Area of certain region within a circle

Question

I need to find the area of this region:

$$\begin{align} y&=1,\dots,y=\sqrt3x&&\implies\left(\frac1{\sqrt3},1\right)\\ y&=1,\dots,x^2+y^2=4&&\implies(\sqrt3,1)\\ y&=\sqrt3x,\dots,x^2+y^2=4&&\implies(1,\sqrt3) \end{align}$$

With Fubini's theorem I get $$\int_{y=1}^{y=\sqrt3}\left (\int_{x=\frac{y}{\sqrt3}}^{x=\sqrt{4-y^2}}\ \mathrm dx\right)\ \mathrm dy=\frac{\pi-\sqrt3}{3}$$

I think this is a good answer.

However how can use polar coordinates to solve this?

Polar coordinates $$x=\rho\times\cos(\theta);\quad y=\rho\times\sin(\theta)$$

I know that $\theta\in\left[\dfrac{\pi}{6},\dfrac{\pi}{3}\right]$ but how can I find $\rho$?

My guess is $\rho\in\left[\dfrac{1}{\sin\theta},\text{something}\right]$. Can somebody explain how to find $\rho$ in this situation?

Answer 1

area in pink + area in blue = $\int_1^2\sqrt{4-x^2}\,dx=\frac{2\pi}3-\frac{\sqrt3}2$

area in blue + area in green = $\frac12\times 2^2\times \frac\pi6=\frac\pi3$

area in green = $\frac12\times1\times\frac1{\sqrt3}=\frac1{2\sqrt3}$

Hence :

$$\mathrm{pink}\,\mathrm{area}=\left(\frac{2\pi}3-\frac{\sqrt3}2\right)-\left(\frac\pi3-\frac1{2\sqrt3}\right)=\boxed{\frac{\pi-\sqrt3}3}$$

Hoping there are not too many mistakes ...

Answer 2

Find the equation of the horizontal line $y=1$ in polar coordinates. Suppose the equation is $\rho = f(\theta).$ Then you need to integrate $2 - f(\theta)$ over the range of angles spanned by the region, that is, from the value of $\theta$ where $y=1$ intersects the circle in the first quadrant to the value of $\theta$ for the line $y=\sqrt3 x.$

Even easier, without calculus at all, construct a line from the origin through the point $(\sqrt3,1)$ where $y=1$ intersects the circle. This and the other line bound a segment of the circle. Find the angles of the two lines, and use the angle between them to find the area of the segment. But the segment includes both your desired region and a triangle below it, so subtract the area of the triangle.