# triple integral- plane/cone

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 $z=\sqrt{x^2+y^2}$

• So you're finding the volume of the cone? – Andrew Li Jun 3 '18 at 2:42
• What did you get when you converted to cylindrical coordinates? – Matthew Leingang Jun 3 '18 at 2:47
• This should work out quite nicely in either cylindrical or spherical coordinates. You had better check your work. @AndrewLi: No, this not the volume; it's the total mass if the density is given by the integrand. – Ted Shifrin Jun 3 '18 at 2:47
• @TedShifrin Ok, what I thought. I was unsure OP never mentioned density (even though they put the function in their post), just wanted to confirm – Andrew Li Jun 3 '18 at 2:50
• When I got to calculating dr part intervals gave me 1/0 at some point. – kenobe Jun 3 '18 at 2:52

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.