微积分 Calculus MAT00001C

This is
$$\left[\left[\begin{array}{c} R \circ g \ -Q \circ g \ P \circ g \end{array}\right] \cdot\left[\left[\frac{\partial g}{\partial u}\right] \times\left[\frac{\partial g}{\partial v}\right]\right]\right] d u \wedge d v$$
(The permutation of the $P, Q, R$ (and the minus sign) come from the way the $d x \wedge d y$ acts on a piece of surface normal to the $(d) z$ direction.)

We can rewrite this as
$$\left|\frac{\partial g}{\partial u} \times \frac{\partial g}{\partial v}\right|\left[\left[\begin{array}{c} R \circ g \ -Q \circ g \ P \circ g \end{array}\right] \cdot \hat{\boldsymbol{n}}[u, v]\right] \quad d u \wedge d v$$
where $\hat{\boldsymbol{n}}[u, v]$ is the unit normal to the surface at $g\left[\begin{array}{l}u \ v\end{array}\right]$, and
$$\left|\frac{\partial g}{\partial u} \times \frac{\partial g}{\partial v}\right|$$
is the “area stretching factor”.
We have that
$$\int_{g\left(I^{2}\right)} \omega$$

BMAT00001C COURSE NOTES ：

is the limit of the sums of values of $\omega$ on small elements of the surface $g\left(I^{2}\right)$. Suppose $g$ takes a rectangle $\triangle u \times \triangle v$ in $I^{2}$ to a (small) piece of the surface. $\omega$ at $g\left[\begin{array}{l}u \ v\end{array}\right]$ is, say,
$$P d x \wedge d y+Q d x \wedge d z+R d y \wedge d z$$
and the unit normal to the surface is $\hat{\boldsymbol{n}}[u, v]$ (located at $\left.g\left[\begin{array}{l}u \ v\end{array}\right]\right)$.
Write $\hat{\boldsymbol{n}}[u, v]$ as
$$\left[\begin{array}{l} \hat{n} x \ \hat{n} y \ \hat{n} z \end{array}\right]$$