# Changing order of integration in cylindrical coordinates

I'm having a problem in changing order of integration in triple integration, in cylindrical coordinates. I would be grateful for a little help.The question is:

Let D be the region bounded below by the cone $z= \sqrt{x^2 + y^2}$ and above by the parabola $z=2-x^2-y^2$. Set up the triple integrals in cylindrical coordinates that give the volume of D using the following orders of integration.

(a) $dzdrd\theta$, (b)$drdzd\theta$, (c) $d\theta dzdr$

I drew the graph, and it looks like an ice cream cone, with radius of 1. I figured out (a), which was $0\leq \theta \leq 2\pi$, $0 \leq r \leq 1$, and $r \leq z \leq 2-r^2$. which would give the equation $\int_0^{2\pi} \int_0^1 \int_r^{2-r^2} dzrdrd\theta$ . And this was the method that our professor taught us.

But I'm not sure what to do for (b) and (c). Which way should I look at the graph? For (b), i guess the interval of integration for $\theta$ is $0\leq \theta \leq 2\pi$ but I'm not sure for $z$ and $r$.....

I thought in double integral with polar coordinates, I can't change the order of $drd\theta$. Why can I do it here?

• I believe you intend $\ z \ = \ 2 \ - \ x^2 \ - \ y^2 \$ for your paraboloid equation. Note, then, that you can write your cone surface as $\ r \ = \ z \$ and the paraboloid as $\ r^2 \ = \ 2 \ - \ z \ \Rightarrow \ r \ = \ \pm \sqrt{2 \ - \ z} \$ (you have two functions here, one for each semi-circle) . That will aid in using other integration orders. Since both surfaces have "axial" symmetry, there is no dependence on $\ \theta \$ , so the integral in angle is separable regardless of integration order. Jul 7 '14 at 6:17
• Oh!! Thank you!!! I'll try that out right now!! Jul 7 '14 at 6:26
• Since you are just computing volume of the enclosed region, you can avoid the sign ambiguity issue for the paraboloid by just integrating in the first octant and multiplying the result by 4 (not 8, since there is nothing below the $\ xy-$ plane). Jul 7 '14 at 6:47

For (b): $\int_0^{2\pi}\int_0^1\int_z^\sqrt{2-z}rdrdzd\theta$
For (c): $\int_0^1\int_r^{2-r^2}\int_0^{2\pi}d\theta dzrdr$
You can bring $d\theta$ anywhere with other limits remaining same due to symmetry of the problem.