## Heat Flow Out of a Cylindrical Surface

The rate of heat flow across a surface
$S$
drawn in a material is
$\int_S \mathbf{\nabla} \cdot \mathbf{n} dS$
. Suppose we have a temperature distribution
$T(x,y,z)=x^2 +y^2$
throughout a cylinder
$x^2 +y^2 =1, -1 \leq z \leq 1$

The total heat flow is the sum of the heat flows through the top, curved surface and bottom.
\begin{aligned} \int_S (\mathbf{\nabla} T) \cdot \mathbf{n} dS &= \int_{S_{TOP}} (\mathbf{\nabla} T) \cdot \mathbf{n} dS_{TOP} \\ &+ \int_{S_{CURVED SURFACE}} (\mathbf{\nabla} T) \cdot \mathbf{n} dS_{CURVED SURFACE} + \int_{S_{BOTTOM}} (\mathbf{\nabla} T) \cdot \mathbf{n} dS_{BOTTOM} \end{aligned}

$T(x,y,z)=x^2 +y^2 \rightarrow (\mathbf{\nabla} T) = 2x \mathbf{i} +2y \mathbf{j}$

Fro the top and bottom surfaces
$\mathbf{n} = pm \mathbf{k}$
respectively, then
$(\mathbf{\nabla} T) \cdot \mathbf{n} = (2x \mathbf{i} _ 2y \mathbf{j} ) \cdot pm \mathbf{k}=0$
and the only contribution to the integral is from the curved surface. For the curved surface
$\mathbf{n} = x \mathbf{i} + y \mathbf{j}$
(note
$|\mathbf{n}| =x^2 +y^2=1$

\begin{aligned} \int_S (\mathbf{\nabla} T) \cdot \mathbf{n} dS &= \int_{S_{CURVED SURFACE}} (\mathbf{\nabla} T) \cdot \mathbf{n} dS_{CURVED SURFACE} \\ &= \int_{S_{CURVED SURFACE}} (2x \mathbf{i} + 2y \mathbf{j}) \cdot (x \mathbf{i} + y \mathbf{j}) dS_{CURVED SURFACE} \\ &= \int_{S_{CURVED SURFACE}} 2x^2 + 2y^2 dS_{CURVED SURFACE} \end{aligned}

Now change to cylindrical polar coordinates. The integral becomes
\begin{aligned} \int^1_{-1} \int^{2 \pi}_0 2r^2 r d \theta dz &= 4 \pi 1^3 \int^1_{-1} dz \\ &= 2 \pi r^3 \times 2 \\ &= 8 \pi\end{aligned}