Volume form on $(n-1)$-sphere $S^{n-1}$

Solution 1:

Hint:

The formula for the volume form of a hypersurface $i:S^{n-1}\rightarrow \mathbb{R}^{n}$ is

$$ds=i^{*}(\iota_\nu dV),$$ where $dV$ is a given volume form on the $\mathbb{R}^{n}$ and $\nu\in T\mathbb{R}^{n}$ is a smooth unit normal field to the surface; with interior product (contraction) and the pull-back (restriction) operations. Intuition: $dV$ measures volumes of n-vectors, "feeding" it a unit vector makes it measure "areas" orthogonal to that unit vector.

Example, $S^2$:

We use $\nu=(x,y,z)$; the linearity of $\iota$ gives

$$ ds=x\ \iota_{\partial_{x}}dV+y\ \iota_{\partial_{y}}dV+z\ \iota_{\partial_{z}}dV. $$

Now remember that $dV=dx\wedge dy\wedge dz=dy\wedge dz\wedge dx=dz\wedge dx\wedge dy$; so contracting $dV$ with $\partial_{x^{i}}$ just removes the $dx^{i}$. Thus $$ ds=x\ dy\wedge dz+y\ dz\wedge dx+z\ dx\wedge dy. $$