Triple Integration in Spherical Coordinates with a paraboloid and plane

Question

I am completely stuck on this question. Any help would be appreciated.

Thanks so much

Answer 1

First we make a sketch. The paraboloid points along the +x axis with its tip at x=10. The paraboloid intersects the x=5 plane in a circle, $C$. (What is the radius of that circle?) We want to integrate from the x=5 plane to the tip.

Consider an area element $rdrd\theta$ within the circle $C$. We are going to integrate over that circle, after we integrate along the x-direction. The area element has coordinates $(5,r,\theta)$. We want to integrate w.r.t. x from 5 to the surface of the paraboloid, which is at $x=10-y^2-z^2$. So those are limits of the x-integration, except we have to write the limits in terms of $r,\theta$, instead of $y,z$.

The next limits of integration are on the variable $r$, which goes from zero to the radius of circle $C$. And, finally, the $\theta$ limits are zero and $2\pi$.

The integrand is $y^2$, which we must write in terms of $r,\theta$. Note that the surface appears in the limits of the volume integral, not part of the integrand.

The final answer is $125\pi/12$, if we work it out, which is half the moment of inertia of the volume about the x-axis. It is half, because the integrand is only $y^2$. If the integrand had been $y^2 + z^2$, we would have gotten the moment of inertia of the paraboloid about the x-axis.