Volume and area form for the circle and sphere using pullback and integration.

Question

I was trying to compute the area of the sphere using calculus and my knowledge of differential form as follow :

Consider the two form $\omega = dx \wedge dy$, we want to use this form to find out the area of a disk. We define a parametrization of the sphere as follows $F(r,\phi) \rightarrow (rcos\phi, rsin\phi)$ So we have :

\begin{align} \int_{\mathbb{S}^2} \omega = \int_{[0,R]}\int_{[0,2\pi]} F^* \omega = \\ \int_{[0,R]}\int_{[0,2\pi]}rdr \wedge d\phi = \\ \int_{[0,R]}\int_{[0,2\pi]} rdrd\phi = \pi R^2 \end{align}

I was hoping to apply the same principle to find the circumference of the circle but I think I ultimately miss some technicalities.

The circumference of the circle is a 1-dimensional manifold so I am trying to define a one-form on it.I do not know how to proceed from here. Do I have to find a 1-d parametrization of the circle ? Is the one form $dx$ the one to integrate ?

I'm trying to compute the circumference of the circle using pullback an forms. So I can understand the mechanics of integrating forms to find volumes ( and hopefully move to more exotic manifolds like the area or volume of a torus ).

EDIT : My reasoning is as follows. To find the circumference of a circle, let's define a mapping $F:\mathbb{R} \rightarrow \mathbb{R}$ that parametrize it. I was thinking of defining $F:[0,\frac{\pi}{2}] \rightarrow \mathbb{R}$ and multiply the result by 4 ( since the mapping parametrize a quarter of the circle ). Then proceed to integrate $\int_{[0,\frac{\pi}{2}]} F^*(dx)$ Unfortunately the few mapping I tried fail to provide me the right answer.

EDIT2 : Using stereo coordinates (attempt)

Let's consider the map $F:[0,\frac{\pi}{4}] \rightarrow \mathbb{R}$ the stereo projection of $\frac{1}{8}$ of the circumference of the circle, defined as $F(\alpha)=Rtan(\alpha)$ Now we have $\int_{\mathbb{\frac{S^1}{8}}} dx = \int_{[0,\frac{\pi}{4}]} F^*(d\alpha) = R\frac{\pi}{4}$ which when multiplied by 8 gives us the result $2\pi R$

Answer 1

$\newcommand{\Reals}{\mathbf{R}}$Stereographic coordinates give a more or less uniform way to approach the calculation of the $n$-volume of the unit $n$-sphere. If $x = (x_{1}, \dots, x_{n})$ denotes the general element of $\Reals^{n}$, the stereographic parametrization of the $n$-sphere is $$ f(x) = \frac{(2x, \|x\|^{2} - 1)}{\|x\|^{2} + 1} = \left(\frac{2x_{1}}{1 + \sum_{j} x_{j}^{2}}, \dots, \frac{2x_{n}}{1 + \sum_{j} x_{j}^{2}}, \frac{-1 + \sum_{j} x_{j}^{2}}{1 + \sum_{j} x_{j}^{2}}\right). $$ A straightforward (and edifying) calculation shows that the induced metric on $\Reals^{n}$ is $$ g = \frac{4(dx_{1}^{2} + \cdots + dx_{n}^{2})}{1 + \sum_{j} x_{j}^{2}}. $$ The volume form is therefore $$ dV = \left(\frac{2}{1 + \sum_{j} x_{j}^{2}}\right)^{n} dx_{1} \wedge \cdots \wedge dx_{n}. $$ The volume of the unit $n$-sphere $S^{n} \subset \Reals^{n+1}$ is the integral of this volume form over $\Reals^{n}$.

For instance,

• The volume (i.e., length) of the unit circle is $$ \int_{-\infty}^{\infty} \frac{2\, dx}{1 + x^{2}}. $$ The value of this integral is

  $2(\arctan \infty - \arctan(-\infty)) = 2\pi$.

• The volume (i.e., area) of the unit $2$-sphere is $$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{4\, dx\, dy}{(1 + x^{2} + y^{2})^{2}}, $$ whose value (by converting to polar coordinates) is

  $\displaystyle\int_{0}^{2\pi} \int_{0}^{\infty} \frac{4r\, dr\, d\theta}{(1 + r^{2})^{2}} = 4\pi \int_{1}^{\infty} \frac{du}{u^{2}} = 4\pi$.

To calculate these for arbitrary $n$, it may be easier to calculate the volume of the ball of radius $r$ (using Cavalieri's theorem and recursion, see for example Where does this formula for the volume of a $n$-dimensional ball come from? or Wikipedia's page on the $n$-sphere), then use the fact that if the $n$-ball has volume $V_{n}(r) = V_{n}(1)r^{n}$, then the $(n-1)$-sphere has $(n-1)$-dimensional volume $$ \frac{d}{dr} V_{n}(r) = nV_{n}(1)r^{n-1}. $$

You may also enjoy attempting the calculation in generalized spherical coordinates, a nice application of the reduction formula for integrating powers of sine.

Answer 2

Area of sphere

You should use spherical coordinate for the sphere: $$ F:(\phi,\psi)\in[0,2\pi]\times\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\mapsto(r\cos\phi\cos\psi,r\sin\phi\cos\psi,r\sin\psi)\in\mathbb R^3 $$ Now you want to integrate the volume form induced from $\mathbb R^3$ by the immersion $F$, which is given by: $$ \Omega=\sqrt{\det g}~\mathrm d\phi\wedge\mathrm d\psi $$ Now you need to compute the induced $2\times 2$ Riemannian metric matrix $g$, $$ g=\begin{pmatrix} g_{\phi\phi} & g_{\phi\psi}\\ g_{\psi\phi} & g_{\psi\psi} \end{pmatrix} $$ where $g_{*\star}=\langle X_{*},X_{\star}\rangle$ and $X_{*}$,$X_{\star}$ are the tangent vectors $\frac{\partial F}{\partial(*)}$, $\frac{\partial F}{\partial(\star)}$. Now $$ \begin{aligned} X_{\phi}&=(-r\sin\phi\cos\psi,r\cos\phi\cos\psi,0)\\ X_{\psi}&=(-r\cos\phi\sin\psi,-r\sin\phi\sin\psi,r\cos\psi) \end{aligned} $$ and therefore: $$ g=\begin{pmatrix} r^2\cos^2\psi & 0\\ 0 & r^2 \end{pmatrix} $$ This leads to $$ \begin{aligned} A_{S^2}&=\int_{S^2}\Omega=\int_{0}^{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\det g}~\mathrm d\phi\wedge\mathrm d\psi\\ &=\int_{0}^{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}r^2\cos\psi\mathrm d\phi\mathrm d\psi\\ &=r^2\int_{0}^{2\pi}\mathrm d\phi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos\psi\mathrm d\psi=r^2\cdot 2\pi\cdot 2=4\pi r^2 \end{aligned} $$

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Circumference of circle

This is nothing different from calculating area of the sphere.

$$ F:\phi\in[0,2\pi]\mapsto(r\cos\phi,r\sin\phi)\in\mathbb R^2 $$

$$ \Omega=\sqrt{\det g}~\mathrm d\phi $$

$$ g=g_{\phi\phi}=\langle X_{\phi},X_{\phi}\rangle $$

$$ X_{\phi}=(-r\sin\phi,r\cos\phi)\Rightarrow g=r^2 $$

$$ A_{S^1}=\int_{S^1}\Omega=\int_{0}^{2\pi}r d\phi=2\pi r $$