Integral over an arbitrary surface

1. Jul 30, 2009

Starproj

Hi,

I am studying a section on mass flux and have come across the following integral over an arbitrary surface. I am told that the integral equals 4pi radians. Can someone direct me to a proof of this and explain why the denominator (r2) drops out?

Thanks!

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2. Jul 30, 2009

John Creighto

3. Jul 30, 2009

arildno

Essentially, that integral, due to the 1/r^2-factor in the integrand becomes independent of the local radius, and equals therefore the value of the solid angle the surface represents with respect to the origin. That again equals 4pi.

4. Jul 30, 2009

daudaudaudau

There is a nice explanation of this in "Mathematical methods for physicists" by Arfken and Weber section 1.14 page 79. This is also known as Gauss' law in electromagnetism.

5. Jul 31, 2009

Starproj

I am not seeing something, for I get the integral to be zero, not 4pi. I have attached my work - can anyone spot where I am going wrong?

Thanks!

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6. Jul 31, 2009

arildno

You are at a SURFACE.

You are required TWO angular variables, not just one.

Go back and read some more..

7. Jul 31, 2009

Starproj

I caught the mistake where the limits of integration for theta are 0 to pi, not 2pi. But that doesn't eliminate the integer value of cosine, giving zero.

This may be a little above my level, which is why I am so frustrated. I think the book just wanted me to "take their word for it." I would really appreciate some help. Is there an error in my substitutions in the calculus? Why can't I get this to work out?

8. Jul 31, 2009

daudaudaudau

You are mixing two different thetas. When you are integrating on a sphere, n is parallel to g, so cos(theta) is one because this theta is zero. To see why this theorem holds for arbitrary surfaces you should check out the reference I gave.