# Thread: Vector Calc. Triple integral help!!!

1. ## Vector Calc. Triple integral help!!!

Alright, so here's the problem:

Evaluate the triple integral with domain of E of ((x^2)yz) where E lies between the spheres p=2 and p=4 (the greek rho) and above the cone phi=pi/3 and in the first octant.

So I got the limits as, the first is from 0 to pi/3 then the middle is 0 to pi/2 and the inside is 2 to 4. That wasn't much of a problem since it's in spherical coordinates. But now doing the inside. Since you replace the x,y,and z with the spherical values, i come up with p^6(sin(phi)^4)(sin(theta))(cos(theta)^2)(cos(phi) ) then dp dtheta dphi. I can do the integral for p, but doing the rest is confusing me. any help? And sorry it's a mess. I'm not good with the math symbols on here

2. Originally Posted by Centara
Alright, so here's the problem:

Evaluate the triple integral with domain of E of ((x^2)yz) where E lies between the spheres p=2 and p=4 (the greek rho) and above the cone phi=pi/3 and in the first octant.

So I got the limits as, the first is from 0 to pi/3 then the middle is 0 to pi/2 and the inside is 2 to 4. That wasn't much of a problem since it's in spherical coordinates. But now doing the inside. Since you replace the x,y,and z with the spherical values, i come up with p^6(sin(phi)^4)(sin(theta))(cos(theta)^2)(cos(phi) ) then dp dtheta dphi. I can do the integral for p, but doing the rest is confusing me. any help? And sorry it's a mess. I'm not good with the math symbols on here
I didn't check your above work but since your limits are constant we have(by Fubini's theorem)

$\displaystyle \iiint \rho^6\sin^4(\phi)\cos(\phi) \sin(\theta) \cos^2(\theta) dV=$

$\displaystyle \left( \int\rho^6d\rho \right)\left( \int \sin^4(\phi)\cos(\phi)d\phi\right)\left( \int \sin(\theta) \cos^2(\theta)d \theta \right)$

each of these can be knocked out with a u sub.

Good luck.

3. Sweet, totally didn't think of that! Thanks for the help!