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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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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).[reply]

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)[reply]

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