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I'm trying to calculate the surface area of an n-dimensional sphere $\mathbb{S}^n\subset\mathbb{R}^{n+1}$ by means of a differential form. For let $$\omega=\sum_{i=1}^{n+1}(-1)^{i-1}x^{i}dx^{1}\wedge\cdots\wedge \hat{dx^{i}}\wedge\cdots\wedge dx^{n+1}$$ be the volume form of the sphere and take the parameterization $F:[0,\pi]^{n-1}\times[0,2\pi]\to\mathbb{S}^n$ where $$F(y^1,\dots,y^n)=(z^1,\dots,z^n,z^{n+1}) $$

and

$$ \begin{align} z^1&=\cos y^1 \\ z^2&=\sin y^1\cos y^2 \\ z^3&=\sin y^1\sin y^2\cos y^3\\ .\\ .\\ .\\ z^n&=\sin y^1\cdots\sin y^{n-1}\cos y^n\\ z^{n+1}&=\sin y^1\cdots\sin y^{n-1}\sin y^n\\ \end{align} $$

So I've tried to compute $F^{*}\omega$ in order to compute $\int F^{*}\omega$ , but the computations began to turn out very complicated. Can anyone help me with such calculations? Is there a easier way to do these computations ?

Jacaré
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    Yes, the easier way is to use Stokes's Theorem. Either way, you're going to need an inductive approach, as both the area of $S^{n-1}$ and the volume of $D^n$ involve powers of $\pi$ and factorials. – Ted Shifrin Aug 28 '22 at 19:04
  • @TedShifrin Can you make a sketch? Because I've tried to find some patterns in order to use inductive arguments, but I didn't find a smart way to do so. – Jacaré Aug 28 '22 at 19:47
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    There are lots and lots of posts here on this. Just search for volume of n-ball. The most efficient way to approach this is to write $$\text{vol}(D^n) = \int_0^{2\pi}\int_0^1\text{vol}\big(D^{n-2}(\sqrt{1-r^2})\big)r,dr,d\theta = \text{vol}(D^{n-2})\int_0^{2\pi}\int_0^1 (1-r^2)^{(n-2)/2}r,dr,d\theta.$$ – Ted Shifrin Aug 28 '22 at 20:05
  • Maybe you find this help – tryst with freedom Sep 04 '22 at 15:07

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