Question

∫∫∫sqrt(x^2+y^2+z^2)dv the region bounded by the plane z=3 and cone z=sqrt(x^2+y^2)

Answer 1

The region of interest (the interior of the cone in the sketch) is given by
\[D=\left\{(x,y,z)~:~-3\le x\le3,~-3\le y\le 3,~\sqrt{x^2+y^2}\le z\le3\right\}\]
\[\int\int\int_D\sqrt{x^2+y^2+z^2}~dV=\int_{-3}^3\int_{-3}^3\int_\sqrt{x^2+y^2}^3\sqrt{x^2+y^2+z^2}~dz~dy~dx\]
You'll want to convert to spherical coordinates for this:
\[D=\left\{(r,\theta,\phi)~:~0\le r\le3,~0\le\theta\le2\pi,~0\le \phi\le\frac{\pi}{4}\right\}\]
and the integral changes to
\[\int_0^3\int_0^{2\pi}\int_0^{\pi/4}r~(r^2\sin\phi~d\phi~d\theta~dr)\]