- #1

Rasalhague

- 1,387

- 2

I'm trying out the technique described in Bachman:

*A Geometric Approach to Differential forms*, § 6.2 "Integrating 2-forms". To take a simple example, I tried to use this method to get the surface area of a sphere, S

^{2}, with radius R, embedded in

**R**

^{3}. I think the 2-form I need to integrate is

[tex]\omega = dy \wedge dz + dz \wedge dx + dx \wedge dz.[/tex]

I've chosen the obvious parameterization

[tex]\psi : \mathbb{R}^2 \rightarrow \mathbb{R}^3 \; \bigg| \; \psi (\zeta, \alpha) = R \; (\sin \zeta \cos \alpha, \sin \zeta \sin \alpha, \cos \zeta)[/tex]

where [itex]0 \leq \zeta \leq \pi[/itex] is the zenith angle, and [itex]0 \leq \alpha \leq 2 \pi[/itex] the azimuth angle. Let [itex]\textbf{x}=\psi(\zeta, \alpha)[/itex]. Then

[tex]\frac{\partial \textbf{x}}{\partial \zeta} = R \; (\cos \zeta \cos \alpha, \cos \zeta \sin \alpha, -\sin \zeta)[/tex]

[tex]\frac{\partial \textbf{x}}{\partial \alpha} = R \; (-\sin \zeta \sin \alpha, \sin \zeta \cos \alpha, 0)[/tex]

Barring copying errors on my part, mathematica confirms this.

[tex]dy \wedge dz\left ( \frac{\partial \textbf{x}}{\partial \zeta}, \frac{\partial \textbf{x}}{\partial \alpha} \right ) = \begin{vmatrix}

R \cos \zeta \sin \alpha & R \cos \alpha \sin \zeta\\

-R \sin \zeta & 0

\end{vmatrix} = R^2 \cos \alpha \; \sin^2 \zeta[/tex]

[tex]dz \wedge dx \left ( \frac{\partial \textbf{x}}{\partial \zeta}, \frac{\partial \textbf{x}}{\partial \alpha} \right ) = \begin{vmatrix}

-R \sin \zeta & 0 \\ R \cos \zeta \cos \alpha & -R \sin \zeta \sin \alpha

\end{vmatrix} = R^2 \sin \alpha \; \sin^2 \zeta[/tex]

[tex]dx \wedge dy \left ( \frac{\partial \textbf{x}}{\partial \zeta}, \frac{\partial \textbf{x}}{\partial \alpha} \right ) = \begin{vmatrix}

R \cos \zeta \cos \alpha & -R \sin \zeta \sin \alpha \\ R \cos \zeta \sin \alpha

& R \sin \zeta \cos \alpha

\end{vmatrix} = R^2 \cos \zeta \sin \zeta[/tex]

But according to Mathematica,

[tex]R^2 \int_0^{2\pi} \int_0^\pi \left [ \sin^2 \zeta (\cos \alpha + \sin \alpha) + \cos \zeta \sin \zeta \right ] d \zeta \wedge d \alpha = 0[/tex]

What went wrong?

Code:

```
x = R {Sin[z] Cos[a], Sin[z] Sin[a], Cos[z]};
tz = D[x, z]; ta = D[x, a];
Integrate[
Simplify[Det[{{tz[[2]], ta[[2]]}, {tz[[3]], ta[[3]]}}] +
Det[{{tz[[3]], ta[[3]]}, {tz[[1]], ta[[1]]}}] +
Det[{{tz[[1]], ta[[1]]}, {tz[[2]], ta[[2]]}}]], {z, 0, Pi}, {a, 0,
2 Pi}]
```

Mathematica gave the same results for each of the matrices and their determinants as I got by hand.