0
$\begingroup$

The volume of a hypercylinder in d-dimensions can be derived in a general way using the Cartesian product.

What is the volume of a hyper cylinder in d - dimension?

I want to find the surface area of a hypercylinder, is there an analogous way to calculate this?

$\endgroup$
1
$\begingroup$

The volume of a $d$-dimensional hyper-cylinder or radius $r$ and length $h$ is the product of the volume of a $d-1$-dimensional sphere times $h$:

$$V = \frac{r^{d-1} \pi^{(d-1)/2}}{\Gamma \left( \frac{d-1}{2}+1 \right)}h.$$

The AREA of such a cylinder is twice the surface area of an $d-1$-dimensional sphere (the "caps") of radius $r$, plus the area of a the perimeter of the $d-1$-dimensional sphere times $h$.

$\endgroup$
3
  • $\begingroup$ Agreed! But is there a way to get the surface area of that shape? $\endgroup$ – user2944352 May 27 '19 at 20:17
  • $\begingroup$ Oh... my mis-reading. You want SURFACE Area... I'll amend my solution in a few minutes. $\endgroup$ – David G. Stork May 27 '19 at 20:35
  • $\begingroup$ Thanks, perfectly clear. $\endgroup$ – user2944352 May 27 '19 at 20:49
0
$\begingroup$

The analysis above is perfectly correct.
There is an ambiguity in the question regarding surface area with regards a 4-dimensional object. This is equivalent to asking the circumference of a 3-dimensional object. There is no single answer. For example, the circumference of a cylinder depends both on its dimensions and where this circumference is taken. It is the locus of points of the intersection of a plane with with the cylinder and has a range of values. Even a sphere has a range of circumferences associated with is but we usually, unambiguously associate the question with the great circle of maximum circumference with it. Unfortunately, we cannot do this with any other shape. In all dimensions only the N-volume and N-surface area have unambiguous meanings. This is probably why no additional formula was forthcoming for this question as the first was a complete as it could be.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.