Double integral in polar coordinates (trouble with boundaries)

Solution 1:

The region that we are talking about here is the upper half of the circle with center $(1,0)$ and radius $1$. So, the possible values of $\theta$ lie in the range $\left[0,\frac\pi2\right]$. For each such $\theta$, let $t=\tan(\theta)$. One must determine where $y=tx$ intersects the circle $(x-1)^2+y^2=1$ (see the picture below). So, we solve the equation$$(x-1)^2+t^2x^2=1\left(\iff(t^2+1)x^2-2x=0\right)$$and we get that $x=0$ or that $x=\frac2{t^2+1}=2\cos^2\theta$. And, when $x=2\cos^2\theta$, then$$y=2\cos^2(\theta)\tan(\theta)=2\cos(\theta)\sin(\theta),$$and therefore $r=\sqrt{x^2+y^2}=2\cos\theta$. So, compute$$\int_0^{\pi/2}\int_0^{2\cos\theta}r^2\,\mathrm dr\,\mathrm d\theta.$$