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I've noticed the following things while studying calculus, and would like experts to tell me if my conclusions are right.

My Observations

If I represent the area of a circle by $A$ and its perimeter by $C$, I can write $$C=\frac{dA}{dr}$$ Similarly, for a sphere, if I represent volume by $V$ and surface area by $S$, I can write $$S=\frac{dV}{dr}$$ I tried doing the same for other 2D and 3D figures, and saw that it worked only in case of the circle and the sphere​.

My questions​

My questions are:

  • Why does this happen only in the case of the circle and the sphere​?

  • Can't I express the surface area ​of a solid in terms of its volume and the perimeter of a closed figure in terms of its area using calculus? Why or why not? I remember reading something of that kind in Jenny Olive's book "Mathematics: A Self-study Guide"

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Myungjin Hyun
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  • I don't think this is true for an ellipse, since the area is given by the elementary formula $\pi ab$, while the perimeter/arc length needs the elliptic integral of the second kind $4aE(e)$ (https://en.wikipedia.org/wiki/Ellipse#Circumference). – Chappers Jul 14 '17 at 17:26
  • @Chappers , thank you, I was mistaken. I'll edit my question. I assumed $C=π(a+b)$. I guess I'll have to edit the ellipsoid thing, too. – Myungjin Hyun Jul 14 '17 at 17:36
  • Yes, the ellipsoid surface area is even worse: https://en.wikipedia.org/wiki/Ellipsoid#Surface_area – Chappers Jul 14 '17 at 17:38

2 Answers2

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This observation holds for all balls, cubes, and simplexes provided they're centered at the origin.

If we look at a square set in $\Bbb{R}^2$ Euclidean space centered at the origin and having a half-length of $s$, then its area would be $A=(2s)^2=4s^2$ and its perimeter would be $P=\frac{dA}{ds}=8s$.

I'm only an undergraduate, but I can offer this for some intuition: if you have this closure $X\subset\Bbb{R}^n$ and a scaling factor $k>0,\;k\in\Bbb{R}$, then you could get this image $kX$ of $X$ through $x\mapsto kx$ which is this uniform scaling of all points $x\in{X}$ about the origin such that the volume $\operatorname{V}_n(kX)=k^n\operatorname{V}_nX$. So, if you let $k=1+h$ and $h\to0$, then there's going to be this uniform shell of thickness $h$ all around the shape you're looking at (ignoring edge behavior which turns out to not matter anyway because $h\to0$ ). The volume of this shell is going to be $h$ multiplied by the surface area which would be $(n-1)-$dimensional.

This article could probably talk more in depth about it than I can: https://www.math.byu.edu/~mdorff/docs/DorffPaper07.pdf

user12386
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  • Thanks, @JellyHops , your answer was useful! I'll think deeper over the matter, too! I'm in my senior year, by the way, and am just curious about calculus among other things. – Myungjin Hyun Jul 15 '17 at 17:26
  • I guess I didn't look deep enough into this matter, but your answer helped me get on the track! Thanks again, @JellyHops ! – Myungjin Hyun Jul 15 '17 at 17:30
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Think about it: what is the first differential of an area of a circle? It's basically how much area will increase when you increase a radius. If we increase a radius, where is the additional area? It's a ring outside of the original circle. As we reduce that ring, the additional area becomes equal to the circumference. Same logic applies to sphere and we can prove it algebraically using limits. Circle and sphere expand equally (2D and 3D) in all directions when the radius expands.

Vasili
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