Question: Let $\Delta$ be the unit disc in $\mathbb{C}$ and $\rho(z)|dz|^2$ be a complete conformal metric on $\Delta$ where $\rho(z)$ is continuous on $\Delta$. Let $a$ be the infimum of $p (p>0)$ such that 
$\iint\_\Delta |\rho(z)|^pdxdy=+\infty.$ 


I guess that $a\leq 1$. Of course, generally $a$ depends on the complete metric $\rho(z)$. 
For example, w.r.t. the Poincare metric, $a=\frac{1}{2}.$ Also, one may consider the infimum of $a$.


Note. We only assume that $\rho(z)$ is continuous and complete on the unit disc.



A question on the area of the unit disc w.r.t. a complete conformal metric
Completeness implies that 
$$\int\_{1/2}^1\sqrt{\rho(r,\theta)}dr=\infty$$
for all $\theta$.
So, for a complete metric,
$$\int\_\Delta\sqrt{\rho}=\int\_0^{2\pi}\int\_0^1\sqrt{\rho(r,\theta)}rdrd\theta=\infty.$$
Thus $a\leq 1/2$.


For Poincare metric $\rho=1/(1-r^2)^2$, so
$\alpha=1/2$, and this is best possible.


