Find the volume of the part of the ball $\rho \le 7$ that lies between the cones $ \phi = \frac \pi 6$ and $\phi = \frac \pi 3$.

Question

My question is simple, when setting up the bounds for this problem, I believed that the bounds would be from $\frac \pi 3$ to $\frac \pi 6$. however its wrong and its actually the other way around, from $\frac \pi 6$ to $\frac \pi 3$. This doesn't make sense to me as the cone made by $\frac \pi 6$ sits on top of the cone made by $\frac \pi 3$. What am I not understanding? Thanks.

Answer 1

In spherical coordinates, when you're integrating with respect to $\phi$ you integrate from the z-axis down, so if you only want the top half of $\mathbb{R}^3$ then you integrate $\phi=0$ to $\phi=\frac{\pi}{2}$. So in this case you would integrate from $\phi=\frac{\pi}{6}$ to $\phi=\frac{\pi}{3}$