# What is the shape that have the biggest surface area with hypervolume of 1 in $N$ dimensional space?

There is a theorm saying that from all shapes in $$N$$ dimensional space with hypervolume of 1 (in arbitrary units) $$N$$ dimensional ball is the one with the smallest surface area ($$N$$ dimensional sphere),

What is the shape from all the shapes in $$N$$ dimensional space with hypervolume 1 (in arbitrary units) have the biggest surface area?

There is no upper bound to the (equivalent to) surface area for $$N \geq 2$$. For instance, with $$N = 2$$, consider what happens as we crinkle the surface more and more vigorously:

The first is a polar plot of $$\frac{1}{2\pi}(2+\sin(20 \theta))$$ for $$0 \leq \theta \leq 20$$. Its perimeter is $$14.6\dots$$.

The second is a polar plot of $$\frac{1}{2\pi}(2+\sin(200 \theta))$$ for $$0 \leq \theta \leq 20$$. It's perimeter is $$127.6\dots$$.

The third is a polar plot of $$\frac{1}{2\pi}(2+\sin(2000 \theta))$$ for $$0 \leq \theta \leq 20$$. It's perimeter is $$1273.2\dots$$. The waving back and forth is so rapid that there aren't enough pixels to show the spaces between the "petals" of the curve.

The last is a polar plot of $$\frac{1}{2\pi}(2+\sin(20\,000 \theta))$$ for $$0 \leq \theta \leq 20$$. It's perimeter is $$12\,732.\dots$$. Even though we cannot see any difference, this plot has ten times as many petals as the previous one.

It should be clear we can do the same sort of thing in higher dimensions, so there is no upper bound for (the analog of) surface area in higher dimensions.

• Wow, amazing illustration, thank you very much! Nov 20, 2019 at 17:45
• What if the body was convex? Is there any result? Nov 21, 2019 at 15:14
• @MichaelHoppe : Consider a (hyper-)cylinder (which is convex) of unit (hyper-)volume. As the radius is decreased, the (hyper-)surface area increases without bound. In 2 dimensions, this is a rectangle with perimeter $4r + 2/r$, which increases without bound as $r \rightarrow 0$. Nov 21, 2019 at 15:43
• Of course, excuse my temporary blindness. Nov 21, 2019 at 16:04
• Oops. rectangle with perimeter $2(2r) + 2(1/2r) = 4r+1/r$. Nov 21, 2019 at 16:30

There is no such shape, for any shape with unit hypervolume, we can create another shape with unit hypervolume but more surface area. Let S be the surface area of our candidate shape. Then we can create a shape with surface area at least 2S by constructing an N-hyperprism with two ends composed of hypersquares of area S, edges between the square ends of length $$\frac 1S$$.