My young son was asked to derive the surface area of a sphere using pure algebra. He could not get to the right formula but it seems that his reasoning is right. Please tell me what's wrong with his logic.

He reasons as follows:

  • 1.Slice a sphere into thin circles (or rings if you hollow them out)
    1. The sum of the circumferences of all the circles forms the surface area of the sphere.
    2. Since the formula of circumference is $2{\pi}R$, the sum of the circumferences would be $2{\pi}(R_1+R_2+R_3+...+R_n)$
    3. My son draws the radii beside each other and concludes that their sum would be equivalent to one-half of the area of the largest circle or $({\pi}R^2)/2$. He appears to be right from his drawing.
    4. Substituting the sum of the radii, he comes up with a formula for the surface area of the sphere as $(R^2)*(\pi^2)$.
    5. He asks me what's wrong with his procedure that he cannot derive the correct formula of $4{\pi}R^2$.
  • $\begingroup$ Could you publish the drawing mentioned? $\endgroup$ – zoli Sep 1 '15 at 22:11
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    $\begingroup$ It seems to me that the primary mistake is in step 4 - he assumes the radii grows linearly as we move through the sphere linearly. This is not true - the radii of the rings of the sphere grows quickly at first and then slows down and shrinks again. (If the radii did grow linearly we would see something like two cones stuck together at their bases) $\endgroup$ – Linus S. Sep 1 '15 at 22:12
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    $\begingroup$ The drawings are of importance to be able to find an error. For instance, I assume that each slice/ring has a given height, which I don't see mentioned here. $\endgroup$ – Scounged Sep 1 '15 at 22:17

In this answer, it is shown geometrically that the area of a strip of sphere between two parallels of latitude is the same as the area of the orthogonal projection of that strip onto the cylinder with the same radius as the sphere and whose axis is parallel to the north-south axis of the sphere.

Finding the area of this cylinder is just multiplying its height, $2r$, by its circumference, $2\pi r$. Thus, the area of the sphere is $4\pi r^2$.

Your son's procedure seems to be summing the circumferences of the green strips on the sphere, but there must be some accounting for the widths of these strips. This seems to be the problem he is encountering.


An interesting proof may go through the following lines:

  1. If $S(R)$ and $V(R)$ are the surface area and volume of a sphere with radius $R$, we have $S(R)=\frac{d}{dR} V(R)$ by triangulating the surface and exploiting $S(\alpha R)=\alpha^2 S(R), V(\alpha R)=\alpha^3 V(R)$ for any $\alpha > 0$, so the problem boils down to computing $V(R)$;
  2. By using Cavalieri's principle and exploiting the fact that the area of a circle with radius $r$ is $\pi r^2$, we may see that the problem of computing $V(R)$ boils down to computing the area of $\{(x,y)\in\mathbb{R}^2 : 0\leq x\leq R, 0\leq y\leq x^2\}.$

  3. The last problem can be solved in many ways: by exploiting the optical properties of the parabola like Archimedes did, or by applying the Cavalieri's principle again and recalling that: $$ \frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^2 = \frac{(n+1)(2n+1)}{6n^2}.$$

A very short proof relying on the Gaussian integral is given by Keith Ball at page $5$ of his notorious book An Elementary Introduction To Modern Convex Geometry, but maybe it is the case to wait for your son to grow a bit :)


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