## Flux Out of a Plane Surface Example

The flux out of a surface is
$\int_S \mathbf{F} \cdot \mathbf{S} = \int_S \mathbf{F} \cdot \mathbf{n} dS$

For a plane surface
$\mathbf{n}$
is a constant vector and this can make the calculation quite simple. Consider the flux of the vector field
$\mathbf{F} = xy \mathbf{i} + y^2 \mathbf{j} + yz \mathbf{k}$
out of the triangular plane surface with vertices
$A(1,0,0),B(0,2,0),C(0,0,3)$
.

We can take a unit normal
$\mathbf{n}$
to the plane as
\begin{aligned} \mathbf{n} &= \frac{(\mathbf{OB} - \mathbf{OA}) \times (\mathbf{OC} - \mathbf{OA})}{|(\mathbf{OB} - \mathbf{OA}) \times (\mathbf{OC} - \mathbf{OA})|} \\ &= \frac{\left\{ \begin{pmatrix}-1\\2\\0\end{pmatrix} , \begin{pmatrix}-1\\0\\3\end{pmatrix} \right\}}{\left| \left\{ \begin{pmatrix}-1\\2\\0\end{pmatrix} , \begin{pmatrix}-1\\0\\3\end{pmatrix} \right\} \right|} = \frac{\begin{pmatrix}6\\3\\2\end{pmatrix}}{\sqrt{49}} = \frac{1}{7} \begin{pmatrix}6\\3\\2\end{pmatrix} \end{aligned}

Thhe equation of the plane is
$6x+2y+3z=d$
. To find
$d$
substitute a point in the plane, say
$A(1,0,0)$
, obtaining
$d=6$
hence
$z= \frac{1}{3} (6- 6x-2y)$
then
\begin{aligned}\int_S \mathbf{F} \cdot \mathbf{n} dS &= \frac{1}{7} \int_S 6xy + 3y^2 + 2yz dS \\ &=\frac{1}{7} \int_S 6xy+3y^2 +y(6-6x-3y) dS \\ & = \frac{1}{7} \int_S 6y dS \\ &= \frac{1}{7} \int^1_0 \int^{2-2x}_0 \frac{6y}{| \mathbf{n} \cdot \mathbf{k}|}dy dx \\ &= \frac{3/7}{2/7} \int^1_ 0 4-8x+4x^2 dx \\ &= \frac{3}{2} [4x-4x^2+\frac{4}{3}x^3]^1_0 \\ &= 2 \end{aligned}