Relation between surface area and volume; and perimeter and area. [duplicate]

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"

marked as duplicate by Andrew D. Hwang, Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 14 '17 at 18:49

• 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 $4aE(e)$ (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. – Harry Weasley Jul 14 '17 at 17:36
• Yes, the ellipsoid surface area is even worse: en.wikipedia.org/wiki/Ellipsoid#Surface_area – Chappers Jul 14 '17 at 17:38

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.