# Integral of f over the surface of a sphere (in spherical polar coordinates)

1. Nov 11, 2008

### sephiseraph

I have a function f(r, $$\phi$$, $$\vartheta$$) = 3cos$$\vartheta$$.

Evaluating the repeated integral of this function over the surface of a sphere, centered at the origin, with radius 5, I have come up with 0 as my result. I'm not sure if this is correct. I've double checked my calculations, and tried subdividing the surface S into smaller subsections and summing the integrals of each section, and I get the same result. Intuitively it makes some sense that the answer would be 0 since cos$$\var{theta}$$ takes on values either side of 0 for $$\vartheta$$ in the intervals [0, pi] and [pi, 2pi].

Still, for some reason I'm uncomfortable with this result. Can anybody shed some light on this?

Last edited by a moderator: Nov 12, 2008
2. Nov 11, 2008

### sephiseraph

To find the repeated integral of f over the surface S with respect to dA, where dA is the limit of the small areas on S, I used the fact that $$dA = R^2 sin\theta \,d\phi \,d\theta$$.

From there I calculated:

$$\int \int_{S} f dA = \int^{\pi}_{0} \int^{2\pi}_{0} (3cos\theta 5^2 sin\theta) \, d\phi \, d\theta = \int^{\pi}_{0} \int^{2\pi}_{0} (75 cos\theta sin\theta) \, d\phi \, d\theta = \int^{\pi}_{0} [ 75 \phi cos\theta sin\theta ]^{2 \pi}_{0} \, d\theta = 150\pi \int^{\pi}_{0} cos\theta sin\theta \, d\theta = 150\pi [(1/2) sin^2\theta]^{\pi}_0$$

... which of course equals 0 as $$sin\pi = sin0 = 0$$.

3. Nov 11, 2008

### Ben Niehoff

It looks correct to me.

4. Nov 12, 2008

### HallsofIvy

Staff Emeritus
For any point on the surface of the sphere, $(\phi, \theta)$ (you are using "engineering" notation here: $\phi$ and $\theta$ are reversed from what would be in a mathematics text) the diametrically opposite point is $(2\pi- \phi, \pi- \theta)$ and [itex]cos(\pi- \theta)= -cos(\theta)[/tex]. Your function is "anti-symmetric" so its integral over the entire surface of the sphere is 0. The integral over any region around a given point is cancelled by the integral over its diametrically opposite point.

Last edited: Nov 14, 2008
5. Nov 14, 2008