Computing $\frac{1}{\pi}\int\int_\mathbb{D}|\phi_\alpha '|^2 dxdy$ 
If $\phi_\alpha(z)=(\alpha-z)/(1-\bar\alpha z)$ for $|\alpha|<1$,
prove that
$\frac{1}{\pi}\int\int_\mathbb{D}|\phi_\alpha '|^2 dxdy=1$ and
$\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=\frac{1-|\alpha|^2}{|\alpha|^2}log\frac{1}{1-|\alpha|^2}$ where in the case $\alpha=0$ the expression on the right is understood
as the limit as $|\alpha|\to 0$.
[Hint: The first integral can be evaluated without a calculation. For
the second, use polar coordinates, and for each fixed $r$ use contour
integration to evaluate the integral in $\theta$]

I did the first question, but I have trouble doing the second question. I don't know how to use polar coordinates method here.
So I computed $\phi'_\alpha(z)=\frac{-1+\alpha\bar\alpha}{(1-\alpha z)^2}$ and by using polar coordinate method I know, I get $\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=\frac{1}{\pi}\int^{2\pi}_0\int^1_0|\frac{-1+\alpha\bar\alpha}{(1-\bar\alpha re^{i\theta)^2}}|rdrd\theta$, but it looks wrong. Can you give me any idea? Thank you.
 A: $\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=\frac{1}{\pi}\int^{2\pi}_0\int^1_0\frac{|-1+\alpha\bar\alpha|}{|1-\bar\alpha re^{i\theta)^2}|}rdrd\theta=(1-|\alpha|^2)\int^1_0r(\frac{1}{\pi}\int^{2\pi}_0\frac{1}{|1-\bar\alpha re^{i\theta)^2}|}d\theta)dr$
Clearly the integral is invariant under a circle rotation $\alpha \to \alpha e^{it_0}$ so we can assume $\alpha=c=|\alpha|>0$
$\frac{1}{|1-cre^{i\theta}|^2}=\frac{P(cr,\theta)}{1-c^2r^2}$ so $\frac{1}{\pi}\int^{2\pi}_0\frac{1}{|1-\bar\alpha re^{i\theta)^2}|}d\theta=\frac{2}{1-c^2r^2}=\frac{2}{1-|\alpha|^2r^2}$ as the integral of the Poisson kernel (divided by $2\pi$) is $1$ hence we get:
$\frac{1}{\pi}\int\int_\mathbb{D}|\phi'_\alpha|dxdy=2(1-|\alpha|^2)\int_0^1\frac{r}{1-|\alpha|^2r^2}dr=\frac{1-|\alpha|^2}{|\alpha|^2}\log\frac{1}{1-|\alpha|^2}$ so we are done!
A: You can directly compute the integral without using the Poisson kernel. We assume that $\alpha$ is real. Using
$$ z=e^{i\theta}, \bar z=e^{-i\theta} $$
we have
\begin{eqnarray}
&&\frac{1}{\pi}\iint_\mathbb{D}|\phi'_\alpha|dxdy\\
&=&\frac{1-|\alpha|^2}{\pi}\int^1_0\int^{2\pi}_0\frac{1}{|1-\alpha re^{i\theta}|^2}r\,d\theta\,dr\tag1\\
&=&\frac{1-|\alpha|^2}{\pi}\int^1_0r\int_{|z|=1}\frac{1}{|1-\alpha rz|^2}\frac{dz}{iz}dr\tag2\\
&=&\frac{1-|\alpha|^2}{\pi i}\int^1_0r\int_{|z|=1}\frac{1}{(1-\alpha rz)(1-\alpha r\frac1z)}\frac{dz}{z}dr\tag3\\
&=&\frac{1-|\alpha|^2}{\pi i}\int^1_0r\int_{|z|=1}\frac{1}{(1-\alpha rz)(z-\alpha r)}dz\,dr\\
&=&\frac{1-|\alpha|^2}{\pi i}\int^1_0r\cdot2\pi i\frac{1}{1-\alpha^2r^2}dr\\
&=&2(1-|\alpha|^2)\pi\int^1_0\frac{r\,dr}{1-\alpha^2r^2}\\
&=&-\frac{1-|\alpha|^2}{\alpha^2}\ln(1-\alpha^2).
\end{eqnarray}
Update:

*

*Since $z=e^{i\theta}$, we have $dz=ie^{i\theta}d\theta=izd\theta$ and hence $d\theta=\frac{dz}{iz}$. So we have (2).


*Since $|z|^2=1$, we have $\bar z=\frac1z$ and hence
$$|1-\alpha r z|^2=(1-\alpha r z)(1-\alpha r\bar z)=(1-\alpha z)(1-\alpha r\frac1z). $$
So we have (3).
