Heat Flow Through a Surface Example

For a material with coefficient of thermal conductivity  
  with a temperature distribution  
  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 
\[T=z^2 +6x\]
\[\mathbf{\nabla} T = 6 \mathbf{i} - 2z \mathbf{k}\]

Take the surface  
  to be the cuboid shown below.

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

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

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

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

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