Check this double integral

Half way down it shows a double integral:

$$\int_{\phi_1}^{\phi_2} d \phi\int_{\theta_1}^{\theta_2}\sin \theta\,d\theta = (\phi_2 - \phi_1)(\cos \theta_2 - \cos \theta_1)$$

I think the $\theta_1$ and $\theta_2$ on the RHS should be swapped. I'm very rusty on integration though. Can someone tell me what the right answer is please?

• Yes, this does appear to be a typo. – user296602 Dec 16 '15 at 21:26

The result is wrong, and you are correct: $\theta_1$ and $\theta_2$ should be swapped.
We have $$\int^{\theta_2}_{\theta_1} \sin\theta\ d\theta = (-\cos\theta)\rvert^{\theta_2}_{\theta_1} = (-\cos\theta_2) - (-\cos\theta1) = \cos\theta_1 - \cos\theta_2.$$
Then $$\int_{\phi_1}^{\phi_2}d\phi\int^{\theta_2}_{\theta_1} \sin\theta\ d\theta = \int_{\phi_1}^{\phi_2}(\cos\theta_1 - \cos\theta_2)d\phi = (\phi_2 - \phi_1)(\cos\theta_1 - \cos\theta_2).$$