I'm trying to evaluate the following integral:
$\int_H x\ dxdy$, where $H:=\{(x,y)\in\mathbb{R^2}:x,y\geq 0, x^2+y^2\leq 4,x^2+y^2-2y\geq 0\}$
i.e. $H$ is the region in the first quadrant inside the circle centered at the origin with radius $2$ and outside the circle centered in $(0,1)$ with radius $1$.
First attempt:
$\int_H x\ dxdy=\int_{y=0}^{y=2}(\int_{x=\sqrt{y^2-2y}}^{x=\sqrt{4-y^2}}x\ dx)dy=\frac{1}{2}\int_{y=0}^{y=2}(4-y^2-(y^2-2y))dy=\int_{y=0}^{y=2}(-y^2+y+2)dy=\frac{10}{3}$.
Second attempt (polar coordinates): $0\leq\theta\leq\frac{\pi}{2}$ and $x^2+y^2-2y\leq 0\Leftrightarrow r^2-2r\sin\theta\geq 0\Leftrightarrow r-2\sin\theta\geq 0$ so $2\sin\theta\leq r \leq 2$ thus $\int_H x\ dxdy=\int_{\theta=0}^{\theta=\pi/2}(\int_{r=2\sin\theta}^{r=2}r^2\cos\theta\ dr) d\theta=\frac{1}{3}\int_{\theta=0}^{\theta=\pi/2}\cos\theta[r^3]_{2\sin\theta}^{3}\ d\theta=\frac{8}{3}\int_{\theta=0}^{\theta=\pi/2}(\cos\theta-\cos\theta\sin^3\theta)d\theta=\frac{8}{3}[\sin\theta-\frac{1}{4}\sin^4\theta]_{0}^{\pi/2}=2$.
As you can see the result is different so I've tried to figure out where I made a mistake but I haven't yet succeeded so I'd appreciate if someone could double-check this.
Thanks a lot.