I devised yesterday a false calculation of the area of the sphere S^2=\{(x,y,z)\mid x^2+y^2+x^2=1\}. The idea is quite natural. Decompose  the sphere in circles \{(x,y)\mid x^2+z^2=1-y^2\} of radius \sqrt{1-z^2}. Thus it seems that the area of a sphere should be 2\pi\int_{-1}^1\sqrt{1-z^2}dz. However, by the change of variable z=\sin x, we easily notice that \int_{-1}^1\sqrt{1-z^2}dz=\int_{-\pi/2}^{\pi/2}\cos^2xdx=\pi/2, so we get \pi^2, instead of 4\pi, the well known value we should find. What’s wrong?

Well why should it be true on the first place? The intuitive reason why this should be true consists in approximating our sphere by a ziggurat of cylinders, of height 1/n and radius \sqrt{1-(i/n)^2}, thus getting something like \frac{2\pi}{n}\sum_{i=-n}^{n-1}\sqrt{1-(i/n)^2} for the area of our tower. This Riemann sum does converge to 2\pi\int_{-1}^1\sqrt{1-z^2}dz, but there’s no reason why it should be the area of the sphere. There’s no reason why it should be the area of anything definite either… Notice by cutting this picture by a plane, it amounts to approximating a (quarter of a) circle by a bunch of vertical segments… whose sum of lengths is of course constantly equal to 1, instead of converging to \pi/2.

In the same vein, take a half circle based on the segment [-1,1],  and then two smaller half circles under it, and split again each arc into two arcs. This yields a sequence of curves \gamma_n :[-1,1]\rightarrow \mathbb{R}^2 which converge in the L^{\infty} norm to the straight line segment, but which have a constant length \pi which doesn’t converge to the length of the limit curve. Indeed, the length of a curve is not a continuous function, only semi-continous.

So we have to think hard about the notion of surface area of a (say convex) body in the space. Certainly the right way to proceed is via Minkowski content (in other words in this case: derivative of the volume of the sphere) or Hausdorff measure.