I am currently studying for the GRE. I came across the following question, and I can't seem to get the correct answer. The question reads:
Compute the area of a unit sphere contained between the meridians $\phi =30^{\circ}$ and $\phi = 60^{\circ}$ and parallels $\theta =45^{\circ}$ and $\theta =60^{\circ}$.
First, I recalled that the area element of the unit sphere is: $dA=\sin(\phi)d\phi d\theta$. From there, I made the following computations:
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\sin(\phi)d\phi d\theta = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}(\frac{\sqrt{3}}{2}-\frac{1}{2})d\theta = \frac{\pi}{12}(\frac{\sqrt{3}}{2}-\frac{1}{2})$$
However, the correct answer is: $\frac{\pi}{12}(\sqrt{3}-\sqrt{2})$. Any guidance on this problem would be greatly appreciated.
Since here $\theta$ denotes the latitude ($\theta=0^{\circ}$ is the equator), the element of area is $dA=\cos(\theta)d\phi d\theta$, and therefore the required area is $$\int_{\theta=\frac{\pi}{4}}^{\frac{\pi}{3}}\int_{\phi=\frac{\pi}{6}}^{\frac{\pi}{3}}\cos(\theta)d\phi d\theta=\left(\frac{\pi}{3}-\frac{\pi}{6}\right)\left(\sin\left(\frac{\pi}{3}\right)-\sin\left(\frac{\pi}{4}\right)\right)=\frac{\pi(\sqrt{3}-\sqrt{2})}{12}.$$