Here’s something to get you started, I’ll leave the creative stuff to you.
Consider a sphere:
V = (4/3)(pi)(r[sup]3[/sup]), SA = (4)(pi)(r[sup]2[/sup]), where r = radius of the sphere.
so the ratio of V to SA (V/SA) = r/3.
If V = 100, SA = 300/r.
Cube:
V = s[sup]3[/sup], SA = 6s[sup]2[/sup], where s = length of one side.
V/SA = s/6.
If V = 100, SA = 600/s.
If you take our sphere and cube where r = s, both of which having a Volume of 100, the cube will have twice the surface area. This should get you started.
A few more fun formulae:
Box (not necessarily a cube):
V = (s[sub]1[/sub])(s[sub]2[/sub])(s[sub]3[/sub])
SA = (2)(s[sub]1[/sub])(s[sub]2[/sub])+(2)(s[sub]1[/sub])(s[sub]3[/sub])+(2)(s[sub]3[/sub])(s[sub]2[/sub])
where s[sub]n[/sub] = the length of one side
Cylinder:
V = (pi)(r[sup]2[/sup])(h)
SA = (2)(pi)(r[sup]2[/sup]) + (2)(pi)®(h)
I think that the derivative of the n-dimensional measure for an n-sphere is the (n-1)th measure of the sphere for all n. I’m not 100% on that, though.
What I do know for sure, and what’s very interesting, is that the limit of the n-dimensional volume of the unit n-sphere goes to zero as n goes to infinity.
hajario, I think you’re on the right track, but there’s a slight problem with your example. If you have a sphere with V = 100, then necessarily r = 2.8794. If you have a cube with V = 100, then necessarily s = 4.6416. So you can’t set both V’s to 100, and then say r = s. What you can do is put both ratios in terms of the volume:
For a sphere: (V / SA) = 0.2068 V[sup]1/3[/sup]
For a cube: (V / SA) = 0.1667 V[sup]1/3[/sup]
From this you can clearly see that if the volumes are equal, the sphere will always have a larger ratio. However, the cube will never have twice the surface area of the sphere.
ultrafilter, you’re right about Anthracite’s formula working in n dimensions. SA = dV/dr. But I believe that it also works for things other than spheres. Consider a cube. If r is the “radius” of the cube, that is, r = s/2, then V = 8r[sup]3[/sup] and SA = 24r[sup]2[/sup]. In n dimensions, V[sub]n[/sub] = 2[sup]n[/sup]r[sup]n[/sup], and SA = 2nV[sub]n-1[/sub] = n2[sup]n[/sup]r[sup]n-1[/sup].
Well said, Achernar. That’s a better explanation. I thought that there might have been a problem with my example but, in my defense, while I was writing it Mrs. H was yelling at me that we were running late for our dinner plans and I had to “get off the damned computer” so I didn’t have a chance to proof.
That’s interesting. Not to hijack too much, but does it work for anything topologically equivalent to a sphere? All of my geometry books are in another city, else I’d look for possible counterexamples myself.
I think that Aerogels may be the state of the art.
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You know, I don’t know. This is really more of something I came up with in the shower than something I learned in a course, so I don’t really have a good handle on its extent. I checked it out with the other four Platonic solids, using formulas I got here, and it seems to work (although the algebra for the dodecahedron was ugly enough that I may have erred). In these cases, r is the radius of the inscribed sphere, which makes sense to me. But I was thinking, how would you define r for an arbitrary shape? I don’t know. Maybe if you or someone can come up with a counterexample, the scope of this “theorem” would become clear.
Well, at least you have an excuse. Typically, I will proof my posts four or five times and still submit them with mistakes.
The condition for that formula is that the shortest line from any “face” to the center is the same length, and that length is what you call the radius (treat curved surfaces like spheres as having infinitesimal faces). A skeleton of a proof: Consider how much paint you’d need to put a coat of thickness [sym]D[/sym]r over the whole object. On the one hand, you can take the surface area and multiply by [sym]D[/sym]r, or on the other hand, you can say that you’ve just made a similar shape of radius r + [sym]D[/sym]r, and take the difference of the two volumes. Since it’s the same amount of paint, this means that the surface area is the derivative of the volume with respect to r.
The condition on “r” being the same for all surfaces arises because if it isn’t, then the “painted” solid won’t be similar to the original.
That’s a very perceptive point, Chronos. You’ve explained why it works for the n-sphere, the n-cube, and all the Pythagorean solids. You’re saying that r is the radius of the inscribed sphere, and an inscribed sphere which is tangent to every face must exist for it to work. However. I don’t think you’ve shown that it won’t work for irregular shapes, such as a spheroid of constant eccentricity. For generalized shapes, making it bigger is not the same thing as painting on more surface. In many cases, I suspect, you can pick an arbitrary distance, say a, and then V [Sym]µ[/Sym] a[sup]3[/sup] and SA [Sym]µ[/Sym] a[sup]2[/sup]. If so, then there has to exist some r [Sym]µ[/Sym] a such that dV/dr = SA. I don’t think r will be the radius of an inscribed sphere, but maybe it’s something else significant.
Typically, I will proof my posts four or five times and still submit them with mistakes.