## Alternative Form for Area of Surface Element for Parametrized Surface

Theorem
For a surface
$S$
defined para metrically with parameters
$u,v$
, we can write
$dS=| \sqrt{(\frac{\partial \mathbf{r}}{\partial u}|^2 \frac{\partial \mathbf{r}}{\partial u}|^2 -\frac{\partial \mathbf{r}}{\partial u} \cdot \frac{\partial \mathbf{r}}{\partial u}}du dv$
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Proof
\begin{aligned} dS &= | \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}| \\ &=|(\frac{\partial x}{\partial u} \mathbf{i} + \frac{\partial y}{\partial u} \mathbf{j} + \frac{\partial z}{\partial u} \mathbf{k}) \times (\frac{\partial x}{\partial v} \mathbf{i} + \frac{\partial y}{\partial v} \mathbf{j} + \frac{\partial z}{\partial v} \mathbf{k})| \\ &= |(\frac{\partial y}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial y}{\partial v} \frac{\partial z}{\partial u} ) \mathbf{i} +(\frac{\partial z}{\partial u} \frac{\partial x}{\partial v} - \frac{\partial z}{\partial v} \frac{\partial x}{\partial u} ) \mathbf{j} +(\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} ) \mathbf{k} | \\ &= \sqrt{(\frac{\partial y}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial y}{\partial v} \frac{\partial z}{\partial u} )^2 + (\frac{\partial z}{\partial u} \frac{\partial x}{\partial v} - \frac{\partial z}{\partial v} \frac{\partial x}{\partial u} ) ^2 + (\frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} )^2 } \\ &= \sqrt{ ((\frac{\partial x}{\partial u})^2 + (\frac{\partial y}{\partial u})^2 + (\frac{\partial z}{\partial u})^2)((\frac{\partial x}{\partial v})^2 + (\frac{\partial y}{\partial v})^2 + (\frac{\partial z}{\partial v})^2) - (\frac{\partial x}{\partial u} \frac{\partial x}{\partial v}+ \frac{\partial y}{\partial u} \frac{\partial y}{\partial v}+ \frac{\partial z}{\partial u} \frac{\partial z}{\partial v})^2 } \\ &= \sqrt{|\frac{\partial \mathbf{r}}{\partial u}|^2 |\frac{\partial \mathbf{r}}{\partial u}|^2 -\frac{\partial \mathbf{r}}{\partial u} \cdot \frac{\partial \mathbf{r}}{\partial u}} \end{aligned}