# Triple Integral in Spherical Coordinates

• Apr 27th 2010, 06:10 PM
wizrd54
Triple Integral in Spherical Coordinates
∫∫∫16x^3 dV

Bounded by the cone and sphere below:

Z = √(x^2 + y^2)

x^2 + y^2 + z^2 = 4

For my bounds of integration I got 0 < phi < pi/3 , 0 < theta < 2pi , and 0 < rho < 4. Is this correct?
• Apr 27th 2010, 07:23 PM
Tinyboss
There are different ways to set up spherical coordinates, but if you use the setup in the Wikipedia article on spherical coordinates, I think you'd want $0\le\theta\le\frac\pi 2$, $0\le\phi<2\pi$, and $0\le\rho\le4$.