1. ## Spherical Coordinate problem

the problem reads "Find the volume of the part of the ball p(rho)=<a that lies between the cones phi=pi/6 and phi=pi/3"

can anyone help at least set up the triple integral? I'm not sure how to go about this

the problem reads "Find the volume of the part of the ball p(rho)=<a that lies between the cones phi=pi/6 and phi=pi/3"

can anyone help at least set up the triple integral? I'm not sure how to go about this
From the question I gather the equation for the sphere(ball) is

$x^2 + y^2 + z^2 \le ( \sqrt{a} ) ^2$

What we're looking for is

$\iiint dV$ where $dV = p^2 sin \phi dp d \phi d \theta$

So what are our bounds for theta, phi and p? Well, for P this is the value of the minimum point and maximum point. In this case, clearly the farthest point P is $\sqrt{a}$. But what is the shortest point/distance it can be? Well, in this case the origin! So,

$0 \le P \le \sqrt{a}$

From the question were are given our bounds of phi!

$\frac{ \pi }{6} \le \phi \le \frac{ \pi }{3}$

What about theta? Well, we are going 360 degress around the way! So our theta bounds become

$0 \le \pi \le 2 \pi$

Therefore,

$\int_0^{2 \pi } d \theta \int_{ \frac{ \pi }{6} }^{ \frac{ \pi }{3} } sin \phi d \phi \int_0^{ \sqrt{a} } p^2 dp$