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With integration you can prove that if a sphere is cut into $n$ paralel slices of equal width, then those slices have the same external area.

It is often presented as "a spherical loaf of bread is cut $n-1$ times with equidistant paralel cuts, thus leaving $n$ slices of equal width. Those slices have the same amount of crust".

Is is possible to get to the same conclusion through a classical geometrical aproach?

Darth Geek
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    Think about Cavalieri's Principle. Maybe it helps :-) – Mikasa Aug 02 '14 at 10:29
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    You may want to check the redundant department of redundancy in the first line. A spherical sphere? By the way, nice question. – Mark Fantini Aug 02 '14 at 10:48
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    @MarkFantini LOL I was originally going to write the "spherical bread" thing, but I changed it to sphere. – Darth Geek Aug 02 '14 at 10:55
  • @BabakS. I see that it's somehow connected but I don't actually see how to apply it? Care to elaborate? Maybe a hint? – Darth Geek Aug 02 '14 at 10:55
  • You'll find a beautiful explanation in this blog post: http://blog.zacharyabel.com/2012/01/spherical-surfaces-and-hat-boxes/ Even in the classical approach, a limit process is neede, though no integration as we use it today or coordinate geometry. – Dario Dec 23 '15 at 18:46

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as a surface of revolution a sphere may be viewed as the disjoint union of infinitesimal frustra created as slices between planes perpendicular to a line passing through its centre. if we make this line our x-axis, and use a circle of radius $r$ centered at the origin, then the radius of the frustrum at a particular value of $x$ is just $y$ whereas the length of the infinitesimal line element is $ds$. thus between suitably chosen values $x=a$ and $x=b$ we have: $$ A = \int_{x=a}^{x=b} dA = \int_{x=a}^{x=b} 2\pi yds \tag{1} $$ however $$ ds = \sqrt{dx^2+dy^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx = \frac{rdx}{y}\tag{2} $$ (because $\frac{dy}{dx}=-\frac{x}{y}$ and $x^2+y^2=r^2$). now substituting (2) into (1) we obtain the all-important cancellation of $y$, leaving $$ A = 2\pi r \int_{x=a}^{x=b} dx = 2\pi r (b-a) $$ although expressed here in the more recent language of calculus and coordinate geometry, the all-important cancellation is a mere matter of classical geometry once the principle of infinitesimals is applied. certainly this principle - e.g. as Archimedes' exhaustions - was known to the later Greek geometers.

David Holden
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  • I feel like this answer is more of a clarification, i.e. how to do the problem using calculus, as it doesn't really add any suggestions for the solution other than those in the comments. Am I missing something? – Thoth19 Aug 02 '14 at 23:36
  • perhaps a reasonable criticism. though my intention was to regard the calculus as a translation of an ancient idea into our contemporary mathematical idiom. i think to Archimedes the area theorem introduced by OP would have been intuitively an almost immediate consequence of the circle's unique property: that the tangent at P is always perpendicular to the line PO. – David Holden Aug 03 '14 at 10:23
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    TBH I'm not sure how to solve it with the approach hinted at in the comments. I thought the question seemed interesting, and I agree that the calc approach is likely better. But the OP asked for the classical solution,I'm not entirely sure of the relevance, though it is clear, and does explain why the statement is true. – Thoth19 Aug 03 '14 at 14:54
  • Although the principle was known to the Greek geometers, I doubt those techniques were rigurously proved... – Darth Geek Aug 13 '14 at 16:49