Show $\iint_{D} f(x,y)(1 - x^2 - y^2) ~dx ~dy = \pi/2$ Suppose $f(x,y)$ is a bounded harmonic function in the unit disk
$D = \{z = x + iy : |z| < 1 \} $
and $f(0,0) = 1$. Show that
$$\iint_{D} f(x,y)(1 - x^2 - y^2) ~dx ~dy = \frac{\pi}{2}.$$
I'm studying for a prelim this August and I haven't taken Complex in a long time (two years ago).  I don't know how to solve this problem or even where to look unless it's just a game with Green's theorem-any help?  I don't need a complete solution, just a helpful hint and I can work the rest out on my own.

Harmonic functions have the mean value property, that is
$$
\frac1{2\pi}\int_0^{2\pi}f(z+re^{i\phi})\,\mathrm{d}\phi=f(z)
$$
If we write the integral in polar coordinates
$$
\begin{align}
\iint_Df(x,y)(1-x^2-y^2)\,\mathrm{d}x\,\mathrm{d}y
&=\int_0^1\int_0^{2\pi}f(re^{i\phi})(1-r^2)\,r\,\mathrm{d}\phi\,\mathrm{d}r\\
&=2\pi\int_0^1f(0)(1-r^2)\,r\,\mathrm{d}r\\
&=2\pi f(0)\cdot\frac14\\
&=\frac\pi2
\end{align}
$$
