# Why do 4 circles cover the surface of a sphere?

Is there a geometric explanation for why a sphere has surface area $4 \pi r^2$ ?
Ie equal to 4 times its cross-section (a circle of radius r).

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This link does not give a full answer, but it may help a little: en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder –  please delete me Mar 27 '11 at 13:16
I would add to the comment of Eivind: the map from the cylinder to the sphere given by orthogonal projection from the axis is area-preserving. It's a nice exercise to show that it shrinks horizontal infinitesimal distances by the same factor as it expands vertical infinitesimal distances. –  user8268 Mar 27 '11 at 13:36
What does cross-section mean here? –  Rasmus Mar 27 '11 at 13:38
Here's a cute interpretation of the problem: On a spherical wedge of angle 90°, the curved outer surface has the same surface area as the two planar semicircular ends put together. One can think of these as two non-minimal surfaces on the same boundary curve. Why do they have the same area? (Of course, the answer may just be that it is a coincidence.) –  Rahul Mar 27 '11 at 16:38

Let $Z$ be a cylinder of height $2r$ touching the sphere $S_r$ along the equator $\theta=0$. Consider now a thin plate orthogonal to the $z$-axis having a thickness $\Delta z\ll r$. It intersects $S_r$ at a certain geographical latitude $\theta$ in a nonplanar annulus of radius $\rho= r\cos\theta$ and width $\Delta s=\Delta z/\cos\theta$, and it intersects $Z$ in a cylinder of height $\Delta z$. Both these "annuli" have the same area $2\pi r \Delta z$. As this is true for any such plate it follows that the total area of the sphere $S_r$ is the same as the total area of $Z$, namely $4\pi r^2$.

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A "plate" here just means a very short cylinder, right? –  LarsH Mar 14 at 20:34
@LarsH: That's right. –  Christian Blatter Mar 15 at 7:17

One geometric explanation is that $4\pi r^2$ is the derivative of $\frac{4}{3}\pi r^3$, the volume of the ball with radius $r$, with respect to $r$. This is because if you enlarge $r$ a little bit, the volume of the ball will change by its surface times the small enlargement of $r$.

So why is the volume of the full ball $\frac{4}{3}\pi r^3$? By slicing the ball into disks, using Pythagoras, you get that its volume is $$\int_{-r}^r \pi (r^2-x^2)\mathrm{d}x$$ which is indeed $\frac{4}{3}\pi r^3$.

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Is this true for all manifolds, dV/dr=S?, where V is the n-volume and S is the n-1-surface and r is the distance from a point in the interior to the surcface ? –  user1708 Mar 27 '11 at 13:58
@solomoan: I think you will have trouble defining "the distance from a point in the interior to the surface" for a general manifold with boundary. I don't see how to generalise the result in a meaningful way to general manifolds. –  Rasmus Mar 27 '11 at 14:02
@solomoan: If $$f: B\to{\mathbb R}^3, \quad (u,v)\mapsto f(u,v)$$ produces a surface $S$ with unit normal $n(u,v)$ then $$x: \ B\times[0,\epsilon]\ \to\ {\mathbb R}^3, \quad (u,v,t)\mapsto f(u,v)+ t n(u,v)$$ produces a plate of thickness $\epsilon$. You can compute the volume $V(\epsilon)$ of this plate by means of the Jacobian of $x$, and calculating the limit $$\lim_{\epsilon\to0}{V(\epsilon)\over\epsilon}$$ you get the formula for the surface area $\omega(S)$. –  Christian Blatter Mar 27 '11 at 14:18
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