# 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?

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.

• If theta represents the azimuthal angle than it needs to be the full $2\pi$ radians around to get the full sphere – Triatticus Oct 5 '17 at 20:33
• Yes, actually I had multiplied the whole by 4 but removed it later and forgot to adjust theta...thanks for pointing it out... is everything else alright? – So Lo Oct 5 '17 at 21:02

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.