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Suppose we have a box (parallelepiped) A completely contained within another box B. Is the surface area of A nessecarily less than the surface area of the B?

Edit: note that the sides of A are not nessecarily parallel to the sides of B.

I happen to know that the answer is yes but the only solution I know of is very hand-wavy.

math_lover
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  • Which one is the larger box and which one is the smaller. – user258250 Aug 15 '15 at 05:38
  • Edited for clarity. – math_lover Aug 15 '15 at 05:56
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    Yes, it is true. In fact, this is true for any convex bodies (i.e. bounded closed set which is compact). If $A$ and $B$ are any two convex bodies such that $A \subset B$, then $\verb/Area/(A) \le \verb/Area/(B)$. – achille hui Aug 15 '15 at 06:37
  • Do you know of a proof? – math_lover Aug 15 '15 at 06:38
  • Imagine the smaller box is a balloon. We can then inflate the balloon until it fills the larger box. This operation increases the surface area. – augurar Aug 15 '15 at 06:48
  • Wow that was the same idea I had. Unfortunately this is not rigorous and increasing the volume does not nessecarily increase the surface area. – math_lover Aug 15 '15 at 06:51
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    The word convex is the key here. Clealry you can create a concave set with large surface area but small volume (think of a ball of small radius with lots of thin spikes) within the larger body. – math_lover Aug 15 '15 at 06:59
  • Might there be some way using determinants? – P i Aug 15 '15 at 10:51

2 Answers2

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One form of Crofton's formula states that the area of a surface in $\mathbb R^3$ is proportional to the integral, over all planes, of the length of the intersection between the plane and the surface. (Reference: Stereology for Statisticians, equation 4.13)

Now, for any plane, its intersection with the inner box is a convex curve which is fully inside the intersection with the outer box. By another application of Crofton's formula in $\mathbb R^2$, the inner curve is shorter. So the inner box's integral is smaller. And so the inner box's area is smaller!

As achille hui noted in the comments, we don't even need the fact that the bodies are boxes. We just need them to be convex.

Chris Culter
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  • Beautiful. This is a fantastic idea, although highly unintuitive. – math_lover Aug 15 '15 at 07:09
  • Well, that's integral geometry! The "integral" part is the insight that you can measure a thing by adding up small pieces. The "geometry" part is the insight that the small pieces can be lines and planes. – Chris Culter Aug 15 '15 at 07:20
  • Come to think of it, equation 4.17 gives an even shorter proof: We can probe the surface with lines and count the number of intersections. That might be easier to grok. – Chris Culter Aug 15 '15 at 07:25
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Oops, I have given a wrong proof before. Here is a corrected version:

Let $K$ be any convex body in $\mathbb{R}^3$, i.e. bounded convex subset of $\mathbb{R}^3$ which is convex (with non-empty interior). Let $u$ be any direction represented as a unit vector in $S^2$. Consider the orthogonal projection of $K$ onto a plane with normal vector $u$ and let $f(K,u)$ be the area of the projected image.

Cauchy surface area formula states that the average of $f(K,u)$ over $u$ is equal to $\frac14$ of the surface area of $K$. If we parametrize $u$ by spherical polar coordinates $$[0,\pi] \times [0,2\pi] \ni (\theta,\phi) \quad\mapsto\quad u = (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) \in S^2$$ This is equivalent to the integral identity: $$\frac{1}{4\pi}\int_0^{2\pi}\int_0^{\pi} f(K,u(\theta,\phi)) \sin\theta d\theta d\phi = \frac14 \verb/Area/(K)$$

When $K$ is a box, it is easy to verify this yourself. After a little algebra, everything comes down to evaluation of a simple integral: $$ \frac{1}{4\pi}\int_0^{2\pi}\int_0^{\pi} \max(\cos\theta,0)\sin\theta d\theta d\phi = \frac14$$

Back to our problem, if we are given two boxes $A, B$ such that $A \subset B$, then for all unit vectors $u$, we have $f(A,u) \le f(B,u)$. Taking averages over $u$ immediately give us $\verb/Area/(A) \le \verb/Area/(B)$.

achille hui
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