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Let be $ \Phi $ a parametrization of a surface $ \in \mathbb{R}^3 $ and $g$ the metric tensor.

As the title says...how to show that How to prove that $ | \Phi_u \times \Phi_v | = \sqrt{\det g} $ ?

I started like this:

$$| \Phi_u \times \Phi_v | = \left| \begin{pmatrix} \frac{\delta \phi_1}{ \delta u} \\ \frac{\delta \phi_2}{ \delta u} \\ \frac{\delta \phi_2}{ \delta u} \end{pmatrix} \times \begin{pmatrix} \frac{\delta \phi_1}{ \delta v} \\ \frac{\delta \phi_2}{ \delta v} \\ \frac{\delta \phi_2}{ \delta v} \end{pmatrix} \right| = \left| \begin{pmatrix} \frac{\delta \phi_2}{ \delta u} \frac{\delta \phi_3}{ \delta v}- \frac{\delta \phi_3}{ \delta u} \frac{\delta \phi_2}{ \delta v}\\ \frac{\delta \phi_1}{ \delta u} \frac{\delta \phi_3}{ \delta v}- \frac{\delta \phi_3}{ \delta u} \frac{\delta \phi_1}{ \delta v}\\ \frac{\delta \phi_1}{ \delta u} \frac{\delta \phi_2}{ \delta v}- \frac{\delta \phi_2}{ \delta u} \frac{\delta \phi_1}{ \delta v}\end{pmatrix} \right|$$

I am not sure how to proceed..If I continue I am coming to a dead end..hope you can help me out a bit :) !

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Recall Lagrange's identity for cross product: $$\|a \times b\|^2 = \|a\|^2\|b\|^2 - \langle a,b\rangle^2, \qquad a,b \in \mathbb{R}^3$$

Therefore $$\|\Phi_u \times \Phi_v\|^2 = \|\Phi_u\|^2\|\Phi_v\|^2 - \langle \Phi_u, \Phi_v\rangle^2 = \begin{vmatrix} \|\Phi_u\|^2 & \langle \Phi_u, \Phi_v\rangle \\ \langle \Phi_u, \Phi_v\rangle & \|\Phi_v\|^2\end{vmatrix} = \det g$$

where $g = \begin{bmatrix} \langle \Phi_u, \Phi_u\rangle & \langle \Phi_u, \Phi_v\rangle \\ \langle \Phi_u, \Phi_v\rangle & \langle\Phi_v, \Phi_v\rangle\end{bmatrix}$ is the metric tensor.

Since $\det g > 0$ we conclude $\|\Phi_u \times \Phi_v\| = \sqrt{\det g}$.

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Call ${\bf r} = \Phi(u, v)$, and recall the unitary vector along the direction $u$ is

$$ h_u \hat e_u = \frac{\partial {\bf r}}{\partial u} = \Phi_u $$

with $h_u$ the scale factor. You have then

$$ |\Phi_u \times \Phi_v| = |h_u h_v| |\hat{e}_u\times \hat{e}_v| = |h_u h_v| = \sqrt{g} $$

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