Proving that the area of a region of a regular surface is independent on parameterization This is the definition of area of a region of a regular surface given by Do Carmo’s ‘Differential Geometry of Curves & Surfaces’:
Let $R\subset S$ be a bounded region of a regular surface contained in the coordinate neighborhood of the parameterization $\mathbf{x}:U \subset \mathbb{R}^2 \rightarrow S$. The positive number
$$\iint_Q |\mathbf{x}_u\times \mathbf{x}_v|\,du \,dv=A(R),\quad Q=\mathbf{x}^{-1}(R)$$
is called the area of $R$.
I’d like to know how to show that this definition is independent of parameterization, i. e., that if $\mathbf{y}:V \rightarrow S$ is another parameterization such that $R=\mathbf{y}(\bar{Q})$, then
$$\iint_Q |\mathbf{x}_u\times \mathbf{x}_v|\,du \,dv= \iint_{\bar{Q}} |\mathbf{y}_{\bar{u}}\times \mathbf{y}_{\bar{v}}|\, d\bar{u} \,d\bar{v}. $$
How can this be done? Do I have to do a change of variables in the integral?
 A: Yes, this can be done using a change of variables.
Sketch
Set $h=\mathbf{x}^{-1}\circ \mathbf{y}:\overline{Q}\to Q$ so that $h(\overline u,\overline v)=\big( u(\overline u,\overline v), v(\overline u,\overline v) \big)$. Then, by change of variables,
$$
\iint_Q |\mathbf{x}_u\land \mathbf{x}_v|\,du\,dv = \iint_\overline Q |\mathbf{x}_u\land \mathbf{x}_v| \cdot \left|\frac{\partial (u,v)}{\partial(\overline u,\overline v)}\right|\, d\overline{u}\,d\overline{v}.
$$
Since $\mathbf{y}=\mathbf{x}\circ h$, by the chain rule we will get
$$
\frac{\partial \mathbf{y}}{\partial \overline u} = \frac{\partial h_u}{\partial \overline{u}}\cdot \frac{\partial \mathbf{x}}{\partial u} + \frac{\partial h_v}{\partial \overline{u}}\cdot \frac{\partial \mathbf{x}}{\partial v}
$$
(and analogously for $\mathbf{y}_\overline v$). Hence
$$
\begin{aligned}
\mathbf{y}_\overline u \land \mathbf{y}_\overline v &= \left( \tfrac{\partial u}{\partial\overline u} \cdot \mathbf{x}_u + \tfrac{\partial v}{\partial\overline u} \cdot \mathbf{x}_v \right) \land \left( \tfrac{\partial u}{\partial\overline v} \cdot \mathbf{x}_u + \tfrac{\partial v}{\partial\overline v} \cdot \mathbf{x}_v \right) \\
&= \left( \tfrac{\partial u}{\partial\overline u}\cdot \tfrac{\partial v}{\partial\overline v} - \tfrac{\partial u}{\partial\overline v}\cdot \tfrac{\partial v}{\partial\overline u} \right) \mathbf{x}_u \land \mathbf{x}_v,
\end{aligned}
$$
and the desired equality follows by taking the norm on both sides.
Details
I will change your notation a bit to avoid causing conflicts later. I will let $\alpha:U\to S$ and $\beta:V\to S$ be parametrizations with components
$$
\begin{aligned}
\alpha(u,v) & =\big( \alpha_x(u,v),\alpha_y(u,v),\alpha_z(u,v) \big) \\
\beta(s,t) &= \big( \beta_x(s,t), \beta_y(s,t),\beta_z(s,t) \big),
\end{aligned}
$$
where $(u,v)$ and $(s,t)$ denote general points in $U$ and $V$ respectively; set $A=\alpha^{-1}(R)$ and $B=\beta^{-1}(R)$. Just like on page 100 of the book you cited, we get a (bijective and differentiable) change of parameters $h=\alpha^{-1}\circ\beta:B\to A$. Note that we can write $h$ in terms of its components $h(s,t)=\big( h_u(s,t),h_v(s,t) \big)$; this will be used in writing down the Jacobian matrix of $h$ and in using the chain rule later on.
If $f:A\to\mathbb{R}$ is defined by $f(p)=|\alpha_u(p)\land \alpha_v(p)|$, then by the formula for change of variables (equation (7) on page 45) it follows that
$$
\iint_A f(u,v)\,du\,dv = \iint_B f\big( h(s,t) \big) \cdot \left| \det Dh{(s,t)} \right| \, ds \, dt,
$$
where
$$
Dh(q)=\begin{pmatrix} \displaystyle\frac{\partial h_u}{\partial s}(q) & \displaystyle\frac{\partial h_u}{\partial t}(q) \\ \displaystyle\frac{\partial h_v}{\partial s}(q) & \displaystyle\frac{\partial h_v}{\partial t}(q) \end{pmatrix}.
$$
Let $q\in B$ be an arbitrary vector and set $p=h(q)$, then
$$
f\big( h(q) \big)\cdot\left| \det Dh(q) \right|=|\alpha_u(p)\land \alpha_v(p)| \cdot |\det Dh(q)|.
$$
It will suffice to show that the right-hand side is equal to $|\beta_s(q)\land \beta_t(q)|$ for all $q$.
Because $\alpha\circ h$ and $\beta$ are equal, their components $\alpha_i\circ g=\beta_i$ will all be equal for $i\in\{x,y,z\}$. Each component is a real-valued function on $B$, so we can apply the chain rule of page 127 to find that for each $i$,
$$
\frac{\partial \beta_i}{\partial s}(q) = \frac{\partial \alpha_i}{\partial u}(p) \cdot \frac{\partial h_u}{\partial s}(q) + \frac{\partial \alpha_i}{\partial v}(p) \cdot \frac{\partial h_v}{\partial s}(q).
$$
Recall that $\partial \beta / \partial s = \big( \partial \beta_x/\partial s, \partial \beta_y/\partial s, \partial \beta_z/\partial s  \big)$ and likewise for $\partial \alpha/\partial s$, which means that we can combine the three equations into the vector equality
$$
\frac{\partial \beta}{\partial s}(q) = \frac{\partial h_u}{\partial s}(q)\cdot \frac{\partial \alpha}{\partial u}(p) + \frac{\partial h_v}{\partial s}(q)\cdot \frac{\partial \alpha}{\partial v}(p).
$$
A completely analogous result holds for $\partial\beta/\partial t$. Using this, we can calculate
$$
\begin{aligned}
\beta_s(q) \land \beta_t(q) &= \left( \frac{\partial h_u}{\partial s}(q)\cdot \frac{\partial\alpha}{\partial u}(p) + \frac{\partial h_v}{\partial s}(q)\cdot \frac{\partial\alpha}{\partial v}(p) \right) \land \left( \frac{\partial h_u}{\partial t}(q)\cdot \frac{\partial\alpha}{\partial u}(p) + \frac{\partial h_v}{\partial t}(q)\cdot \frac{\partial\alpha}{\partial v}(p) \right) \\
&= \left(\frac{\partial h_u}{\partial s}(q) \frac{\partial h_v}{\partial t}(q) - \frac{\partial h_u}{\partial t}(q) \frac{\partial h_u}{\partial t}(q) \right) \cdot \frac{\partial \alpha}{\partial u}(p)\land \frac{\partial \alpha}{\partial v}(p) \\
&= \left. \det\begin{pmatrix}
\displaystyle\frac{\partial h_u}{\partial s}(q) & \displaystyle\frac{\partial h_u}{\partial t}(q) \\
\displaystyle\frac{\partial h_v}{\partial s}(q) & \displaystyle\frac{\partial h_v}{\partial t}(q)
\end{pmatrix} \right. \cdot \alpha_u(p)\land \alpha_v(p) \\
&= \det Dh(q) \cdot \alpha_u(p)\land \alpha_v(p).
\end{aligned}
$$
Using the definition in the book, in general, $|\lambda u\land v|^2=(\lambda u\land v) \cdot (\lambda u\land v) = \lambda^2 |u\land v|^2$ so $|\lambda\cdot u\land v|=\sqrt{\lambda^2}\cdot |u\land v|=|\lambda|\cdot|u\land v|$. In this case, then, we get $|\beta_s(q)\land\beta_t(q)| = |\det Dh(q)|\cdot |\alpha_u(p)\land\alpha_v(p)|$.
Using this, we can finally conclude that
$$
\iint_A |\alpha_u\land \alpha_v|\,du\,dv = \iint_B \big(|\alpha_u\land\alpha_v|\circ h\big)\cdot |\det Dh|\, ds\,dt = \iint_B |\beta_s\land \beta_t|\,ds\,dt,
$$
so the area of $R\subset S$ is invariant under a change of parametrization.
