Cone with spherical top triple integration

[Digitalism]

Homework Statement

Homework Equations

∫∫∫dV

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?

 

[LCKurtz]

Homework Statement

Homework Equations

∫∫∫dV

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?

Set it up in spherical coordinates.

 

[Digitalism]

sorry, I haven't learned 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 :/

 

[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.