Integral over a ball Let $a=(1,2)\in\mathbb{R}^{2}$ and $B(a,3)$ denote a ball in $\mathbb{R}^{2}$ centered at $a$ and of radius equal to $3$.
Evaluate the following integral:
$$\int_{B(a,3)}y^{3}-3x^{2}y \ dx dy$$
Should I use polar coordinates? Or is there any tricky solution to this?

Note that $\Delta(y^3-3x^2y)=0$, so that $y^3-3x^2y$ is harmonic. By the mean value property, we get that the mean value over the ball is the value at the center. Since the area of the ball is $9\pi$ and the value at the center is $2$, we get
$$
\int_{B(a,3)}\left(y^3-3x^2y\right)\mathrm{d}x\,\mathrm{d}y=18\pi
$$
