I'm solving this integral $$\iint_Dxy \,dA$$ where $$D=\{(x,y)\in\mathbb{R}\,/\,x^2+y^2\leq1,\,y\leq\sqrt3x\}$$
From the plot I know it is a circumference and a line
I'm switching to polar coordinates, however I'm not sure how to find the angle of $\theta$
$$\begin{aligned} r\sin\theta&=\sqrt3r\cos\theta \\ \tan\theta&=\sqrt3\\\theta&=\frac{\pi}{3}\end{aligned}$$
I know this is my upper bound from $\theta$ but how do I find my lower bound?
$$? \leq\theta\leq\frac{\pi}{3}$$
On
The given integral is clearly zero. $D$ is given by half a circle, i.e. a domain admitting a (polar) parametrization with $\theta\in[a,a+\pi]$ and $\rho\in[0,1]$. Switching to polar coordinates
$$ I = \int_{0}^{1}\int_{a}^{a+\pi}\rho^3 \sin\theta\cos\theta\,d\theta \,d\rho = \frac{1}{8}\int_{a}^{a+\pi}\sin(2\theta)\,d\theta=0$$ since the integral of a sine function over a full period equals zero: $a$ is irrelevant.
The line has an angle $\pi$, then the angle is $−(\pi−\pi/3)=−2\pi/3$.