## Heat Flow Through a Surface Example

For a material with coefficient of thermal conductivity
$\kappa$
with a temperature distribution
$T(x,y,z)$
the rate of heat flow through a surface drawn in the material is
$H = \int_S - \(\mathbf{\nabla} T) \cdot \mathbf{n} dS$

Suppose that
$T=z^2 +6x$
then
$\mathbf{\nabla} T = 6 \mathbf{i} - 2z \mathbf{k}$

Take the surface
$S$
to be the cuboid shown below.

FThe normals to the front, surface 2 and the back are
$\mathbf{j}$
and
$- \mathbf{j}$
respectively.
$(\mathbf{\nabla} T) \cdot \mathbf{n} =(6 \mathbf{i} - 2z \mathbf{k}) \cdot pm \mathbf{j} =0$

From surface 2 =, the top and the bottom surface, the normals are
$pm \mathbf{k}$
respevtively
$(\mathbf{\nabla} T) \cdot \mathbf{n} =(6 \mathbf{i} - 2z \mathbf{k}) \cdot pm \mathbf{k} =-2z$

On the bottom surface
$z=0$
so only the top surface contributes to the integral.
On the leftmost and rightmost surface 3, the normals are
$+- \mathbf{i}$

$(\mathbf{\nabla} T) \cdot \mathbf{n} =(6 \mathbf{i} - 2z \mathbf{k}) \cdot pm \mathbf{i} =6$

Then
\begin{aligned} H &=\kappa \int_S - (\mathbf{\nabla} T) \cdot \mathbf{n} dS \\ &=\kappa \int_{S_{TOP}} - (\mathbf{\nabla} T) \cdot \mathbf{n} dS_{TOP}+ \int_{S_{FRONT}} - (\mathbf{\nabla} T) \cdot \mathbf{n} dS_{FRONT} + \kappa \int_{S_{BACK}} - (\mathbf{\nabla} T) \cdot \mathbf{n} dS_{BACK} \\ &= -\kappa \int^1_0 \int^1_0 (6 \mathbf{i} + 2z \mathbf{k}) \cdot \mathbf{k} dx dy -\kappa \int^1_0 \int^5_0 (6 \mathbf{i} + 2z \mathbf{k}) \cdot -\mathbf{-} dx dz -\kappa \int^1_0 \int^5_0 (6 \mathbf{i} + 2z \mathbf{k}) \cdot \mathbf{-} dx dz \\ &= -\kappa \int^1_0 \int^1_0 2z dx dy -\kappa \int^1_0 \int^5_0 -6 dx dz -\kappa \int^1_0 \int^5_0 6 dx dz \\ &- -10\kappa \end{aligned}