Double integration, faulty result (?)

Question

Region $D$ is given by $x^2+(y-2)^2=4$ and $x^2+(y-1)^2=1$.

$$\iint _D xy^2dxdy $$

I THINK I can compute this by $$\iint _{D_1\setminus D_2}xy^2dxdy $$, where $D^1$ is the large circle and $D_2$ the smaller circle. If we convert to polar coordinates,either circle could (?) be traced by $r(\theta) = a\sin\theta, 0\leq\theta\leq\pi$. Then we get $$\int _0^\pi\int_0^{a\sin\theta}r^4\cos\theta\sin ^2\theta \,drd\theta = \frac{a^5}{5}\int_0^\pi \sin ^7\theta\cos\theta\, d\theta = \frac{a^5}{40}\sin ^8\theta\vert _0^\pi = 0$$ By the above the entire double integral would be $0$, which is incorrect. What am I doing wrong? Answer is given as $\frac{\pi}{128}$

Answer 1

In my opinion you are correct. Note that the region $D$ is symmetric with respect to the line $x=0$ and the integrand function is odd with respect to $x$ (i.e. $f(-x,y)=-f(x,y)$). Hence $$\iint_D f(x,y)dxdy=\iint_{D\cap\{x>0\}} f(x,y)dxdy+\iint_{D\cap\{x<0\}} f(x,y)dxdy\\ =\iint_{D\cap\{x>0\}} f(x,y)dxdy+\iint_{D\cap\{x>0\}} f(-x,y)dxdy=0$$ the integral of $f$ over $D$ is zero.

Answer 2

$0$ is the right answer. For we know $D$ is symmetric about $x$, and $xy^2$ is an odd-function of $x$，so the double integration must be $0$.