Integral between two circles and two lines

Question

I have a Problem with the following task: I have to calculate the integral $\iint_Barctan(\frac{y}{x})d(x,y)$ And $B \in R^2$ is the area which is bounded through two circles $x^2+y^2=1, x^2+y^2=9$ and the two lines $ y=\frac{1}{3}\sqrt3x, y=\sqrt3x$

I divied $B$ into two areas with $x=\frac{3}{2}$ so i got $\iint_Barctan(\frac{y}{x})d(x,y)= \int_\frac{1}{2}^\frac{3}{2}\int_\frac{\sqrt3x}{3}^{\sqrt3x}arcan(\frac{y}{x})dydx+\int_\frac{3}{2}^\frac{3\sqrt3}{2}\int_\frac{\sqrt3x}{3}^{\frac{3}{2}}arcan(\frac{y}{x})dydx$

I guess i have to use polar coordinates but i dont know what do to with the bounds for the integral then

A picture of the problem

Thanks!

Answer 1

You can do a Change of coordinate:

$x=rcos(\theta) $

$y=rsin (\theta)$

and you get

$arctan(\frac{y}{x})=\theta$

and

$d(x,y)=rd(r,\theta)$

So the integral $I$ becames

$I=\int_{arctan(\frac{\sqrt{3}}{3})}^{arctan(\sqrt{3})}\int_{1}^3 r\theta d(r,\theta)=$

$=(\frac{9}{2}-\frac{1}{2})\int_{arctan(\frac{\sqrt{3}}{3})}^{arctan(\sqrt{3})} \theta d\theta=...$