How to find an area between $x =\frac 12$ on the right, $y = \frac x{\sqrt3}$ above and by the circle $x^2 + y^2 = 2x$ with polar coordinates?

Question

The problem is: Find the area of the region that is bounded by $x =\frac 12$ on the right and $y = \frac x{\sqrt3}$ above, and by the circle $x^2 + y^2 = 2x$.

I have a general idea of how to find the bounds of the radius and angles, but I'm not entirely sure. As of right now, I calculated that the radius bounds were $0$ to $2\cos(\theta)$ and the theta bounds are $\frac{5π}3$ to $\frac{π}6.$ Am I going in the right direction?

Answer 1

You need to calculate $$\displaystyle\int_0^{1/2}\int_{-\sqrt{1-(x-1)^2}}^{\frac{x}{\sqrt3}}dydx$$ to find the area of the region.

Notice that there's no need to use polar coordinates.

Answer 2

With polar coordinates you can not solve it with one integral.

You need to connect the origin to the point on the circle where the circle and the line $x=1/2$ intersect.

For the bottom part your $\theta$ goes from $\pi /6$ to $\pi /3$, while your $r$ goes from $0$ to $(\sec \theta )/2$

For the upper part your $\theta$ goes from $\pi/3 $ to $\pi/2$, while your $r$ goes from $0$ to $2\cos \theta$

Good Luck