Integrating with polar coordinates, can't find the proper integration limits.

Question

The task is to evaluate $$\int_{0}^{\frac{\sqrt3}{2}} ~dx \int_{1 - \sqrt{1-x^2}}^{\sqrt{1-x^2}} \sqrt{x^2 + y^2 }~dy$$ using polar coordinates.

I don't have any attempts here because I don't get how to find the limits. The lower limit of inner integral doesn't even look like a circle and I don't know what to do.

Answer 1

A diagram should convince you the region $1-\sqrt{1-x^2}\le y\le\sqrt{1-x^2},\,0\le x\le\tfrac{\sqrt{3}}{2}$ is equivalent to $0\le\theta\le\tfrac{\pi}{2},\,0\le r\le\min\{2\sin\theta,\,1\}$. (In particular, the curve $1-y=\sqrt{1-x^2}$ satisfies $2y=r^2$ after you square and rearrange.) This is a union of two disjoint regions: one is $0\le\theta\le\tfrac{\pi}{6},\,0\le r\le2\sin\theta$, while the other is $\tfrac{\pi}{6}\le\theta\le\tfrac{\pi}{2},\,0\le r\le1$. Now just evaluate $\int r^2drd\theta$ over each and add.

Answer 2

Note that in polar coordinates $x= R \cos \theta.$ But even better, notice that from the top limit, $x^2 + y^2 \leq 1\dots$