I need to calculate this double integral using change of variables
$$\iint_D\sqrt{x}\ dx\ dy $$
with $D=\{(x,y) \mid x^2+y^2 < x\}$.
I thought of putting $x=r\cos\theta$ and $y=r\sin\theta$. So $x^2+y^2<x$ will become $r<\cos\theta$ (should I suppose here that $r \ne 0$?). I'll just rewrite the integral as
$$\int_0^{2\pi}\int_0^{\cos\theta}r\sqrt{r\cos\theta}$$
which is double with a bit of calculation.
I would be very thankful if someone can tell me if my method is right. If not, how can I solve it?
Thank you very much.
Your method is correct but your limits for $\theta$ is incorrect.
The correct integral is:
$$\int_{-\pi/2}^{\pi/2}\int_0^{\cos\theta} r \, \sqrt{r\cos\theta} \, dr \, d\theta$$
Circle $r = \cos \theta$ with center at $(\frac{1}{2}, 0)$ and with radius of $\frac{1}{2}$ is between $ - \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.