# Proof of $\cos(\theta) da=r^2 d\Omega$

I'm reading through Classical Electrodynamics by Jackson and he makes use of a particular identity that I've not actually seen before. I can't seem to find a proof for it anywhere. It is just as the title says:

$\cos(\theta)da=r^2 d\Omega$

where $d\Omega$ is $\sin\theta d\theta d\phi$, the solid angle.

As an example for context consider Gauss' law. If the electric field E at a point on a Gaussian surface due to the charge q within the surface makes an angle $\theta$ with the unit normal, then the normal component of E times the area element is:

$\vec{E} \cdot \hat{n} da=\frac{q}{4\pi\epsilon_{0}}d\Omega$

Since E is directed along the line from the surface element to the charge q, $\cos\theta da=r^2 d\Omega$, where $d\Omega$ is the element of solid angle sutended by da at the position of the charge. Therefore,

$\vec{E} \cdot \hat{n} da=\frac{q}{4\pi\epsilon_0}d\Omega$.

A simple geometric proof is all I need, I just want to know where this comes from.

• I believe this would be more suitable for a physics forum. That said, how do your variables-$\theta, r, \Omega$-relate? Jan 10, 2017 at 21:08
• Contextualize the question. Can you show us a an equation/step where he uses this equality? Jan 10, 2017 at 21:08
• also what is da? Jan 10, 2017 at 21:11
• And now there is also a $\phi$ :-).. Jan 10, 2017 at 21:12
• edited with all that in mind. da is not specified in the book, that's part of the problem.
– Karl
Jan 10, 2017 at 21:24 