# Calculating the area of region of the sky

I am given the right ascensions ($\alpha_1$ and $\alpha_2$) and declinations ($\delta_1$ and $\delta_2$) of a specific region of the sky. How can we find the area of this region? I know that there must be a cosine term in the calculation, but cannot figure it out.

• Hi I'm not saying this is off topic but would you consider posting this on the astronomy and/or maths forums? imo it's more likely to be answered fully there. Regards – user74893 Mar 18 '15 at 12:58
• How accurate do you need it and how big approximately is the area? Is is it like a hemisphere or a square degree. – ProfRob Mar 18 '15 at 13:06
• can you post a diagram? – danimal Mar 18 '15 at 13:31
• @irishphysics Astronomy is on-topic here, and the answer might not be of purely mathematical character. – Thriveth Mar 19 '15 at 7:31
• @Thriveth. Hi....Newbie error on my part, I should have looked at the tag list first but I learned from the answer given. Thanks and regards – user74893 Mar 19 '15 at 10:52

## 1 Answer

I'm assuming you mean the region $\{(\alpha,\delta) : \alpha_1\leq \alpha\leq \alpha_2, \delta_1\leq \delta\leq \delta_2\}$. Measuring $\alpha$ and $\delta$ in radians, we can find the desired formula by integrating in spherical coordinates: \begin{align*} \int_{\alpha_1}^{\alpha_2} \int_{\delta_1}^{\delta_2} \cos \delta\ d\delta\ d\alpha &=\int_{\alpha_1}^{\alpha_2} \sin\delta|_{\delta_1}^{\delta_2} \ d\alpha \\ &= \int_{\alpha_1}^{\alpha_2} (\sin{\delta_2}-\sin{\delta_1})\ d\alpha\\ &= (\alpha_2-\alpha_1)(\sin{\delta_2}-\sin{\delta_1}) \end{align*} This gives the area of the region on the unit sphere (where the entire unit sphere has area $4\pi$).

• This is correct only on distances so short that we can neglect cosmological expansion. – Thriveth Mar 19 '15 at 7:32