Curve integral of exponent of superharmonic function. 
Let $\phi$ be a real smooth superharmonic function on unit disc $D$ in $\mathbb C$; i.e. $\triangle \phi\le 0$.
  Then there is a curve $\gamma$ from the center of $D$ to its boundary such that 
  $$\int\limits_\gamma e^\phi<\infty.$$

The question came from my failed answer to this question.
I know that the answer is YES, but I do not see a direct proof.
 A: I'll assume that $\varphi=-\psi$ where $\psi$ is subharmonic and not too weird (say, with isolated non-degenerate critical points; it seems like you can always add something bounded to achieve it but I haven't checked the details). Take any piece of some level curve of $\psi$ inside the disk parameterized by length $\ell$ and start the gradient accent from each point parameterized by the level $t$ of $\psi$. All but countably many of those escape to the boundary. Let $v(t)$ be the absolute value of the gradient and $S(t)$ be the "cross-section factor". Then $S(t)v(t)$ is non-decreasing (divergence of the gradient field is positive), the length element is $1/v(t)dt$ and the area element is $S(t)/v(t)d\ell dt$. Note that the total area of the disk is finite and the gradient curves cannot meet (we use the non-negativity of the divergence again here). Thus $\int S(t)/v(t)dt<+\infty$ most of the time. But then, since $Sv$ is non-decreasing, we also have $\int 1/v(t)^2dt<+\infty$ while we need just $\int e^{-t}/v(t)<+\infty$ and Cauchy-Schwartz ends the story.
A: Here is a simple answer in a special case when $\phi$ is harmonic. 
 Geometric proof: Let $g_0$ be the (incomplete) metric on the unit disc induced from $\mathbb R^2$. Then the metric $e^\phi g_0$ is complete if and only if the integral
of $e^\phi$ over $\gamma$ is infinite for each $C^1$-curve $\gamma$ from the center 
to the boundary of the disk. Now the sectional curvature of $e^\phi g_0$ is $-\frac{1}{2}e^{-\phi}\Delta\phi$, which is zero since $\phi$ is assumed harmonic. Thus if $e^\phi g_0$ were complete, it would be isometric to the standard $\mathbb R^2$, but isometries are conformal, so we would get that the unit disk is conformal to the plane.
 Sketch of complex analysis proof: If by a "direct proof" you mean then one that uses only complex analysis, then Huber's proof quoted in my question does just that. However,
Huber's argument simplifies a lot when $\phi$ is harmonic. Indeed, harmonicity of $\phi$ allows you to find its harmonic conjugate, and since the disk is simply connected, $\phi$ becomes the real part of an analytic function $\tau$ on the disk. 
Thus $e^\phi=|e^\tau|$, and $e^\tau$ is analytic and nowhere vanishing. Now I think the simple argument in the note "Paths of rapid growth of entire functions" by Kaplan does the job (see middle of the first page). 
 Remark. Kaplan's note is in fact an elaboration of the Huber's proof, and it gives a simple answer to your question for all superharmonic functions of the form $\phi=-\log|f|$ such that $f$ does not vanish (and this is what you seem to care about by restricting to smooth superharmonic functions). For general superharmonic functions I do not see how to do better than Huber.
