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?
Best Answer
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.