$u$ is a harmonic function in a domain $\Omega \subset \mathbb{C}$ and $ u : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ suppose $\bar{D{(a,R)}}$$ \subset \Omega $ then To show that $u(a) = \frac{1}{\pi R^2}\int \int_{\bar{D{(a,R)}}}u(x,y)dxdy$
My attempt. I took a function $f$ and parametrized the boundary of the disc. Integrating over the boundary by using Cauchy integral formula, I get
$f(a) = \frac{1}{2\pi}\int_{0}^{2\pi} f(a+Re^{1\theta})d\theta$
How should I proceed further. I know this is mean value theorem but I am stuck here.
Please help !
You're on the right track, but haven't quite been general enough. You used the Cauchy integral formula to get to $$ 2\pi f(a) = \int_0^{2\pi} f(a+Re^{i\theta}) \, d\theta, $$ but this is also true for any disc of radius $0<r<R$, i.e. $$ 2\pi f(a) = \int_0^{2\pi} f(a+re^{i\theta}) \, d\theta, \tag{1} $$ Now, we have a series of integrals over circles, and want to turn them into an integral over a disc. The area element is $r \, dr$, because each circle has length $r$, so the area of the small annulus between them is $$ \pi (r+h)^2-\pi r^2 = \pi(2rh+h^2) \sim 2\pi rh $$ as $h \to 0$. (we divide this by $2\pi$ now as we've already integrated over the angles) So the integral over the disc will be found by multiplying both sides of (1) by $r$ and integrating from $0$ to $R$: $$ 2\pi f(a) \int_0^R r \, dr = \int_0^R \int_0^{2\pi} f(a+re^{i\theta}) r \, d\theta \, dr. $$
The left hand side is $\pi R^2 f(a)$, and the right hand side is the integral over the disc $|z-a|<R$, which is equivalent to what you wanted.