# Surface integral of half sphere

1. Dec 21, 2008

### JamesMay

1. The problem statement, all variables and given/known data

I am trying to sort out surface integrals in my head, and have become more confused when attempting to calculate the surface integral of a hemisphere. I am getting confused about which values to use as boundaries.

2. Relevant equations

da=R^2 sinθdθdφ

3. The attempt at a solution

To get half a sphere I integrated dφ between 0 and 2pi, and integrated sinθdθ between 0 and pi/2.

This gives R^2*0*2pi=0

Whereas if I integrate dφ between 0 and pi, and integrate sinθdθ between 0 and pi i get

R^2*1*pi

which is the answer I am looking for but shouldn't I get the same using both sets of boundary values? Confused...

2. Dec 21, 2008

### HallsofIvy

Staff Emeritus
??? Not if you integrate correctly! $d\phi$, integrated from 0 to $2\pi$ is, of course, $2\pi$ and $sin\theta d\theta$ integrated from 0 to $\pi/2$ is $-cos\theta$ evaluated between 0 and $2\pi$: -(0- (1))= 1, not 0. The half sphere area is $2\pi R^2$.

If, instead, you integrate $d\phi$ from 0 to $\pi$ and $sin\theta d\theta$ from 0 to $\pi$ you get $\pi$ for the first integral and $-cos(\theta)$ evaluated from 0 to $\pi$, which is -(-1- (1))= 2 for the second, again getting $2\pi R^2$.

3. Dec 21, 2008

### JamesMay

Oh right, I always forget that cos(0) isn't just 0. You would think I'd have learnt that by now. Thanks!