TITLE: Is it possible to uniquely identify a shape in $\mathbb{R}^3$ knowing its surface area and volume?
QUESTION [7 upvotes]: More precisely, can an orientable closed compact 2-manifold embedded in $\mathbb{R}^3$ be, up to an isometry, uniquely identified by its surface area and the volume it encloses? Conversely, can a counter example be given to disproof this claim?
In the answer to an earlier question related to the isoperimetric inequality a proof for the statement that "sphere is the only closed surface in $\mathbb{R}^3$ that minimizes the surface area to volume ratio" has been outlined. That and the beginning of this post on Brunn-Minkowski inequality on Terence Tao's What's New made me curious about the existence (or lack thereof) more stringent constraints akin to the isoperimetric inequality.
As a first step towards constructing a counterexample, I examined the simple case of rectangular cube having the same surface area and volume as a cube. I ended up with the following equation:
$$ 
(\underbrace{a b c}_{volume})^\frac{2}{3} = \frac{1}{6}\times\underbrace{(2ab+2bc+2ac)}_{\text{surface area}}\,, 
$$
where $a$, $b$, and $c$ are the dimensions of the rectangular cube. (For a cube of the sidelength $d$, $(d^3)^{2/3}=\frac{1}{6}\,6\,d^2$; hence the above equation.) I have no idea how to proceed with the search for any positive definite solutions to the above problem. I suppose it can be simplified by invoking (isotropic) scaling to set one of the dimensions of the rectangular cube to $1$ and the following equation:
$$ 
(a b)^2 = \left(\frac{2}{6}\right)^3(ab+b+a)^3\,. 
$$

REPLY [8 votes]: Your simple example is in a good direction but too simple.  We know the cube is the most efficient parallepiped in terms of volume to surface ratio so you won't get another parallelepiped to match both.  But we should be able to find two parallelepipeds, one a flat pancake and one a long needle that have the same surface and volume.  Let one be $1 \times 1 \times 100$ and the other $a \times a \times b$ with $a \gt b$.  Then we can write two equations and solve them(thanks Alpha)
$$a^2b=100\\2a^2+4ab=402\\2a^2+\frac {400}a=402\\a=\frac 12(3\sqrt{89}-1)\approx 13.651$$