Calculate $\int\int\int_D(x^2+y^2)dV$ where D is the smallest area of $z=\frac{x^2+y^2}{a}$ and $x^2+y^2+z^2=6a^2$

Question

I have to calculate $\int\int\int_D(x^2+y^2)dV$ where D is the smallest area of $z=\frac{x^2+y^2}{a}$ and $x^2+y^2+z^2=6a^2$ and I figured i should use spherical coordinates but i can't quite figure out where to go from there. any hints would be appreciated.

Answer 1

First we must look for the set where the paraboloid and the sphere mets,

\begin{align*} a^2(x^2+y^2)+(x^2+y^2)^2&=6a^4\\ u^2+a^2u-6a^4&=0,\qquad\text{where }u=x^2+y^2\\ (u+3a^2)(u-2a^2)&=0 \end{align*}

Then $x^2+y^2=2a^2$, $z=2a$ .

It seems that cylindrical coordinates fits well:

$D$ is the set of points $(r,\theta,z)\in\mathbb{R}^3$ such that $\frac{r^2}a\le z\le\sqrt{6a^2-r^2}$, $0\le r\le \sqrt{2}a$, $0\le \theta\le 2\pi$. Then

\begin{align*} \iiint_D(x^2+y^2)dV&=\int_{0}^{2\pi}\!\!\int_0^{\sqrt{2}a}\!\!\int_{r^2/a}^{\sqrt{6a^2-r^2}}r^2\,rdz\,dr\,d\theta \end{align*}