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

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
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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
    
@40Plot The bounds in the first integral are already wrong, since $(\pi/2)^2 > 2$. Could you either update the question with the relevant information, or delete it if that has been lost? I'm afraid it's insolvable in its current state. –  Lord_Farin May 24 '13 at 21:54
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