Volume form on a smooth hypersurface in $\mathbb{R}^n$

Question

This is an exercise from chapter 16 of Loring Tu's "Differential Geometry" textbook.
Let $f:\mathbb{R}^n\to\mathbb{R}$ be a smooth function whose partial derivatives do not simultaneously vanish on the zero locus $Z(f)$. I have shown that the gradient vector field $\text{grad}f=\sum\frac{\partial f}{\partial x^i}\frac{\partial}{\partial x^i}$ is a unit normal vector field to the surface, when divided by its norm. I am then asked to compute the volume form on $Z(f)$, with the following in mind:
an orthonormal frame $(e_2,...,e_n)$ at $p\in Z(f)$ is said to be positively oriented if and only if $(X,e_2,...,e_n)$ is positively oriented for $T_p\mathbb{R}^n$.
The main way that I know of computing the volume form, is to find an orthonormal frame, take the dual frame, and then compute the interior multiplication with a unit normal vector field on the surface. In fact, other methods have not been presented in the book thusfar. However, it doesn't seem to me that I have enough information to be able to compute the volume form, explicitly. Am I mistaken? If yes, what am I missing, and if not, what else could the question be asking?

Answer 1

You should know the following convenient formula, which follows immediately from what you've said. If $\vec A = (a^1,\dots,a^n)$ is the unit normal to oriented hypersurface [I would usually use $\vec n$, but that doesn't work too well with our being in $\Bbb R^n$.], then the volume form for the hypersurface is $$\sigma = \sum_{i=1}^n (-1)^{i-1} a^i dx^1\wedge\dots\wedge\widehat{dx^i}\wedge\dots\wedge dx^n.$$