Deriving the Formula for the Surface Area of a Sphere

Question

I was trying to derive the surface area of a sphere, and I was hoping someone could possibly explain why this one method I am trying does not work. I've seen and done other ways of deriving the surface area so I'm not asking for a different way, but I'm specifically wondering why this one wouldn't work.

I started with a sphere centered at the origin with radius $r$. I then imagined taking narrow, verticle slices out of the sphere (that are parallel with the z-axis and perpendicular to the x-axis) to create rings. I would then unroll each ring and find its area. The area would be $2\pi \sqrt{r^{2}-x^{2}} * dx$ (where $dx$ is the width of each ring) It seems to me that if I did this for all x values on $[-r, r]$, and summed them, I should get the surface area of the sphere. I wrote this as the integral :

$ \int_{-r}^{r} 2\pi\sqrt{r^{2}-x^{2}} \,dx $

This integral is equal to $\pi^{2}r^{2}$, which is obviously not the right answer.

I'm thinking that maybe I need to account for the curvature in some way that I'm not, but I'm really not sure. If anyone could explain why this doesn't work, I would appreciate it greatly.

Answer 1

The fundamental problem with your approach is that you've been tripped up by the staircase paradox:

and see also https://en.wikipedia.org/wiki/Schwarz_lantern.

To be fair, I remember in high school being asked to do a whole bunch of textbook surface area and volume calculations using slicing arguments, and it was never explained to me why approximating volumes with shells and slices works or when and why the same thing could be done with surface area.

It turns out that unlike for volume, in order for the surface area of a sequence of approximations to a smooth surface to converge to the surface area of that smooth surfaces in the limit of refinement, you need both the surface positions and the surface normals to converge to those of the smooth surface. Pointwise convergence of the positions is not enough (as you've discovered).

So if you repeat your calculation using e.g. a sequence of piecewise truncated cones instead of annular rings, you should recover the correct surface area of the hemisphere.

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By the way, for some more intuition about why the normal convergence condition is needed: one way of defining surface area is as the first variation of volume with respect to normal inflation. If $M$ is a volume and if we abuse the notation $M + \epsilon \hat{n}$ to mean "take all points on the boundary of $M$, and move them outwards a small distance $\epsilon$ along the surface normal at that point" (similar to inflating a balloon), then

$$\mathrm{area}(M) = \lim_{\epsilon\to 0} \frac{\mathrm{vol}(M+\epsilon \hat{n}) - \mathrm{vol}(M)}{\epsilon}.$$

Or, equivalently,

$$\mathrm{vol}(M+\epsilon \hat{n}) = \mathrm{vol}(M) + \epsilon\, \mathrm{area}(M) + o(\epsilon^2):$$ the volume of the inflated region is the volume of the original region, plus the volume of a slab with base area equal to the surface area of $M$, and thickness $\epsilon$, gift-wrapped around $M$; plus some higher order terms that depend on curvature.

If you have a sequence of approximations $M_i$ and want the surface area of $M_i$ to approach the area of $M$, you not only need that the $M_i$ increasingly well-approximate the volume of $M$, but also that the $M_i$ converge to having the same inflated volume as $M$, which will be true if the normals converge also.

Answer 2

It should be $ \displaystyle \int_{-R}^{R} 2\pi\sqrt{R^{2}-x^{2}} \, ds \ ,$ where $ds$ is width element along the surface (arc length element). Your mistake is in taking width as $dx$, which does not account for the arc.

If $r$ is the radius of the circle parallel to yz-plane at $x, \ r = \sqrt{R^2 - x^2}$

$ds = \sqrt{(dx)^2 + (dr)^2}$

$dr = d (\sqrt{R^2 - x^2}) = - \frac{x}{\sqrt{R^2 - x^2}} dx$

So, $ds = \frac{R}{\sqrt{R^2 - x^2}} dx$

Plugging into the integral,

$\displaystyle \int_{-R}^{R} 2\pi\sqrt{R^{2}-x^{2}} \, ds = \int_{-R}^R 2 \pi R \ dx = 4 \pi R^2$