1
$\begingroup$

I was wondering about the following:

Consider a sphere with radius $r$, then the volume equals $\frac{4}{3}\pi r^3$. Now consider to cover this sphere with a shell of thickness $h$. Then the new volume becomes $\frac{4}{3}\pi(r+h)^3$. I was wondering if we can compute the value of $h$ for which the new sphere has twice the volume of the original sphere. This leads to the following expression

$$\frac{4}{3}\pi(r+h)^3 = \frac{8}{3}\pi r^3\\ h^3 + 3h^2r + 3hr^2+r^3 = 2 r^3\\ h^3 + 3h^2r + 3hr^2-r^3 = 0$$ This is a cubic polynomial, with coefficients $(1,3r,3r^2,-r^3)$. I was wondering if there exists an easy expression for these kind of cubics, for easiness, it may be assumed that $r$ is integer.

Seeing the origin of this question I am also interested in finding the real root of the cubic. It would be nice if this is an relative easy expression, which does not need for the use of, for example, Cardano's method.

Edit: Thanks for all the answers. I was a bit vague on what I meant with easy expression, to me the answer is rather simple. I was afraid that Cardano would lead to nested roots, or maybe roots of $r$, which is not the case. I am interested in being able to do some "on the back of an envelope" calculations, so since $2^{1/3}-1$ is just a constant factor, this is easy enough for me.

$\endgroup$
3
  • 1
    $\begingroup$ I think you can assume $r=1$ here. If you do, does the resulting cubic have a rational root [it could only be 1 or -1]? If no rational root you need Cardano or the like. $\endgroup$
    – coffeemath
    Commented Dec 4, 2018 at 9:15
  • $\begingroup$ In this case, it has no real solution $p(1)=6$ and $p(-1)=-2$. Then we do need Cardano indeed, but I do not see how this holds for every choice of $r$. $\endgroup$ Commented Dec 4, 2018 at 9:21
  • $\begingroup$ Regarding your edit: I think the "missing complication" might be hidden away in showing rigorously that using a scale factor is valid—eg by filling an object with progressively smaller cubes and showing that their total volume converges to that of the object, and finding the requirements for it to work. (For example there might be weird fractal shapes where it goes wrong.) $\endgroup$
    – timtfj
    Commented Dec 4, 2018 at 12:28

3 Answers 3

2
$\begingroup$

$(r+h)^3 = 2r^3$

$\Rightarrow \left(1+\frac{h}{r}\right)^3=2$

$\Rightarrow 1 + \frac{h}{r} = \sqrt[3]{2}$

$\Rightarrow h = r(\sqrt[3]{2} -1)$

$\endgroup$
1
$\begingroup$

It is easy to see that $h=(\sqrt[3]2-1)r.$ In order for the 'thickened' sphere to have double the volume of the original one, its radius must be $\sqrt[3]2$ times as big as the original, as the volume is a homegenous function of degree 3 of the radius.

I don't think there is an 'easy' way to find that solution from the equation, although applying Cardano should of course get that result.

$\endgroup$
0
$\begingroup$

Easy method: doubling the volume while preserving the shape means multiplying linear dimensions by the cube root of 2. So $$h+r=\sqrt[3]{2}r$$ and so $$h=(\sqrt[3]{2}-1)r$$ by a couple of lines of algebra.

The key to keeping it simple is realising that you don't need the volume formula at all, just a scale factor.

For the other two solutions, just put the non-real values of $\sqrt[3]{2}$ in the result above. Then you'll at least have the solutions to your cubic, and that might help in thinking how to get them from it.

The non-real roots of 2 can be found easily if you know that the $n$th roots of a positive number $a$ are equally spaced on a circle of radius $\sqrt[n]{a}$ on the complex plane. A bit of geometry, or knowing the sine and cosine of 120°, gives the two non-real cube roots of 2 as $$\frac{\sqrt[3]{2}}{2}\left(-1\pm i\sqrt 3\right).$$

Edit: I think the "missing complication" is probably in proving rigorously that using a scale factor is valid for problems like this. For example you might end up having to define what volume is, in a way that still works for weird fractal shapes.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .