# Thread: help with triple integral spherical coordinates

1. ## help with triple integral spherical coordinates

Set up a triple integral that represents the volume of part of the ball x^2 + (y-1)^2 + z^2 </= 1 that is outside the cone rho = pi/6 and inside the cone rho = pi/3

In the answer, the dTheta part of the triple integral goes from 0 to pi. My question is how do you find that? Normally theta goes from 0 to 2pi.

2. Originally Posted by messianic
Set up a triple integral that represents the volume of part of the ball x^2 + (y-1)^2 + z^2 </= 1 that is outside the cone rho = pi/6 and inside the cone rho = pi/3

In the answer, the dTheta part of the triple integral goes from 0 to pi. My question is how do you find that? Normally theta goes from 0 to 2pi.
If you look at the sphere from the top it's a circle of radius 1 and centered at $\displaystyle (0,1)$. The values of $\displaystyle \theta$ for this is $\displaystyle [0,\pi]$. If it was centerd at $\displaystyle (0,0),$ then it would be $\displaystyle [0,2 \pi]$ as you say.

3. Perhaps I'm missing something, but cones in spherical coordinates are usually given by $\displaystyle \phi=c$ and not $\displaystyle \rho$. If the question was find the volume between the two cones $\displaystyle \phi=\pi/6$, $\displaystyle \phi=\pi/3$ and the sphere $\displaystyle x^2+(y-1)^2+z^2=1$, then that looks pretty tough to me. It's the part of the sphere in the plot below between the two cones. Note a small part of the sphere goes through the smaller cone so would have to take that into account. Maybe though this isn't what you want.