$D=\{(x,y)\mid x^2+y^2 \le x\}$ , by changing to polar coordinate, find this integral: $$\iint_{D}\sqrt x\,dx\,dy$$
So I got this by changing to polar coordinate
$$\int_{-\pi/2}^{\pi/2}\int_{0}^{1/2}\sqrt{r\cos \theta}\ rdr\,d\theta$$
First is my limit for integral true?
And I don't know how to integrate $\sqrt {r\cos\theta} $ here.
Thank you!
Converting to polar we see $D= \{(r, \theta) | r^2 - r\cos\theta \le 0\} =\{(r,\theta) | r\le \cos\theta\}$.
Not forgetting to multiply by the Jacobian (which I suspect you did originally) we have that
$$\int \int_D \sqrt{x}dxdy = \int_{-\pi/2}^{\pi/2}\int_0^{\cos\theta}r\sqrt{r}\sqrt{\cos\theta}drd\theta$$ $$ = \int_{-\pi/2}^{\pi/2}\frac{2}{5}\cos^3\theta d\theta = \frac25 \int_{-\pi/2}^{\pi/2}\cos\theta (1 -\sin^2\theta)d\theta$$ $$ = \frac25 \left(\sin\theta - \frac13 \sin^3\theta \biggm|_{-\pi/2}^{\pi/2}\right)= \frac25 \cdot \frac43 = \frac{8}{15}.$$