Surface area changes Take an arbitrary shape in space. Double its length but halve its width. Does the total surface area stay the same?
Intuitively, this seems right to me. Obviously the comment above holds for a rectangle. And the total area of the arbitrary shape can be thought of as an infinite sum of rectangles. 
But my comment above is only an intuitive approach. I am looking for something more rigorous. Any help appreciated.
 A: In case of a rectangular solid in three dimensions with dimensions $l,w,h$ the surface area is $$2lw + 2lh + 2 hw.$$ Doubling the length but halving the width gives you surface area $$2lw + 4lh + hw.$$ The result is obvious, but obvious that these are in general not the same.
A: An ellipse with constant product of its axes having an Astroid envelope satisfies the condition for same extension/shrink factor = k.
$$ \frac{A}{\pi} = r\cdot r = {r/k}\cdot{rk} =   {a}\cdot{b} $$
The same happens for a central ellipse of  arbitrarily positioned major axis but placed between its two touching orthogonal parallel line sets making up its "length" and "height, constrained to pass through points $ (x,y)= (\pm 1, 
\pm 1)$

A: If you have a planar shape (i.e., a shape in $\mathbb{R}^2$) your intuition is correct.  We can use calculus to prove this.  Suppose that \begin{equation} T: \mathbb{R}^2 \to \mathbb{R}^2 \end{equation}
is the mapping that stretches by a factor of $\alpha\in\mathbb{R}\setminus 0$ in one direction and by $\frac{1}{\alpha}$ in a perpendicular direction.  For simplicity's sake, let's take $T$ to be the mapping $T(x, y) = (\alpha x, \frac{1}{\alpha}y)$.   Then the Jacobian of $T$ is the matrix \begin{equation}J(T) =\begin{pmatrix}\alpha & 0 \\ 0 & \frac{1}{\alpha}\end{pmatrix},
\end{equation}
and we note that $\det(J(T)) = 1$.  Let $R\subset\mathbb{R}^2$ be the region enclosed by your planar shape, and let $R^* = T(R)$, or in other words, let $R^*$ be the 'stretched out' version of your region $R$.  Then
\begin{eqnarray}
Area(R^*) &=& \iint_{R^*} dA \\
&=& \iint_{R} |\det(J(T))|dA \\
&=& \iint_R dA \\
&=&Area(R),
\end{eqnarray}
which proves that the two shapes have equal area.
