Area under circle of center $(0, 1)$ and radius $1$ Compute $$\int\int_A(3x^2y-y^3)\, dx\, dy$$ where $$A=\{(x, y)\ | \ x^2+(y-1)^2\leq 1\}.$$
I put $x=r\cos \theta$ and $y=1+r\sin \theta$ to get $$\int\int_A(3x^2y-y^3)\, dx\, dy=\int_{\theta=0}^{2\pi}\int_{r=0}^1(3r^2\cos^2\theta(1+r\sin\theta)-(1+r\sin\theta)^2) rdrd\theta.$$ But I am not getting it equal to $-\pi$. Is my approach correct or is there any easier way to solve this problem?

There is a wrong exponent in your transformed integral; it should be
$$\int_0^{2\pi}\int_0^1(3r^2\cos^2\theta(1+r\sin\theta)-(1+r\sin\theta)^{\mathbf3})r\,dr\,d\theta=-\pi$$
Alternatively, use the normal polar coordinate transformation $x=r\cos\theta,y=r\sin\theta$, under which $A$ is parametrised by $0\le\theta\le\pi,0\le r\le2\sin\theta$:
$$\int_0^\pi\int_0^{2\sin\theta}r^4(3\cos^2\theta\sin\theta-\sin^3\theta)\,dr\,d\theta=-\pi$$
