# Quick question - Flux through a sphere

1. Feb 13, 2014

### Zatman

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

Evaluate ∫∫r.ndS where r=(x,y,z) and n is a normal unit vector to the surface S, which is a sphere of radius a centred on the origin.

2. The attempt at a solution

I decided to use polar coordinates. The radius of the sphere is clearly constant, a. So a surface element is dS = a2dθdø.

r = a(sinθcosø, sinθsinø, cosθ)

n = (sinθcosø, sinθsinø, cosθ)

r.n = a

therefore, ∫∫r.ndS = ∫∫a3dθdø

where θ varies from 0 to 2π, and ø varies from 0 to π.

This gives an answer of 2π2a3. Is this correct? I'm not so sure.

2. Feb 13, 2014

### vanhees71

You forgot that
$$\mathrm{d} S=a^2 \sin \theta \mathrm{d} \theta \mathrm{d} \phi.$$

3. Feb 13, 2014

### Zatman

Good point, thank you.

But then I get ∫∫a3sinθdθdø

Which evaluates to zero. Are the limits incorrect?

4. Feb 13, 2014

### pasmith

Yes: If $z = r\cos\theta$ (there are two exactly opposite conventions for the angle coordinates) then $dS = r^2 \sin\theta\,d\theta\,d\phi$ and the sphere is given by $0 \leq \theta \leq \pi$ and $0 \leq \phi \leq 2\pi$.

Also: once you have that $\mathbf{r} \cdot \mathbf{n} = a$ you know that the answer must be $a$ times the surface area of a sphere of radius $a$.

5. Feb 13, 2014

### Zatman

Oh, I had theta and phi mixed up. So that gives

∫∫r.ndS = ∫(2a3)dø

from 0 to 2π, hence

= 4πa3