# Math Help - Double Integral In Polar Coordinates

1. ## Double Integral In Polar Coordinates

Hi all,

I'm trying to solve the following by converting it to polar coordinates.

$\int_{0}^{\sqrt{2}}\int_{y}^{\sqrt{4-y^2}} xy dxdy$

I know that the x*y term will become:

$(rcos(\theta))(rsin(\theta))r drd\theta$

However I'm not sure how to get the integration boundaries. I've looked around but I can't find an example that is similar. Any help would be greatly appreciated. Thanks.

$\displaystyle{\int_{0}^{\sqrt{2}}\int_{y}^{\sqrt{4-y^2}}\ 1\ dx\ dy}$
$\displaystyle{\int_{0}^{\sqrt{2}}\sqrt{4-y^2}}\ -\ y\ dy}$
So it's between the lines/curves for $\displaystyle{x = \sqrt{4-y^2}}$ and $x = y$ and $y = 0$ and $y = \sqrt{2}$. So it's a nice sector of a circle, which gives you the polar limits.