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

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.

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 $$0 \le r \le \frac{\sin \theta}{\cos^2 \theta}, \quad 0 \le \theta \le \frac{\pi}{4}.$$
• Oh, I can see it now. I was messing up with lower $r$ since I was writing $rcos(\theta)=rsin(\theta)$, and I know that it works for $\pi/4$ but was trying to "get" $r$ from an equation. Thanks. Oct 12, 2020 at 23:03