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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.

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marked as duplicate by Sharkos, rschwieb, Start wearing purple, Tom Oldfield, Rick Decker May 26 '13 at 12:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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 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

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 $\Delta$

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I hope I am not assuming too much when i say that the circular segment for phi is a simple argument. I can always attempt a hand written response if it seems a little confusing. – Triatticus May 26 '13 at 10:41

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