I'm studying Do Carmo's differential geometry.
Let $Q$ be a compact region in $\mathbb{R^2}$ that is contained in a coordinate nbd $X : U \rightarrow S$, where $S$ is regular surface. I want to get that the integral \begin{equation} \int\int_Q |X_u \wedge X_v|dudv \end{equation} does not depend on the parametrization $X$.
Next is what the author demonstrate.
Let $\bar{X} : \bar{U} \rightarrow S$ be another parametrization with $X(Q) \subset \bar{X}(\bar{U})$ and set $\bar{Q} = \bar{X}^{-1}(X(Q))$. Let $\frac{\partial(u,v)}{\partial(\bar{u},\bar{v})}$ be the Jacobian of the change of parameters $h=X^{-1}• \bar{X}$. Then \begin{equation} \int\int_{\bar{Q}} |\bar{X}_{\bar{u}} \wedge \bar{X}_{\bar{v}}|d\bar{u}d\bar{v} \\ =\int\int_{\bar{Q}} |X_u \wedge X_v| | \frac{\partial(u,v)}{\partial(\bar{u},\bar{v})} | d\bar{u}d\bar{v} \\ =\int\int_Q |X_u \wedge X_v|dudv \end{equation}
I can't catch the first equality. Please help. I'm so confused.
You should carefully think about change of coordinates. The partial derivatives of $\bar{X}$ and $X$ are related via the change of parameters diffeomorphism. We have to first assume that the compact $Q\subset U$ lies entirely within the intersection $U\cap X^{-1}(\bar{X}(\bar{U}))$, or more visually, that the compact set $X^{-1}(Q)$ lies in the common image of both parametrizations.
Then, I will avoid using explicit coordinates. Let $X_1,X_2:U\rightarrow \mathbb{R}^3$ be the partial derivatives of the first parametrization, and let $\bar{X}_1,\bar{X}_2:\bar{U}\rightarrow \mathbb{R}^3$ be the partial derivatives of the other parametrization. Recall that by the very definition of regular surface, the map
$$X^{-1}\circ \bar{X}:\underbrace{\bar{U}\cap \bar{X}^{-1}(U)}_{\bar{G}}\rightarrow \underbrace{U\cap\bar{X}^{-1}(U)}_G$$
is a diffeomorphism between open sets of $\mathbb{R}^2$. The chain rule tells us therefore that $\bar{X}|_{\bar{G}} = X|_{G} \circ (X^{-1}\circ \bar{X})$ has partial derivatives (dropping the restriction $|_G$ to the corresponding open subsets)
$$ \bar{X}_1 = X_1 \cdot \partial_1(X^{-1}\circ \bar{X})^1 + X_2 \cdot \partial_1 (X^{-1}\circ \bar{X})^2 $$ $$ \bar{X}_2 = X_1 \cdot \partial_2(X^{-1}\circ \bar{X})^1 + X_2 \cdot \partial_2 (X^{-1}\circ \bar{X})^2 $$
Here $\partial_i (X^{-1}\circ \bar{X})^j$ is the function that denotes the i-th partial derivative of the j-th component of the change of coordinates map. Now we compute the vector product of the two partial derivatives above expanding from the right-hand-side, taking into account that $X_1\times X_1 = 0 = X_2\times X_2$ and $X_2\times X_1 = - X_1\times X_2$: $$ \bar{X}_1\times \bar{X}_2=\left(\partial_1(X^{-1}\circ \bar{X})^1 \cdot \partial_2(X^{-1}\circ \bar{X})^2 - \partial_2(X^{-1}\circ \bar{X})^1 \cdot \partial_2 (X^{-1}\circ \bar{X})^1\right)X_1\times X_2 = \det (D(X^{-1}\circ \bar{X})) \cdot X_1\times X_2 $$ where we have carefully identified the determinant of the Jacobian of the change of coordinates map. Then
$$ \|\bar{X}_1\times \bar{X}_2\| = |\det(D(X^{-1}\circ \bar{X}))| \cdot \|X_1\times X_2\| \circ (X^{-1}\circ \bar{X})\qquad (\star) $$
I am here being pedantic ad nauseam writting explicitly every composition. Any person already accustomed to change of variables would simply write
$$ \|\bar{X}_1\times \bar{X}_2\| = |\det(D(X^{-1}\circ \bar{X}))| \cdot \|X_1\times X_2\| $$
Thus we get from the change of variables theorem for $\psi = X^{-1}\circ \bar{X}$, $Q=\psi(\bar{Q})$
$$\int_{Q=\psi(\bar{Q})} \|{X}_1 \times {X}_2\| d{u}d{v} = \int_{\bar{Q}} \|X_1\times X_2\| \circ (X^{-1}\circ \bar{X}) \cdot |\det(D\psi)|d\bar{u}d\bar{v} \overset{\star}{=} $$ $$ = \int_{\bar{Q}} \frac{1}{|\det(D\psi)|} \|\bar{X}_1\times \bar{X}_2\| |\det(D\psi)| d\bar{u}d\bar{v} =\int_{\bar{Q}} \|\bar{X}_1\times \bar{X}_2\| d\bar{u}d\bar{v} $$