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.
 A: There is a geometrical argument for it, all you need to do is construct a coordinate system in spherical coordinates in the Cartesian system, in order to move along a line of constant radius we must change the angular dependence in $\phi$ and $\theta$ (assuming the usual definition that $\theta$ is the azimuthal angle and $\phi$ is the polar angle), to move in the  $\phi$ direction requires that surface traversed is simply $ dl = R d\phi$, but to move in the theta direction we have to use a bit of trig, a circular segment in the $\theta$ direction has a radius $ r = R \sin \phi $ ( we can construct a triangle with the z axis as one leg, the hypotenuse is the radius R, and the horizontal remaining leg is given by the preceding formula), in order to get the length of a line in the $\theta$ direction its the same relationship as before ($s = r \times \text{ angle}$) so that $dl = r \sin \phi d \theta$. thus multiplying these to get the surface area element gives $dA = R^2 \sin \phi d \theta d \phi$ When $\theta$ encompasses a complete unit circle ( $R= 1$) this yields $ 2 \pi \int \sin \phi d \phi $. If you want a more non-calculus representation replace every "d" with a $\Delta$
