Smooth parameterized surface/area. 
$S\subseteq R^3$ a smooth parameterized surface.
1.1.:) The area of a surface is invariant under euclidean movement. If $y=Ax+b$ is an euclidean movement in $R^3$ ($A\in R_{n\times n}$ is orthogonal and $b\in R^n$ is arbitrary), show that
$A(S')=A(S)$
for $S':=${$Ax+b|x\in S$}.
1.2.:) For $\lambda \neq 0$ let $\lambda \cdot S:=${$\lambda \cdot x|x\in S$}. Show that $A(S')=\lambda^2A(S)$.

I've found this question while browsing in Math.SE. Here is the original link: Smooth parameterized surface.
We had a similar exercise in class a couple of weeks ago, but I wasn't there while it was discussed (I fell sick). I'm kind of curious on how you would solve this type of exercise. I tried using the following formula I found in my class's lecture script:
$A(S)=\int_I\|F_u(u,v)\times F_v(u,v)\|d^2(u,v)$
But how would I find $F_u(u,v)$ and $F_v(u,v)$. I only got information about $S'$ in this case, which is $Ax+b$. I mean, if I were to take the derivative I would get just $A$, right? But it's not like there two variables in the equation.

$F(u,v)$ denoted the point on the surface $S$ which is mapped to by $(u,v)$ in the parameter space. I usually call $F: D \subseteq \mathbb{R}^2 \rightarrow S \subseteq \mathbb{R}^3$ a parameterization of the surface $S$ or a patch of $S$. To calculate surface area of the subset of $S$ covered by the patch we can integrate $dS = || \partial_u F \times \partial_v F || du \, dv$ over $(u,v) \in D$. I'll assume $S$ is completely covered by the patch ($F(D)=S$) in what follows.
If we consider the Euclidean motion $y=Ax+b$ then the surface $S$ is transported to the new surface $S'$ defined by $y = Ax+b$ for each $x \in S$. Naturally, the patch on $S'$ is given by $G(u,v) = AF(u,v)+b$. Notice,
$$ \partial_u G = \partial_u (AF+b) = A\partial_u F $$
and
$$ \partial_v G = \partial_v (AF+b) = A\partial_v F $$
as $A$ is constant in $u,v$. (it has nothing to do with the Euclidean motion)
Finally, to calculate the surface area element of $S'$ we face:
$$ \partial_u G \times \partial_v G = (A\partial_u F) \times (A\partial_v F) $$
the product of $A$ and $\partial_u F$ and $\partial_v F$ is a matrix-column vector product. However, we know $det(A) = \pm 1$ since $A^TA=I$. Calculate,
\begin{align} || (Av) \times (Aw) ||^2 &= ||Av||^2||Aw||^2-[(Av) \cdot (Aw)]^2 \\
&= ||v||^2||w||^2-(v \cdot w)^2 \\
&= || v \times w||^2
\end{align}
using that $A$ is an orthogonal matrix. Apply the identity above to $\partial_u F$ and $\partial_v F$ and we obtain that
$$ \partial_u G \times \partial_v G = \partial_u F \times \partial_v F $$
and as the domain of the patch for the transformed surface is still $D$ we integrate to find that $S$ and $S'$ share the same surface area.
