# Evaluate the integral by converting to polar coordinate

$$\int^{\pi/2}_{\pi/4} \int^{\sqrt{2-y^2}}_y 3(x-y) dx dy$$

I attempted the following:

$$\int_{\pi/4}^{\pi/2} \int_{0}^{1} 3r^2 (\cos\theta - \sin\theta) dr d\theta$$ which is wrong apparently. I think I might have got the wrong drawing of the curve.

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Yes your domain of integration is wrong. In the polar integral you have a rectangle, which would be an arc of a circular disk in the $xy$-plane – AD. Nov 4 '12 at 7:30
Is there a reason why to use polar coordinates? There is certainly a circular bound on the $x$-limit - but that is it. – AD. Nov 4 '12 at 7:40