Calculating Surface Area using Differential Forms

Question

I'm trying to reconcile the definition of surface area defined using manifolds vs the classic formula in $\mathbb{R^3}$, but it seems like I'm off by a square.

In Spivak's Calculus on Manifolds, the surface area is defined as $\int_M dA$ where $dA$ is a 2 form, namely$dA = n_1 dy \wedge dz + n_2 dz \wedge dx + n_3 dx \wedge dy$, and $n= (n_1, n_2, n_3) $ is the outward pointing normal. You compute the integral of the manifold by "pulling" the 2 form back to the parameterization coordinates.

The "classic formula" I'm referring to is $Area = \int_s |\frac{\partial G}{\partial u} \times \frac{\partial G}{\partial v}| dA$ where $G(u,v)$ is a parameterization of the surface, and $dA$ means $du dv$ (not to be confused with the definition above). This formula can be found at the bottom of this page: https://tutorial.math.lamar.edu/Classes/CalcIII/ParametricSurfaces.aspx

The problem is, for a simple surface like a sphere, and other surface I've tried, I'm always off by a square. Here is the attempt to find the surface area of a unit sphere using both methods:

Parameterizing with $G(\theta, \phi) = (\cos \theta \sin \phi, \sin \theta \sin \phi, \cos \phi)$ for $\theta \in [0, 2\pi]$ and$\phi \in [0, \pi]$ we calculate: $$dx = -\sin\theta \sin \phi d\theta + \cos \theta \cos \phi d\phi$$ $$dy = \cos \theta \sin d \theta + \sin \theta \cos \phi d \phi$$ $$dz = -\sin \phi d\phi$$

Therefore $$dx \wedge dy = - \sin \phi \cos \phi d \theta d\phi$$ $$dz \wedge dx = - \sin \theta \sin^2 \phi d\theta d \phi$$ $$ dy \wedge dz = - \cos \theta \sin ^2 \phi d \theta d \phi$$

The normal $n$ is calculated to be:

$$n_1 = - \cos \theta \sin ^2 \phi$$ $$n_2 = - \sin \theta \sin^2 \phi $$ $$n_3 = - \sin \phi \cos \phi $$

$n_3$ just happens to be identical to the coefficient function of $dx \wedge dy$.

$n_2$ just happens to be identical to the coefficient function of $dz \wedge dx$.

$n_1$ just happens to be identical to the coefficient function of $dy \wedge dz$.

So then $\int_M dA = \int_{[0,\pi] \times [0,2\pi]} G^*(dA) = \int_0^{\pi} \int_{0}^{2\pi} (\sin^2 \phi \cos^2 \phi + \sin^2 \theta \sin^4 \phi + \cos^2 \theta \sin ^4 \phi )d \theta d\phi $.

This result is off by a square. The "classic formula" has an additional square root since we're taking the norm of $\frac{\partial G}{\partial u} \times \frac{\partial G}{\partial v}$. So under the classic formula $Area = \int_s |\frac{\partial G}{\partial u} \times \frac{\partial G}{\partial v}| dA = \int_0^{\pi} \int_{0}^{2\pi} \sqrt{\sin^2 \phi \cos^2 \phi + \sin^2 \theta \sin^4 \phi + \cos^2 \theta \sin ^4 \phi} d \theta d\phi$.

The solution with the square root is correct. What am I doing incorrectly with respect to the manifold method?

Answer 1

My mistake is that in Spivak's definition $n$ should be the unit outwards pointing normal. Once $n$ is normalized, the two answers coincide.

Answer 2

Ok, I'll try at answer. Suppose we have a parameterization of the surface by a position vector $R(x,y,z) = (x,y,z(x,y) ) $, then we can find the tangent vectors as $\partial_x R $ and $\partial_y R$. We can associate this to two one forms as $r_x$ and $r_y$ and we may think that the surface area is given by the quantity:

$$ \int_D r_x \wedge r_y$$

Over the whole surface. However, it is wrong to evaluate the tangent vector on this because of the following reason: When we evaluate a two form on the tangent vector, it tells you the projection of the area spanned by two vectors on the area element of the two form. If the area spanned by two vector is in same plane as the two form, then you will get square of the area. The way to fix it is to divide the exp by it's magnitude:

$$ \int_D \frac{r_x \wedge r_y}{|r_x \wedge r_y|}$$

One can check that if they try integrating now, they'll get the right answer.