Mathematiques d'Escalier

[jack]

Many people have observed that, given the volume of a sphere radius r is V(r)=(4/3).pi.r³ and its surface area is S(r)=4.pi.r², that, very conveniently:

dV(r)/dr = 3.(4/3).pi.r² = 4.pi.r² = S(r)

In fact, I use it to remember the surface area, since I only usually remember the volume. It seems to make sense sort of, but many people are not quite sure why.

Today someone pointed out, that the same thing works for a cube, if you take the side length as the diameter and half the side length as the radius.

V = (2r)³
S = 6.(2r)²
dV/dr = 2.3.(2r)² = 6.(2r)² = S

In fact, if you do the trigonometry, it turns out the same thing works for the tetrahedron and other platonic solids, although if you take the side length as d, r is no longer half d.

In fact, if you take d = ar for some unknown constant a, so long as the volume is something times r cubed, and the surface area is something times r squared, there's always some value of a that makes the derivative dV/dr = S exact.

It may or may not be immediately obvious what that value is, but in fact, it's the distance from the centre to the middle of one of the faces aka the radius of the largest sphere which fits inside.

At this point we were puzzled why, and it wasn't until I was at home that I saw the obvious way of thinking about it.

What does dV/dr mean? It means [V(r+δr)-V(r)]/δr (as δr->0).

That is, "imagine a solid with a slightly larger r, and subtract the original r" and ask what's left. What's left is thin slab covering each face, plus some neglible rods at each edge which have an extra factor of δr in so effectively vanish to zero. What's the volume of all those thin slabs? The areas of the faces, times the width of the slab. What's the width? The distance from one side to the other perpendicular to the face, ie. parallel with a line through the centre only at the centre of the face, ie. it is δr, so the volume of the slab is S.δr, and [V(r+δr)-V(r)]/δr is approximately S.

The same diagram apparently works for a sphere, if you imagine δV to be a thin shell covering the original sphere. What's the volume of the shell? Well, it's approximately "surface area times width", but the inner or outer surface area? Well, one's too small and one's too big, so the goldilocks answer is somewhere between S(r).δr <= δV <= S(r+δr).δr. But all the extra terms in the second one all have δr^2 in so they all vanish, and δV ~ S(r).δr.

I think I've probably seen that before but forgotten.

Comments

[simont]

So in fact this ought to work for any polyhedron whose faces all have the same perpendicular distance from the centre point. (That is, there exists some point which we can declare to be the centre point such that this is the case.) For instance, any of the irregular solids I constructed by taking a bunch of tangent planes to the same sphere (though not necessarily the ones constructed by taking the convex hull of some points on the sphere). Because if you expand a solid by some multiplicative factor (1+ε) keeping the centre point fixed, then the perp. distance d from the centre to the face also changes by a factor of (1+ε), and so the face moves by a distance dε. And if that distance is the same for all faces, then subtracting the smaller solid from the larger will give you a shell with the same thickness on all faces, and hence the surface area will be the derivative of the volume, up to a constant. But if the faces aren't all at the same perpendicular distance, then differentiating the volume will instead give you a sort of "weighted surface area" in which some faces are multiplied by a larger factor than others.

I noticed an even sillier thing about the surface area of a sphere the other week, for which I haven't yet spotted the "oh, of course, that makes it obvious that it has to be that way" justification. Take a sphere with total surface area 1. Slide a horizontal plane gradually through the sphere, starting tangent to the north pole and ending tangent to the south pole, and let f(x) be the amount of the surface area of the sphere which is north of the plane when the plane has travelled a proportion x of the distance between the poles. What does the graph of f look like? Certainly f(0)=0, f(1/2)=1/2 and f(1)=1, but do you expect an s-shaped curve in between those points, or a reverse s-shape, or what? I found myself wanting to work this out in the course of a BOTEC, and I was slightly startled to find that the answer is that f(x)=x for all x – that curve is actually a straight line.

[jack]

For instance, any of the irregular solids I constructed by taking a bunch of tangent planes to the same sphere (though not necessarily the ones constructed by taking the convex hull of some points on the sphere).

Yeah, the concept of faces as tangent surfaces came up, and I remembered you did something with them but couldn't remember what, but you're right, that's a very good test for whether my reasoning is actually right (and if it's right might even be useful in calculating one or the other :))

I told people the four-glasses-on-a-turntable problem too, and several people got fairly far though no-one finished it.

[obandsoller]

I think when I first thought about it I visualised the volume of the ball being the sum of all the surface areas of the spheres within it. Obviously this is just the same as your argument, just the otherway round.

In fact, if you do the trigonometry, it turns out the same thing works for the tetrahedron and other platonic solids, although if you take the side length as d, r is no longer half d.

I imagine you've already realised this, but this also applied to shapes of different dimensions. e.g. A=pi r^2, C = 2 pi r.

[obandsoller]

It's late and I should be asleep so sorry if some of this is wrong/muddleheaded.

I'm uncertain whether what you state in the final paragraph is correct or not. But it reminded me of solid angles (which I know basically nothing about!), so looking at the wiki article on solid angles I found the following. "Over 2200 years ago Archimedes proved, without the use of calculus, that the surface area of a spherical cap was always equal to the area of a circle whose radius was equal to the distance from the rim of the spherical cap to the point where the cap's axis of symmetry intersects the cap."

If I've understood that correctly (it's v late, so that's a big if) then that either contradicts your final paragraph or is basically the same as your final paragraph.

tl;dr Google Spherical Cap, that'll probably help you more than I can.

[simont]

If I've understood that correctly (it's v late, so that's a big if) then that either contradicts your final paragraph or is basically the same as your final paragraph.

It confirms it, but not completely obviously. Suppose the cap is centred on the north pole of a unit sphere (discarding my previous measurements), and 0 < y < 1 is the distance from the north pole to the plane cutting off the cap. Let x be the radius of the circle which is the cross-section of the sphere at the point where the plane cuts it. I'm claiming that the surface area A of the cap is a fraction y/2 of the whole surface area 4π, that is, A = 2πy. Whereas Archimedes is saying that A = π(x²+y²).

But the two are equivalent, it turns out, because x and y are related by applying Pythagoras to the radius of the sphere: x² + (1-y)² = 1, which expands out to give x²+y² = 2y.

In other words, I agree with Archimedes even though we've chosen to visualise and phrase our claims very differently; but precisely because of that very different visualisation, Archimedes' statement doesn't constitute an intuitive justification of my one :-)

[liv]

Yay maths! I could never have thought of this myself, but I feel good about being able to follow your reasoning.