[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?