$
I then solved for both the values of r at which the planes intersect the sphere, then added the two fluxes together.

8. Jan 31, 2013

### Dick

I get the antiderivative to be 7*pi*r^4/16.

9. Jan 31, 2013

### sandy.bridge

Nvm, got it. Thanks!

10. Jan 31, 2013

### Dick

Ooopsa. Sorry, I did overlook something. r=sqrt(16-z^2), not sqrt(4-z^2), right?

11. Jan 31, 2013

### sandy.bridge

Yeah, I caught that as well. I am still getting the wrong answer, though.
They both share r=4.
r1=3.819, r2=3.904.
so we have
14*pi*4^4/16 - 7*pi*(3.819^4)/16 - 7*pi*(3.904^4)/16
which evaluates to be not the right answer

12. Jan 31, 2013

### Dick

Hmm. Seems to be the same thing I'm getting. If there's an error we are both making the same one. Do you know what the correct answer is supposed to be?

13. Jan 31, 2013

### sandy.bridge

I have multiple choice answers, and my answer is not one of them.

I'm going to try one more time. The question has z=-1.1912, and z=0.8768 (I simplified it for the question).

Using these values I get r1=3.818513, r2=3.90272
I get the flux to be 92.642, which is not right.

14. Jan 31, 2013

### Dick

Just out of curiousity, what are the choices?

15. Jan 31, 2013

### sandy.bridge

Alright, so the final answer is attained by taking the sine of the total flux. For example, let F be the flux, then the value of sin( F ) =,
a) -0.489343, b) 0.306737, c) -0.540973, d) -0.528764, e) -0.499915, f) 0.552286, g) 0.775165, h) 0.0326167

16. Jan 31, 2013

### Dick

Nah. Doesn't help. Taking the sin of a flux is a pretty strange thing to do. But it sure makes it hard to guess what they are doing. For one thing, just changing the z values from the approximate ones you have before to the more accurate ones changes sin(F) from -0.7978 to -0.9994.

17. Jan 31, 2013

### sandy.bridge

I agree that it's a weird thing to do... hmmm

18. Jan 31, 2013

### Dick

Yeah. Fluxes of fields (like an electric field) usually have dimensions associated with them. You basically never take the sin of a dimensionful quantity.

19. Jan 31, 2013

### sandy.bridge

Either way, I wonder where there is an error. The department has been using these questions for 3 years now, so I would assume if there was an error in the answers itself that it would be known by now.

20. Jan 31, 2013

### Dick

If there is an error, I'm sure not seeing where it is. Lemme ask somebody else to look at this and get a third opinion.