# Math Help - Triple Integral w/ Spherical Coordinates

1. ## Triple Integral w/ Spherical Coordinates

I am having trouble with the following question:

Evaluate the triple integral over E, where E lies between the spheres rho = 2 and rho = 4, and above the cone phi = pi/3. The density function is given by xyz dV.

When I change variables I get the following as my density function:
(rho^5)*(sin^3(phi))*cos(phi)*cos(theta)*sin(theta ). Is this right? This doesn't seem like the easiest integral to work with, since after you integrate with respect to rho, you are left with a bunch of trig. Maybe there is another way to write the density function that I'm not seeing?

Any ideas would be great. Thanks.

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Originally Posted by eigenvector11
I am having trouble with the following question:

Evaluate the triple integral over E, where E lies between the spheres rho = 2 and rho = 4, and above the cone phi = pi/3. The density function is given by xyz dV.

When I change variables I get the following as my density function:
(rho^5)*(sin^3(phi))*cos(phi)*cos(theta)*sin(theta ). Is this right? This doesn't seem like the easiest integral to work with, since after you integrate with respect to rho, you are left with a bunch of trig. Maybe there is another way to write the density function that I'm not seeing?

Any ideas would be great. Thanks.
You are left with a bunch of very easy trig functions! your integral is
$\int_{r= 2}^4 r^5 dr\int_{\theta= 0}^{2\pi}cos(\theta)sin(\theta)d\theta \int_{\phi= 0}^{\pi/3}cos(\phi) d\phi$

I would make either of the two obvious substitutions in the $\theta$ integral and then not bother with the other integrals!

3. Originally Posted by HallsofIvy
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You are left with a bunch of very easy trig functions! your integral is
$\int_{r= 2}^4 r^5 dr\int_{\theta= 0}^{2\pi}cos(\theta)sin(\theta)d\theta \int_{\phi= 0}^{\pi/3}cos(\phi) d\phi$

I would make either of the two obvious substitutions in the $\theta$ integral and then not bother with the other integrals!
that middle integral is zero, $\sin^2 \theta$ at $2\pi$ and 0 is zero.