The area of thin ring of shell of a sphere

Question

In a proof of the Shell Theorem, it is stated that the area of a thin ring of shell of a sphere is $2\pi R^2 \sin(\theta) d\theta$.

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**With shallow search, I could not find any proof of this. I should mention that this question was already asked in here but no good answer was given and furthermore, I have a follow-up question as well.

Therefore, the question is as follows:

Is this just an approximation? If so, if we use another approximation, would we get the same result (I don't think so)? If not, why am I getting a different result:

method 1: Using spherical cap area. here - no further explanation.

method 2:

area$ \approx \sum_{i=1}^{n} 2\pi r_i^* \Delta x \rightarrow \text{area}= \lim_{n \rightarrow \infty} \sum_{i=1}^{n} 2\pi r_i^* \Delta x$

The limit exists, therefore:

area= $\displaystyle \int_{R\cos(\theta+d\theta)}^{R\cos(\theta)} 2\pi \sqrt{R^2 - x^2} \ dx$

For simplification, let $0 \le x \le R$ (instead of $-R \le x \le R$.) Now substitude $x$ with $R\sin(\psi)$ where $0 \le \psi \le \frac{\pi}{2}$ holds. Then

area= $\displaystyle \int_{\pi/2-(\theta+d\theta)}^{\pi/2-\theta} 2\pi R^2 \cos^2(\psi)\ d\psi =\ \left. 2\pi R^2 (\frac{\psi}{2} + \frac{\sin(2\psi)}{4}) \right ]_{\pi/2-(\theta+d\theta)}^{\pi/2-\theta} \neq 2\pi R^2 \sin(\theta) d\theta$

Am I wrong?

Thank you very much for reading this. :)

Answer 1

Method 1 is consistent with approximating the area of a thin ring $2πR^2sin(θ)dθ$, which you can check by integration. But I think the formula for spherical cap area is usually obtained by integration, so this is some sort of circular reasoning.

Method 2 is incorrect. You are missing an infinitesimal length when you substitute $dx$ for $ds = \sqrt{(dx)^2+(dy)^2}$, but after integration the difference will be finite. The correct way to do it is to use $ds$ in your formula.

More generally and rigorously we define the area of some surface by double integration in Cartesian coordinate system, and use a Jacobian to convert an infinitesimal area expressed in $dx,dy$ to $dr,d\theta$.

If we assume $x=rsin(\theta)cos(\phi), y =rsin(\theta)sin(\phi)$ then $J = r^2sin(\theta)cos(\theta)$. So to integrate on a sphere, we write $$I = \int_{-R}^{R}dx\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}dy f(x,y)\frac{1}{cos(\theta)}=\int_{0}^{\pi}d\theta\int_{0}^{2\pi}d\phi f(\theta, \phi) r^2sin(\theta)$$ In shell theorem, $f$ is independent of $\phi$, so we can do the integral for $\phi$ and get our partial result $$I =\int_{0}^{\pi}d\theta f(\theta) 2\pi r^2 sin(\theta)$$