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

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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 DeckerMay 26 '13 at 12:03

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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

## 1 Answer

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. – Dan May 26 '13 at 10:41