∫∫∫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?
∫∫∫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?
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 $\displaystyle 0\le\theta\le\frac\pi 2$, $\displaystyle 0\le\phi<2\pi$, and $\displaystyle 0\le\rho\le4$.