Assume that the temperature is circularly symmetric: $u=u(r,t)$, where $r^2= x^2+y^2$. Consider any circular annulus $a\le r\le b$. It can be shown that the total heat energy is $2\pi \int_a^b c\rho u rdr$.
The next step is to show that the flow of heat energy per unit time out of the annulus at $r=b$ is $-2\pi b K_0 \partial u / \partial r\vert_{r=b}$. That's where I'm stuck. To show this, I do the following:
$$\text{flux}=-K_0\nabla u\cdot \hat{n}$$
But $\hat{n}$ is the outward normal to the surface, so $\nabla u\cdot \hat{n}=0$. What am I missing?
The next step is to use the $2\pi \int_a^b c\rho u rdr$ and the $-2\pi b K_0 \partial u / \partial r\vert_{r=b}$ expressions to show that the circularly symmetric heat equation without sources comes to be
$$\frac{\partial u}{\partial t}=\frac{k}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)$$
But this part is clear to me. Would appreciate clarification of the part in the yellow colour.
Since $u$ depends on $r$ and $t$, but not on $\theta$, we see that
$$\nabla u(r,t)=\hat r \frac{\partial u(r,t)}{\partial r}\tag 1$$
Using $(1)$, we see that the flow of heat energy (per unit time) out of the annulus at $r=b$ is given by
$$\begin{align} \oint_{r=b} \color{blue}{\hat n}\cdot \color{red}{(-K\nabla u(r,t))}\,\color{orange}{d\ell}&=\int_0^{2\pi} \underbrace{\color{blue}{\hat r}\cdot \color{red}{\left.\left(-K\hat r \frac{\partial u(r,t)}{\partial r}\right)\right|_{r=b}}}_{\text{Independent of}\,\,\theta}\,\color{orange}{b\,d\theta}\\\\ &=-2\pi b K \left.\left( \frac{\partial u(r,t)}{\partial r}\right)\right|_{r=b} \end{align}$$
as was to be shown!