Wikipedia:Reference desk/Archives/Mathematics/2017 September 9

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September 9

Volume of spherical caps and hemispheres....

Hi,

1. I am recalling an old exam question from years ago, which involved spheres. On the side of a spherical ball, place a series of markers indicating various volumes, if the ball is filled with powder. (I am of course ignoring any thickness, or indeed the size of the powder granules. ) The ball is of a given radius. From reading Wikipedia I've already found the article on Spherical caps , but the equation given there seemed to be applicable to caps in distinct hemispheres. Does anyone have a reference to an equation that would give volume in relation to height for the entire sphere?

2. Extending the above does anyone have a reference on how to generalise the spherical cap equation for 'partial' elipsoid shapes, like a bowl. I can draw a diagram if it helps? ShakespeareFan00 (talk) 09:20, 9 September 2017 (UTC)

For the first question, it appears to me that the volume formulas given by the linked article are valid for a single cap in either hemisphere. In particular,

$V={\frac {\pi h^{2}}{3}}(3r-h)$.

Loraof (talk) 19:49, 9 September 2017 (UTC)

which for the benefit of those following will re-arranges to :-

$3rh^{2}-h^{3}-{\frac {3V}{\pi }}=0$ solving for h. which isn't as easy as to solve as I remember. (the original question was on an ancient A-level paper IIRC but can't recall the board.).

The next step is naturally solving the cubic equation for various values of V, for which various methods exist see Cubic_function#General_solution_to_the_cubic_equation_with_real_coefficients

Would it be reasonable to post the next steps here or would it be more appropriate to take something like this to Wikiversity?

In respect of 2, I have a very strong feeling a general solution will be at college level. ShakespeareFan00 (talk) 21:51, 9 September 2017 (UTC)

Sorry, when you said you want "volume in relation to height" I assumed you wanted volume as a function of height. Solving your cubic for height involves an expression in terms of cube roots; since there's only one height that can give a particular volume, the cubic must have only one real root. It's probably not a good idea to go through the solution steps here. Loraof (talk) 00:01, 10 September 2017 (UTC)

I'll consider taking it over to Wikiversity then. Thanks for the response. ShakespeareFan00 (talk) 08:27, 10 September 2017 (UTC)

• Let me revise something I said before. Actually by the discriminant test, there are three real roots, but only one is in the geometrically meaningful range $h\in (0,2r).$ Loraof (talk) 15:55, 10 September 2017 (UTC)

There is a decent derivation at Sphere#Enclosed volume using disk integration. If you change the limits of integration from (-r to r) to (-r to h-r), then you get the same formula above. For an ellipsoid, you would need to change the disk that you integrate over (example). The closed-form solution for h is incredibly messy, so maybe you didn't remember that part of the problem correctly. C0617470r (talk) 06:42, 10 September 2017 (UTC)

The elipsoid extension was not part of the original problem, it was an extension I was wanting to look at personally.ShakespeareFan00 (talk) 08:27, 10 September 2017 (UTC)

Even for the sphere, the cubic is not really solvable, assuming that A-level students aren't required to memorize the general cubic formulas. The formulas aren't hard to apply, just incredibly messy. Here's the solution. C0617470r (talk) 17:45, 10 September 2017 (UTC)

Thanks for the link and soloution, but Wolfram Alpha's terms are not necessarily compatible with "free" licenses (suppresses desire to have an ideological rant about the need for "open math" :( )ShakespeareFan00 (talk) 10:55, 11 September 2017 (UTC)

I'd personally consider use of the trignometric solution suggested below. I have a strong feeling that in the original answer to 1. The A-level student would have been expected perhaps to consider the ball as 2 domes, and work from there, by computing the volume of a single 'dome' first and if V was larger than a single one adjust it accordingly. That said I did ask for a general formulae that has been provided. However thanks to the answers given here I've learnt about some math topic I did not at A-level. Congratulations. ShakespeareFan00 (talk) 10:55, 11 September 2017 (UTC)

And notice that the expression under the square root sign, $3V^{2}-4\pi r^{3}V,$ is negative since V is less than the volume $4\pi r^{3}/3$ of the entire sphere. Thus the given algebraic expression involves the cube root of a non-real complex number. This is casus irreducibilis. You can get a solution expression only involving real quantities using trigonometric functions—see Cubic function#Trigonometric solution for three real roots. Loraof (talk) 18:00, 10 September 2017 (UTC)

Let $x=h2^{-1}r^{-1}$ and $y=6V\pi ^{-1}$ such that $y=3x^{2}-2x^{3}$. For $0<y<1$ the solution $x$ is computed by newton iteration: $x=\lim x_{n}$ where $x_{0}=y$ and $x_{n+1}=x_{n}-(3x_{n}^{2}-2x_{n}^{3}-y)6^{-1}x_{n}^{-1}(1-x_{n})^{-1}$. Bo Jacoby (talk) 20:23, 10 September 2017 (UTC).

Quintic equation problem

Take any quintic integer polynomial equation:

ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0

...and what you want to know is whether the solutions are related to the integers by a finite number of algebraic operations. (For polynomial degrees 1-4, all solutions are.)

The only requirements I know are that a can't b 0, f can't be 0, and at least one of b through e can't be 0. Any complete list of requirements?? Georgia guy (talk) 14:09, 9 September 2017 (UTC)

See Quintic function#Solvable quintics. --Deacon Vorbis (talk) 14:20, 9 September 2017 (UTC)