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

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- May 3rd 2010, 11:38 AMRoadRaider12Spherical 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 - May 3rd 2010, 01:17 PMAllanCuz
From the question I gather the equation for the sphere(ball) is

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

What we're looking for is

$\displaystyle \iiint dV $ where $\displaystyle 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 $\displaystyle \sqrt{a} $. But what is the shortest point/distance it can be? Well, in this case the origin! So,

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

From the question were are given our bounds of phi!

$\displaystyle \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

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

Therefore,

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