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Sketch the region represented in the following double integral: $I =∬ln(x^2+y^2)dxdy$

In the region:

$x:ycotβ$ to $\sqrt(a^2+y^2)$

$y:0$ to $asinβ$

where a > 0 and 0 < β < π/2.

By changing to polar coordinates or otherwise, show that I = $a^2β(ln a − 1/2)$.

So far I have found that the sketch yields a triangle inscribed in the circle of radius $a$ in the first quadrant. I am having trouble converting to polar coordinates.

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Hint: In the polar coordinate system $x^2+y^2 \Rightarrow r^2$ and $dxdy \Rightarrow r\cdot drd\theta$. Also, we have that $y=r\cdot \sin{\theta}$ and $x=r\cdot \cos{\theta}$.

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  • $\begingroup$ My problem is finding the limits for $r$ and theta. $\endgroup$ – mat Feb 3 at 12:09

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