Coordinate transformation of boundary condition Let us suppose a heat transfer problem inside a cylinder of radius $r_a$. If we neglect changes along $z$ and $\theta$ directions, i.e. only a cross section of the cylinder, the problem can be formulated as follow
$$
\rho \; c_p \frac{\partial T}{\partial t} = \frac{1}{r}\frac{\partial}{\partial r}\left( k\; r \frac{\partial T}{\partial r}\right) + q(r)
$$
Considering a convective heat transfer with the surrounding environment at temperature $T_n$ gives us the following boundary condition:
$$
\left. k\frac{\partial T}{\partial r}\right|_{r=r_a} = h\left(T_n - T\right)
$$
So far not a problem at all. Now, I want to solve the same problem, but in cartesian coordinates, it turns into a 2D heat transfer problem, given by:
$$
\rho \; c_p \frac{\partial T}{\partial t} = \frac{\partial}{\partial x}\left(k \frac{\partial T}{\partial x} \right) + \frac{\partial}{\partial y}\left(k \frac{\partial T}{\partial y} \right)+ q(x,y)
$$
My question: What about the boundary condition? How can I transform the given boundary condition to cartesian coordinates?
 A: The temperature gradient expressed in cartesian coordinates is $$\nabla T=\frac{\partial T}{\partial x}\hat{i_x}+\frac{\partial T}{\partial y}\hat{i_y}$$The unit vectors in the x- and y directions are related to the radial- and circumferential unit vectors by:
$$i_x=\frac{x}{\sqrt{x^2+y^2}}\hat{i_r}-\frac{y}{\sqrt{x^2+y^2}}\hat{i_\theta}$$
$$i_y=\frac{x}{\sqrt{x^2+y^2}}\hat{i_\theta}+\frac{y}{\sqrt{x^2+y^2}}\hat{i_r}$$So, $$\nabla T=\left[\frac{x}{\sqrt{x^2+y^2}}\frac{\partial T}{\partial x}+\frac{y}{\sqrt{x^2+y^2}}\frac{\partial T}{\partial y}\right]\hat{i_r}+\left[-\frac{y}{\sqrt{x^2+y^2}}\frac{\partial T}{\partial x}+\frac{x}{\sqrt{x^2+y^2}}\frac{\partial T}{\partial y}\right]\hat{i_\theta}$$
At $\sqrt{x^2+y^2}=r_0$ this reduces to:
$$\nabla T=\left[\frac{x}{r_0}\frac{\partial T}{\partial x}+\frac{y}{r_0}\frac{\partial T}{\partial y}\right]\hat{i_r}+\left[-\frac{y}{r_0}\frac{\partial T}{\partial x}+\frac{x}{r_0}\frac{\partial T}{\partial y}\right]\hat{i_\theta}$$So, we have:
$$\frac{\partial T}{\partial r}=\frac{x}{r_0}\frac{\partial T}{\partial x}+\frac{y}{r_0}\frac{\partial T}{\partial y}$$and$$-\frac{y}{r_0}\frac{\partial T}{\partial x}+\frac{x}{r_0}\frac{\partial T}{\partial y}=0$$The latter represents the condition that, at $\sqrt{x^2+y^2}=r_0$, T is constant.
