I have to find the area in the circle $x^{2}+y^{2}=4$ such that it's under the line $y=x\sqrt{3}$ and above $y=1$.
Since in polar coordinates $x=r\cos(t), y=r\sin(t)$, then if $y=1$:
$$r\sin(t)=1\Rightarrow r=\frac{1}{\sin(t)}$$
And due to the fact that the radius of the circle is 2, so the area should be given by
$$\int_{0}^{\pi/3}\int_{1/sin(t)}^{2}rdrdt$$
but the integral above diverges. What's wrong?
Best Answer
$$\begin{align}A&=\iint_{1\le y\le\sqrt3\atop y/\sqrt3\le x\le\sqrt{4-y^2}}dxdy\\&=\int_1^{\sqrt3}\left(\sqrt{4-y^2}-\frac y{\sqrt3}\right)dy\\&=\int_{\pi/6}^{\pi/3}\sqrt{4-(2\sin t)^2}\;2\cos tdt-\left[\frac{y^2}{2\sqrt3}\right]_1^{\sqrt3}\\&=2\int_{\pi/6}^{\pi/3}(1+\cos(2t))dt-\frac1{\sqrt3}\\ &=2\left(\frac\pi6+0\right)-\frac1{\sqrt3}\\ &=\frac\pi3-\frac1{\sqrt3}. \end{align}$$