Flux through top part of sphere

[intkfmr]

Homework Statement

What is the flux of $\vec F=(xz, -yz, y^2)$ through the surface given by $x^2+y^2+z^2=2,\ z>1$?

Relevant Equations

Flux=$\iint \vec F\cdot \hat n dA$

Flux=$$\iint(xz, -yz, y^2)\cdot(x,y,z)/\sqrt{2} dA=\int_0^{2\pi}\int_0^1 r^2\cos^2\theta \sqrt{1-r^2/2} rdrd\theta$$. Integrating this doesn't give the correct answer of $\pi/4$.

 

[andrewkirk]

Your inner integral should not be with respect to r. You are integrating over a surface not a volume. Change your inner integration to $\int_0^{\pi/2}...d\phi$, where $\phi$ is the angle between the point's location vector and the x-y plane (the "elevation").

 

[Orodruin]

*cough* *cough* curl *cough* theorem *cough*

Edit: Divergence theorem also works ...

 

[vela]

Homework Statement:: What is the flux of $\vec F=(xz, -yz, y^2)$ through the surface given by $x^2+y^2+z^2=2,\ z>1$?
Relevant Equations:: Flux=$\iint \vec F\cdot \hat n dA$

Flux=$$\iint(xz, -yz, y^2)\cdot(x,y,z)/\sqrt{2} dA=\int_0^{2\pi}\int_0^1 r^2\cos^2\theta \sqrt{1-r^2/2} rdrd\theta$$. Integrating this doesn't give the correct answer of $\pi/4$.

Where did the factor $\sqrt{1-r^2/2}$ in the last integral come from? I didn't get that when working out the integral.

 

[Charles Link]

The simplest approach I found is to compute the divergence, which turns out to be zero, and thereby the integral of the divergence is zero, so that it allows you to alternatively use the circular face at $z=1$ as a surface of integration.

 

[Orodruin]

The simplest approach I found is to compute the divergence, which turns out to be zero, and thereby the integral of the divergence is zero, so that it allows you to alternatively use the circular face at $z=1$ as a surface of integration.

Well … I tried to say it in not such an explicit form …

*cough* *cough* curl *cough* theorem *cough*

Edit: Divergence theorem also works ...

Edit: It is a good exercise and instructive to consider why both the curl and divergence theorem works for arguing that only the boundary curve of the surface is relevant for the result if the field is divergence free.

 

[Charles Link]

I also have now worked the calculation with the surface being the cap of the sphere, using spherical coordinates. The OP needs to put in the correct $dA=r^2 \sin{\theta} \, d \theta \, d \phi$. Doing it with the spherical coordinates is fairly straightforward, but making use of the divergence theorem (Gauss' law) is easier.