Hi, I'm having problems with converting to polar coordinates when evaluating integrals. Here is an example, it comes down to writing the following equality. After that the evaluation of the intergral is straight forward.(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

\int\limits_0^2 {\int_{ - \sqrt {4 - y^2 } }^{\sqrt {4 - y^2 } } {x^2 y^2 dxdy = } } \int\limits_0^\pi {\int\limits_0^2 {\left( {r^4 \cos ^2 \theta \sin ^2 \theta } \right)} } rdrd\theta

[/tex]

Looking at the original integral I can see that if I equate "x" to the upper or lower limit of the inner integral then I get a x^2 + y^2 = 4, which is the equation of a circle of radius 2. But I'm not sure how to convert to polar coordinates the right way.

Firstly I see that dy corresponds to the outer integral(of the LHS) from y = 0 to y = 2 but what can I do with that? Not much, at least now anyway so I then observe that dx corresponds to the upper and lower limits of the inner integral. So I figured that means a point on the circle would travel from x = -2 to x = 2, then the outer part of the 'polar' integral would have limits 0 to pi/2. But that's wrong.

Actually...I can sort of see why the second integral is correct now. The main problem is that I don't completely understand how the conversions of the upper and lower limits of integration proceed. I kind of understand the r = 0 to r = 2 part but I don't really understand outer integral with zero and pi. Can someone please explain it to me?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Polar coordinates conversions

**Physics Forums | Science Articles, Homework Help, Discussion**