Using spherical coordinates,
$$\iiint_D (x^2 + y^2 +z^2) dV, $$
where $D$ is the region above the cone $z^2 = x^2+ y^2$ and below the plane $z=a$, $a>0$.
Over what region do we have to integrate this?
$\theta$ lies between $0$ and $2\pi$. But for $\rho$ it cannot be zero as well as $\phi$. I have tried to draw a diagram but I am not able to visualize the cone and the plane in the same place.
Can Someone please guide me on this problem?
In spherical coordinates, the condition $z^2\geqslant x^2+y^2$ becomes $\rho^2\cos^2\phi\geqslant\rho^2\sin^2\phi$. Since $\rho\sin\phi=z\geqslant0$, $\phi\in\left[0,\frac\pi2\right]$, and then $\cos^2\phi\geqslant\sin^2\phi\iff\phi\in\left[0,\frac\pi4\right]$. The condition $z\leqslant a$ becomes $\rho\cos\phi\leqslant a$. So, compute$$\int_0^{2\pi}\int_0^{\pi/4}\int_0^\frac a{\cos\phi}\rho^2\sin\phi\,\mathrm d\rho\,\mathrm d\phi\,\mathrm d\theta.$$