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How can the surface area of a sphere be exactly 4 times the area of it's cross-sectional area? This seems too clean; why 4?

If I sliced 4 thin circles from the middle of a sphere and stuck them together (and was able to morph them) would I end up with another hollow sphere?

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  • $\begingroup$ Related: math.stackexchange.com/questions/1833867 $\endgroup$ – Watson Nov 2 '16 at 16:09
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    $\begingroup$ Note your first sphere was solid and your constructed sphere was hollow. So the "magic" of making a new sphere out of a teeny cross-section of a sphere is no more paradoxical as creating a huge pyramid out of 4 flat triangles. As for why 4. Well... calculus... $\endgroup$ – fleablood Nov 2 '16 at 16:36
  • $\begingroup$ @fleablood, but a cube from 6 squares ... $\endgroup$ – z100 Nov 2 '16 at 16:41
  • $\begingroup$ Right, I was not addressing the "why 4" question but the "if I slice a tiny section for the center I can recreate the sphere" issue. I assumed this seemed paradoxical to the OP (else why would s/he have asked it). $\endgroup$ – fleablood Nov 2 '16 at 17:04
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Here the $4\pi$ should be viewed not as $4$ times $\pi$ but rather as $2$ times $2\pi$ and then there is a reasonably nice geometric explanation. Namely, if you project the sphere to, say, the $z$-axis, then the inverse image of each small subsegment of $[-1,1]$ will have area $2\pi h$ where $h$ is the length of the subsegment. You see this particularly clearly in the case of a small subsegment around $0$ in which case you get a thin strip which is roughly the product of the equator (of length $2\pi$) by a segment of length $h$. Therefore the total area is $2\times 2\pi=4\pi$.

Four unit disks do have the same area as the unit sphere, but you can't patch them together without distorting them so as to form the sphere. One way of seeing it is in terms of a metric invariant called Gaussian curvature (this invariant vanishes for the disk but is equal to 1 for the sphere).

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  • $\begingroup$ I believe this property of a sphere fascinated many SE community members. What about a more intuitive answer which does not include a more or less explicit use of integrals? Maybe more like Gauss's law. $\endgroup$ – z100 Nov 2 '16 at 16:57
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If you consider a cross-section of the sphere at a height $h$ above the equatorial plane, you get a circle of radius $\sqrt{r^2-h^2}$, and circumference $2\pi\sqrt{r^2-h^2}$.

Then you can be tempted to conclude that the area of the sphere is the average perimeter from $-r$ to $r$, which would yield $\pi^2r^2$.

But you must take into account the fact that the surface is not vertical but oblique and there is a correction factor equal to the secant of the angle of the normal, or $\dfrac r{\sqrt{r^2-h^2}}$.

So after simplification you are just integrating $2\pi r$ from $-r$ to $r$, which explains the factor $4$.

A geometric interpretation is by saying that any slice of small height $\delta h$ in a sphere has the same area (!)

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  • $\begingroup$ "So after simplification you are just integrating 2πr from −r to r" can you explain this in greater detail. If r is constant that would be the surface area of the sides of a cylander, wouldn't it. If you meant 2pi h that... be only a factor of 2 as int(h) is h^2/2. But you can't mean that as that's be the surface of a cone. $\endgroup$ – fleablood Nov 2 '16 at 17:53
  • $\begingroup$ @fleablood: the factors $\sqrt{r^2-h^2}$ do simplify, and the area is indeed that of the cylinder that tangents the sphere, as counterintuitive as it may sound. For instance, take the last slice close to the pole, from $r-h$ to $r$: it approximately yields a circle of area $\pi(r^2-(r-\delta)^2)\approx2\pi r\delta$, like all other slices. $\endgroup$ – Yves Daoust Nov 2 '16 at 18:24
  • $\begingroup$ I don't think it was intuitive to me to "average" the circles (which honestly never even occured to me) so much as to integrate the constantly changing circumferences. A common mistake would be to assume as h and $\sqrt {r^2-h^2} $ appear interchangeable (they aren't) their integrals are the same. But that'd yield a cone. It boils down to $\int_{0}^1\sqrt {1-x^2}dx $=1, which is where I decided my answer was no longer "intuitive" so I quit writing my answer. Normalizing averages seems an interesting approach, but it wasn't intuitive to me at any rate. $\endgroup$ – fleablood Nov 2 '16 at 19:13
  • $\begingroup$ @fleablood: I don't see how $h$ and $\sqrt{r^2-h^2}$ appear interchangeable and what relation this discussion has with my solution. $\endgroup$ – Yves Daoust Nov 2 '16 at 19:21
  • $\begingroup$ I'm just wondering if you can give greater detail. "Then you can be tempted to conclude that the area of the sphere is the average perimeter" I don't see why I would think that. "But you must take into account the fact that the surface is not vertical but oblique and there is a correction factor equal" A correction factor of what exactly. I don't see what we are calculating. "So after simplification you are just integrating 2πr" after simpligying what exactly and how so? and.... $\endgroup$ – fleablood Nov 2 '16 at 20:27

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