∫∫∫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?

Printable View

- April 27th 2010, 06:10 PMwizrd54Triple 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? - April 27th 2010, 07:23 PMTinyboss
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 , , and .