Currently I am working on this problem that requires me to calculate this triple integral when I am given cone/plane intersection. The issue is that integrals in both cases (when using spherical and cylindrical coordinates) lead to, even more, complicated ones. Only hints required. Thank You.
Calculate $$\int \int \int_E \sqrt{x^2+y^2+z^2}\space dV$$
where $E$ is the is region bound by the plane $z=3$ and the cone $z=\sqrt{x^2+y^2}$
On
In cylindrical coordinates, the cone has equation $z=r$; it intersects the plane in a circle of radius $3$. So the integral is equal to $$ \int_0^{2\pi} \int_0^3 \int_r^3 \sqrt{r^2+z^2}r\,dz\,dr\,d\theta = 2\pi \int_0^3 \int_r^3 \sqrt{r^2+z^2}r\,dz\,dr $$ As written the integral on the right needs a trigonometric substitution; it would be easier to rearrange the limits: $$ 2\pi \int_0^3 \int_r^3 \sqrt{r^2+z^2}r\,dz\,dr = 2\pi \int_0^3 \int_0^z \sqrt{r^2+z^2}r\,dr\,dz $$ Can you take it from there?
When you intersect the cone and the plain you get a circle. The region of integration is the interior of that circle. Watch for the limits of the integration. You want to describe the volume between $z=3$ and $z=\sqrt {x^2+y^2}$ nothing more and nothing less. Drawing a graph is very helpful to find the correct limits.