Connecting the formula for the surface area of a sphere to differential forms 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. 
 A: 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.
A: 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.
$$A=R^2\int_0^\pi\int_0^{2\pi}\sin(\phi)\,d\theta\,d\phi=2\pi R^2\int_0^\pi \sin(\phi)\,d\phi=4\pi R^2.$$
