Area inside the circle $x^2+y^2=4$, under $y=x\sqrt{3}$ and above $y=1$

Question

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?

Answer 1

If you want to use Cartesian coordinate:

Intercept of $x^2+y^2=4$ and $y=1$ is $(\sqrt3,1)$.

Intercept of $y=x\sqrt 3$ and $y=1$ is $(\frac 1{\sqrt 3}, 1)$

Intercept of $x^2+y^2=4$ and $y=x\sqrt 3$ is $(1, \sqrt 3)$

$$A=\int^{\sqrt3}_1 \sqrt{4-x^2}dx+\int^1_{\frac 1{\sqrt 3}}x\sqrt 3 dx-1(\sqrt 3-\frac1{\sqrt3}=\frac {2\sqrt 3}3)$$

$$\int \sqrt{a^2-x^2}dx=\frac x2 \sqrt{a^2-x^2}+\frac {a^2}2\sin^{-1} (\frac xa)$$

$$\int x\sqrt 3 dx=\frac{\sqrt 3}2 x^2 $$

You finally get:

$$A=\frac 13(\pi-\sqrt 3)$$