2
$\begingroup$

I'm learning solid angles from this page among others.

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?

$\endgroup$
  • 4
    $\begingroup$ Yes, this does appear to be a typo. $\endgroup$ – user296602 Dec 16 '15 at 21:26
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.