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.