Question

Can someone please compute this triple integral with steps that I can follow?

Answer 1

Integral(Integral(IntegralD (x^2 + y^2 + z^2 - xz) dxdydz

Here D is the solid bounded below by the xy-plane and above by the cone phi = pi/3 and inside the sphere rho = 3

Answer 2

the domain is not quite clear, can you draw it?

Answer 3

They didnt give me a drawing for it so im not sure. All that is given is that it is the solid bounded below by the xy-plane and above by the cone phi = pi/3 and inside the sphere rho = 3

Answer 4

If you're solving this in spherical coordinates
\[\rho^{2} = x^{2} + y^{2} + z^{2}\]\[x = \rho \sin \phi \cos \theta, z = \rho \cos \theta\]
And when solving in spherical coordinates you have to add \[\rho^{2} \sin \phi\]
So you end up with
\[\int\int\int\ (\rho^{2}-p^{2}\sin \phi \cos^{2} \theta)(p^{2}\sin \phi) d \rho d \phi d \theta\]
... I think. To be honest, I just finished Calc III and triple integrals gave me a tough time. You'll have to give me a minute on the limits though.

Answer 5

That's all i wouldn't know are the limits. But I haven't taken calc III yet D:

Answer 6

It seems like rho is 0 to 3 and theta will be 0 to 2pi. Phi is giving me issues. If its bound below the xy-plane but inside the sphere but above the cone it'd be something from 0 to pi/3, I think. But that doesn't really take into account the amount below the xy-plane...

I wish I could spend some more time on this, but I have to go.

Like I said, triple integrals gave me a hard time, but I hope this could point you in the right direction.

Answer 7

http://tutorial.math.lamar.edu/Classes/CalcIII/TripleIntegrals.aspx

If you need spherical or cylindrical those are there, too. I mean, I can try and study itand see, but I dont know off the top of my head.

Answer 8

I will give a medal to anyone who can help me!