Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question
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

3 Answers 3

up vote 10 down vote accepted

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 $4\pi r^2$.

share|improve this answer

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 $\frac{4}{3}\pi r^3$.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.