# How do you explain the appearance of a sine in the integral for calculating the surface area of a sphere?

Let's say I want to calculate the surface area of a sphere. For simplicity, let's just use the unit sphere. A naïve argument might go like this. Let's say I mark the north and south "poles" and draw half of a great circle, which has length $\pi$. I could say that since I need to go all the way around the sphere, I need to multiply this by $2\pi$ (the circumference of the equator). Therefore, the surface area of the unit sphere is $2\pi^2$.

Now, as we all know it should be $4\pi$. Let's say we do an integral, using the following parametrization:

$$T(\theta, \phi) = \begin{pmatrix} \sin \phi \cos \theta \\ \sin \phi \sin \theta \\ \cos \phi \end{pmatrix},$$

with $0 \le \phi \le \pi$ and $0 \le \theta \le 2\pi$. If we work out all the formulas, we get that $$Area(S^2) = \int_0^{2\pi} \int_0^\pi \sin \phi\ \mathrm{d}\phi \ \mathrm{d}\theta = 2\pi \int_0^\pi \sin \phi\ \mathrm{d}\phi.$$

The $2\pi$ is there all right, but it multiples not $\pi$ but $\int_0^\pi \sin \phi\ \mathrm{d}\phi$, which equals $2$. Where does this come from? In other words, why is it wrong to just multiply $2\pi$ by half the length of a great circle? It would be great if there was a geometric explanation, with as little calculus as possible involved.

-
 Because the strip represented by the half great circle is not constant in width. It is wider at the equator than at either pole. Your approach would be correct for a cylindrical tube of height $\pi$ and radius 1. – Tpofofn Sep 23 '12 at 3:01 @Tpofofn: But where does the $2$, or, if you want, the $\sin \phi$ come from? – Javier Badia Sep 23 '12 at 3:19 The question is not so much where does the 2 come from, rather it is why is the area of the tube equal to $2\pi\cdot\pi$ and the area of the sphere equal to $2\pi\cdot 2$. In both cases the area of a thin strip extending from N to S is $\pi d\theta$ and $2d\theta$ respectively. – Tpofofn Sep 23 '12 at 11:52