Question

I am looking for help setting up a triple integral. See the attached photo.

Answer 1

E: \[\pi/3 \le phi \le \pi/2 \], \[0\le \theta \le 2\pi \]\[4 \le p \le5\], ?

Answer 2

x=pcos[theta]sin[phi], y=psin[theta]sin[phi], z=pcos[phi], dV=p^2sin[phi]dp dtheta dphi ?

Answer 3

convert to Spherical coordinates not polar

Answer 4

*convert to spherical
noticed that as soon as I hit post

Answer 5

its all good

Answer 6

the work above is all good?

Answer 7

=0

Answer 8

\[\int\limits{}\int\limits{}\int\limits{}xyz dV\]
\[4\le \rho \le5\]
\[0 \le \phi \le \frac{\pi}{3}\]
\[0 \le \theta \le 2\pi\]

The Jacobian:
\[\frac{\Delta x,y,z}{\Delta \rho,\theta,\phi} = \rho^2 \sin \phi\]
Now:
\[x = \rho \sin \phi \cos \theta, y = \rho \sin \phi \sin \theta, z = \rho \cos \phi\]
So your integral should look like:
\[\int\limits^{\frac{\pi}{3}}_{0} \int\limits^{2 \pi}_{0} \int\limits^{5}_{4} \rho^5 \sin^3 \phi \cos \phi \sin \theta \cos \theta d \rho d \theta d \phi\]
This can be split into three integrals:
\[\left( \int\limits^{\frac{\pi}{3}}_{0} \sin^3 \phi \cos \phi d \phi \right)\left( \int\limits^{2\pi}_{0} \sin \theta \cos \theta d \theta \right) \left( \int\limits^{5}_{4} \rho^5 d \rho \right)\]
use u sub on the first two integrals:
\[u = \sin \phi, u = \sin \theta\]
and the third integral uses the power rule, which I assume you can do
Hope this helps, let me know if you get stuck

Answer 9

This evaluates to zero. Thanks for your help, nadeem

Answer 10

you bet