Find the volume for the region that remains in the spherical solid $\rho \leq 4$ after the solid cone $\phi \leq \pi/6$ has been removed?

Question

NOTE:Spherical coordinate system has been used in question.

Find the volume for the region that remains in the spherical solid $\rho \leq 4$ after the solid cone $\phi \leq \pi/6$ has been removed?

DIRECT WAY: I set up the following integral,

$\int_{0}^{2\pi} \int_{\pi/6}^{\pi} \int_{0}^{4} \rho^2sin(\phi)d\rho d\phi d\theta$

Is this set up correct or I missed something? I am asking particularly for $\theta$ limits

Indirect:Volume of cone could be subtracted from that of sphere.

Answer 1

This is correct. Your indirect way would be the same except the $\phi$ integral would go from $0$ to $\frac \pi 6$ to represent the removed volume.