# Cone with spherical top triple integration

1. May 1, 2014

### Digitalism

1. The problem statement, all variables and given/known data

2. Relevant equations

∫∫∫dV

3. The attempt at a solution

Ok so I started by setting my bounds equal to √(200-x^2-y^2) ≥ z ≥ √(x^2+y^2), √(100-x^2) ≥ y ≥ -√(100-x^2), 10 ≥ x ≥ -10

which I got from solving z^2 = (200-x^2-y^2) = x^2+y^2 => x^2+y^2 = 100 but it seems to be a very messy integral. Should I try integrating differently? Is there a way to do this using polar coordinates?

2. May 1, 2014

### LCKurtz

Set it up in spherical coordinates.

3. May 4, 2014

### Digitalism

sorry, I haven't learnt that yet and I think this problem was explicity set up in rectangular so I could practice learning these in rectangular I am simply unsure how to place pi in there also :/

4. May 4, 2014

### LCKurtz

If you have had polar coordinates but not spherical coordinates, take your xyz integral (which is set up correctly) and integrate the dz integral. That will leave you a double integral over the circle $x^2+y^2 = 100$ in the xy plane. Change that integral to polar coordinates using $r^2=x^2+y^2,~dydx=rdrd\theta$ and put $r,\theta$ limits for the circle on the integrals.