# Double integral change of variables to polar coordinates.

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.

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 $$x=r\cdot \cos{\theta}$$.

• My problem is finding the limits for $r$ and theta. – mat Feb 3 '19 at 12:09