Polar Coordinates – How to Deduce the Area of a Sphere

polar coordinates

$A = r \int_{0}^{\pi}\int_{0}^{2\pi} e^{i (\alpha+\theta)} d\alpha d\theta
= r \int_{0}^{\pi} [\frac{-i}{\alpha+\theta} e^{i(\alpha + \theta)}]_{0}^{2\pi} d\theta
= …
= r \int_{0}^{\pi} e^{2i\alpha} ( \frac{1}{\alpha} – \frac{1}{\alpha+2\pi} ) d\theta$

and according to wolfram it is wrong, here.
So how do you deduce the area of a sphere in polar coordinates?

[Edit] Trevor gave us formula $r^2 \int_{2π}^{0}\int^{π}_{0}sin(\phi) d\phi d\theta$ without deduction, by deduction, I mean things on which it is based on.

  1. Why is it odd function $sin(\phi)$ and not instead an even function $cos(\phi)$? (I know that it is and I can calculate it but some proof here missing)
  2. Why no term with $\theta$ but with $\phi$?
  3. Why $r^{2}$? -Okay it is area and by symmetry it sounds logical but why $r^{2}$ and not for example $rd$ where $d$ is some lenght or even a path?
  4. What about an arbitrary case?
  5. What determines the order in integration?

Many things open to really deduce it.

[Edit] Zhen Lin has an excellent answer where s/he noted that 2-sphere, i.e. the sphere in three dimensional euclidean space, is two-dimensional. Next deductions should be on 0-sphere and then incrementally on others. More about n-sphere here.

  1. How do you deduce area in 0-sphere, 1-sphere, 2-sphere, and n-sphere?
  2. Why does the Jacobian matrix in 2-sphere-in-3-dim-euclidean-case have only two two variables $(\phi, \theta)$?
  3. Are $\theta$ and $\phi$ non-free-variables because they are bounded somehow by the manifold (maybe misusing terminology)?
  4. What is $r$ in the sphere? For example in the 2-sphere?
  5. I can sense that the var $r$ is somehow different to $\phi$ and $\theta$ because what is it if you derivate it, zero so perhaps it is a trivial case (again probably misusing terminology, sorry)?
  6. What about if $r$ depends on $t$, time, what is it then? $\frac{\partial r}{\partial t}$ is now something. But would the extra dimension be now $t$ instead of $r$?

Please, use the term dimension in your replies, I feel it is very important to even understand what a sphere is, let alone a dimension.

[Conclusion] The general solution is still open. The $r$ is a homothetic transformation. The terms $(\theta, \phi)$ are apparently tied to the manifold, (ALERT! perhaps misusing terminology). The acceptance of a question does not mean that all things appeared here are solved. Zhen`s solution is really the solution to look at, big thanks for it!

Best Answer

Here is a diagram that might help you to visually understand that the element of area $\text{d}A = r^2 \sin \phi \, \text{d} \phi \, \text{d} \theta.$

Note that in the diagram $0 \le \phi \le \pi $ and $ 0 \le \theta \le 2 \pi.$

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