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I have a formula for computing the area of the surface of any object in $\Bbb R^3$, namely given some parametrization $\Phi(s,t)$, I take the cross product of the partial derivative with respect to each variable and norm it.

This looks suspiciously like a wedge product, especially when you think of the cross product as the determinant of the $3\times 3$ with the top row as the $e_i$ vectors.

How do I connect the two concepts? Concretely, I tried to integrate the surface area of a sphere using the differential 2-form $dx\wedge dy$ on the obvious parametrization and kept getting stuck, finding the area was zero.

janmarqz
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operatorerror
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2 Answers2

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Use Riemannian geometry. Put spherical coordinates $(\theta,\phi)$ on the unit sphere $S^2 \in \mathbb{R}^3$, given by

$$F(\phi, \theta) = (x,y,z) = (\sin\phi \cos\theta, \, \sin\phi \sin\theta, \, \cos\phi) $$

The standard Riemannian metric on $\mathbb{R}^3$ is $g = dx \otimes dx + dy \otimes dy + dz \otimes dz$. To get the metric in the above coordinate system, pull-back this tensor by the coordinate map $F$:

$$ F^*g = d\phi \otimes d\phi + \sin^2(\phi) \,\, d\theta \otimes d\theta $$

Writing $g$ (or rather, $F^*g$) as a matrix, we have

$$ g = \left( \begin{array}{cc} 1 & 0 \\ 0 & \sin^2(\phi) \end{array} \right) $$

The Riemannian volume form in coordinates is given by $dV = \sqrt{\mathrm{det}(g)} d\phi \wedge d\theta$. In this case, that is given by $\sin(\phi) d\phi \wedge d\theta$. So you can integrate this over the region $(0,\pi) \times (0,2\pi)$ in the $(\phi,\theta)$-plane to get the area of the sphere.

$$ \int_0^{2\pi} \int_0^\pi \sin(\phi) \, d\phi \wedge d\theta $$

This gives $4\pi$, which is the surface area of the unit sphere.

Nick
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Here's the calculus method:

Begin with parametrization of a sphere with radius R. $$S=\big(R\sin(\phi)\cos(\theta),R\sin(\phi)\sin(\theta), R\cos(\phi)\big)$$ For $\theta\in[0, 2\pi], \phi\in[0, \pi]$.
The equation for the surface area $A$ of a parametric surface $S$ is: $$A=\iint_D \big|\,S_\phi \times S_\theta\big|\,dA$$ Order doesn't matter because you are taking the magnitude.
$\big|\,S_\phi\times S_\theta\big|=R^2\sin(\phi).$ - This part sucks, but just use $\sin^2(x)+\cos^2(x)=1$ a few times and you'll get there. $$A=R^2\int_0^\pi\int_0^{2\pi}\sin(\phi)\,d\theta\,d\phi=2\pi R^2\int_0^\pi \sin(\phi)\,d\phi=4\pi R^2.$$

Eli Berkowitz
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