Element of Surface Area in Terms of Intrinsic Coordinates
Suppose we have a surface defined in terms of intrinsic coordinates\[u,v\]
, \[\mathbf{r} =\mathbf{r}(u,v)\]
.An element of surface area is
\[\begin{equation} \begin{aligned} dS &= \left| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \right| du dv \\ &= \sqrt{(\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}) \cdot (\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v})} du dv \end{aligned} \end{equation}\]
\[(\mathbf{a} \times \mathbf{a}) \cdot (\mathbf{b} \times \mathbf{b}) = (\mathbf{a} \cdot \mathbf{a})(\mathbf{a} \cdot \mathbf{a}) - (\mathbf{a} \cdot \mathbf{b})(\mathbf{a} \cdot \mathbf{b}) \]
to obtain
\[\begin{equation} \begin{aligned} dS &= \sqrt{(\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}) \cdot (\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v})} du dv \\ &= \sqrt{(\frac{\partial \mathbf{r}}{\partial u} \cdot \frac{\partial \mathbf{r}}{\partial u}) (\frac{\partial \mathbf{r}}{\partial v} \cdot \frac{\partial \mathbf{r}}{\partial v}) -(\frac{\partial \mathbf{r}}{\partial u} \cdot \frac{\partial \mathbf{r}}{\partial v}) (\frac{\partial \mathbf{r}}{\partial u} \cdot \frac{\partial \mathbf{r}}{\partial v})} dudv \end{aligned} \end{equation}\]