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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.

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    is space = $\mathbb R^2$? What is length or width in $\mathbb R^n$? – 2017-02-24
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    What is the length or width of a star in $\mathbb R^2$? – 2017-02-24
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    This is not a question for $\mathbb R^n$. I'm not expected to think things through further than $\mathbb R^3$. – 2017-02-24
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    I think what the question I've been asked to think about is trying to say is that the shape is being stretched by a factor of 2 in one direction and by a factor of 1/2 in an orthogonal direction. – 2017-02-24

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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.

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    I think this answers my question. I cannot say the area will stay the same. Thanks. – 2017-02-24
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    I am curious though, why is my intuition wrong? – 2017-02-24
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    As far as intuition goes, if you are considering changing two dimensions separately you should think of cases where one change will be drastic and the other not so much. Think about a rectangle in the plane whose length is 1 kilometer and whose width is 1 centimeter. Halving the width doesn't change the perimeter very much. Doubling the length on the other hand... – 2017-02-24
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    That was how I first thought about the question but I don't think the reasoning holds for areas? – 2017-02-24
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    Of course it does. Instead of a rectangle consider a box whose cross section is $1$ cm$^2$ and whose length is $1000$m. – 2017-02-24
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    Of course, that's absolutely right! Thanks. – 2017-02-24
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    When you say a "shape" in space that is not well defined. If you mean a solid, no, the surface area will change. If you men a two-dimensional bounded surface, for example surface of a hemisphere or of a cone between two levels or of a function z = f(x,y) given certain bounds, and then you dilate by a factor of 2 in one direction and contract by a factor of 2 in the orthogonal direction, given no part of the surface is perpendicular to either direction, *then* the area should stay the same. – 2017-02-24
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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)$

ElliSEM

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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.