Defining region $D*$ for double interal in polar coordinates.

Question

I need to compute $\int_0^1\int_{x^2}^x\frac{1}{\sqrt{x^2+y^2}}dydx$.

Switching to polar coordinates, I get $0\leq\theta\leq\pi/4$ (from 0 to 1). Then, for $x^2=y$, $r=\tan(\theta)/cos(\theta)$ since $x=rcos(\theta)$ and $y=rsin(\theta)$. However, I can't figure out how to get the upper extreme of the integral for $r$, since $y=x \iff rcos(\theta)=rsin(\theta)$ and $r$ just cancel. I'm not even sure if the lower extreme of the integral is right.

Answer 1

For the curve $y = x^2$, we simply substitute $$(x,y) = (r \cos \theta, r \sin \theta)$$ to obtain $$r \sin \theta = r^2 \cos^2 \theta,$$ hence $$r(r \cos^2 \theta - \sin \theta) = 0.$$ Since $r = 0$ corresponds to the origin, the desired relationship is $$r = \frac{\sin \theta}{\cos^2 \theta}, \quad 0 \le \theta \le \frac{\pi}{4}.$$ When $\theta = \pi/4$, we get $r = \sqrt{2}$ which corresponds to $(x,y) = (1,1)$ as expected. Thus the region of integration in polar coordinates satisfies the inequalities $$0 \le r \le \frac{\sin \theta}{\cos^2 \theta}, \quad 0 \le \theta \le \frac{\pi}{4}.$$