Why does $ \int_{\Omega} f\, dA = 2\pi \,f(0) $ and not zero? I have been experimenting with quadrature domains.  The most obvious one is a circle.  Let $f(z)$ be holomorphic on a large enough region and $\Omega= \{|z| < 1 \}$ then:
$$ \int_{\Omega} f\, dA = \pi \,f(0) $$
I almost reasoned the answer should be zero.  We could change the area to polar coordinates:
$$ \int_{\Omega} f\, dA = \int_0^1 \int_{|z| = r } f(x+iy) \,dx\, dy 
=  \int_0^1 \int_{|z| = r } f( r \, e^{i\theta}) \,r \, dr \, d\theta = 0$$
since the integral around the circle should be zero.  Assuming there is no   pole in side the circle $\{ |z| < 1\}$:
$$ \oint_{|z| = 1} f(z) \, dz = \int_{|z| = 1} f ( r \, e^{i\theta})\, r e^{i\theta} d\theta = 0 \cdot f(0) = 0$$
 A: Short Version: $dz\ne d\theta$.
In Detail:
Forget $r$ for the moment - it's irrelevant to the problem.
The notation $\int_{|z|=1}f(e^{i\theta})\,d\theta$ doesn't really make sense, because there's no $z$ in the integrand. Assumng $f$ is hoolomorphic in a neighborhood of the closed unit disk both of the following hold:
$$\int_{|z|=1}f(z)\,dz=0$$(Cauchy's Theorem)
$$\frac1{2\pi}\int_0^{2\pi}f(e^{i\theta})\,d\theta=f(0)$$
(mean-value property for holmorphic (hence harmonic) functions).
The two integrals are simply not the same at all. If we parametrize the unit circle via $z=e^{i\theta}$, $0\le\theta\le2\pi$ we get $$dz=ie^{i\theta}\,d\theta,$$hence
$$\int_{|z|=1}f(z)\,dz=i\int_0^{2\pi}f(e^{i\theta})e^{i\theta}\,d\theta
\ne\int_0^{2\pi}f(e^{i\theta})\,d\theta.$$
Now when you do the original integral in polar coordinates you get $d\theta$, not $dz$. A correct version of your polar-coordinate calculation is this: $$\int_{\Omega} f\, dA = \int_0^1 \int_0^{2\pi} f(x+iy) \,dx\, dy 
=  \int_0^1 \int_0^{2\pi} f( r \, e^{i\theta}) \,r \, dr \, d\theta = 
2\pi\int_0^1 f(0)r\,dr=\pi f(0).$$The integral $\int_{|z|=r}f(z)\,dz$ simply doesn't come up there.
(Look up the polar coordinates formula. You don't see $\int_{|z|=r}$ there.)
A: $\int_{|z|=1} f(r e^{i \theta}) e^{i \theta}\,d\theta$ is zero as you say, but $\int_{|z|=1} f(r e^{i \theta}) \,d\theta$ doesn't have to be.  You could think about the constant function $f=1$, for instance.
A: Note that $$\frac{1}{2\pi}\int_0^{2\pi} f(re^{it}) dt = \frac{1}{2\pi i}\int_{\partial B_r(0)} \frac{f(z)}zdz$$
and use Cauchy's theorem.
