Basically, one of the problems stated to use the second law of thermodynamics to derive the laplace equation: $$\nabla^2_{r,\theta} = \frac{\delta^2}{\delta r^2} + \frac{1}{r}\frac{\delta}{\delta r} + \frac{1}{r^2}\frac{\delta^2}{\delta\theta^2}$$ is a differential operator operating on the temperature function $T = T(r,\theta)$
EDIT: The exact question is something along the lines of: "Using the conservation of energy principle, derive the Laplace equation for a steady-state heat transfer problem in a polar-cylindrical coordinate system when the temperature function only varies with the radial $r$- and the angular $\theta$- directions. In other words, derive $\nabla^2_{r,\theta} T = 0$.
Basically, what I can understand from this is that it is a 2D heat conduction problem on polar coordinates, so a cylinder of length L, in that $\dot{Q_r} - (\dot{Q_r} + \frac{\delta \dot{Q_r}}{\delta r}dr) + \dot{Q_\theta}- (\dot{Q_\theta} + \frac{\delta \dot{Q_r}}{\delta r}d\theta)$ = 0 ; that is, I know how to do it from this equation to the laplace equation. My problem, however, is doing it the other way around.
If I were to do it from Q to the laplace, it might look something like this: $$\frac{\delta \dot{Q_r}}{\delta r}dr + \frac{\delta \dot{Q_\theta}}{\delta \theta}d\theta$$ $$\vec{Q} = -k\nabla T$$ $$\dot{Q_r} = -k_rA_r\frac{\delta T}{\delta r};\; \; \; \dot{Q_\theta} = -k_\theta A_\theta\frac{\delta T}{r\delta \theta}$$ $$A_r = r d\theta L; \; \; \; A_{\theta} = drL$$ The reason the areas are as such is because, when taking a small portion from a cylinder of length $L$, it would form a rectangle with sides $dr$ and $rd\theta$ (actually, one side is $(r+dr)d\theta$, but since $drd\theta$ is really small anyway, I just ignored it) $$\frac{\delta}{\delta r}(-k_r (rd\theta L)\frac{\delta T}{\delta r})dr - \frac{\delta}{\delta \theta}(-k_\theta drL \frac{\delta T}{r \delta \theta})d\theta$$ $$-kd\theta drL(\frac{\delta}{\delta r}(r\frac{\delta T}{\delta r}) + \frac{1}{r}\frac{\delta}{\delta \theta}(\frac{\delta T}{\delta \theta})) = 0$$ $$\frac{\delta}{\delta r}(r\frac{\delta T}{\delta r}) + \frac{1}{r}\frac{\delta ^2T}{\delta \theta^2}$$ $$\frac{\delta r}{\delta r}(\frac{\delta T}{\delta r}) + r\frac{\delta ^2T}{\delta r^2} + \frac{1}{r}\frac{\delta ^2T}{\delta \theta^2}$$
I'm confused in doing it the other way around maybe because something went wrong in the process from Q to the laplace, where I'm pretty much basing my derivation. Methinks the biggest problem for me is how I would explain to them in introducing the factor $-kd\theta drL$ to the equation if I were to do it the other way around.