I am asked to compute $$\int_D x^3-3xy^2\,{\rm d}x\,{\rm d}y,$$where $D = \{ (x,y) \mid (x+1)^2+y^2 \leq 9, \text{and }(x-1)^2+y^2 \geq 1 \}$. Granted, $u(x,y) = x^3-3xy^2$ is harmonic (it is the real part of $z^3$, which is holomorphic, say), so we have the mean value property here. Let $D_1$ be the bigger disk and $D_2$ the smaller one. Then $$\int_D x^3-3xy^2\,{\rm d}x\,{\rm d}y = \int_{D_1} x^3-3xy^2\,{\rm d}x\,{\rm d}y -\int_{D_2} x^3-3xy^2\,{\rm d}x\,{\rm d}y .$$We should have have that: $$u(-1,0) = \frac{1}{9\pi}\int_{D_1}x^3-3xy^2\,{\rm d}x\,{\rm d}y \implies \int_{D_1} x^3-3xy^2\,{\rm d}x\,{\rm d}y = -9\pi.$$But wolfram alpha gives the other sign. I don't know how to set up the double integral there, so I passed it to (shifted) polar coordinates and threw it there - I don't think I made mistakes in this part. I got so paranoid that I actually did it again by hand and got $9 \pi$ too.
For the second one, we'd get: $$u(1,0) = \frac{1}{\pi}\int_{D_2}x^3-3xy^2\,{\rm d}x\,{\rm d}y \implies \int_{D_2}x^3-3xy^2\,{\rm d}x\,{\rm d}y = \pi.$$
Wolfram doesn't seems to understand this second integral, so I couldn't double-check - anyway, I'm not trusting anything here.
My computation would give that the initial integral is equal to $-10 \pi$.
Can someone please find out what I'm missing here, explain what is going on here and find out what the integral evaluates to?
There was a mistake when setting up the integral in WA. The transformation of coordinates should be
$$x+1=r\cos(t) \implies x=r\cos (t)-1$$
rather than $x=r\cos(t)+1$.
When making this correction SEE HERE the answer is indeed $-9\pi$ as guaranteed by the mean-value theorem!