Assume for $\xi\in S^{n1}$ the parametrization of a closed hypersurface is given by $x(\xi)=R(\xi)\xi\in \mathbb R^n$. Here $R: S^{n1}\to \mathbb R$ is a positive function. Is there a reference for a proof of the formula \begin{eqnarray*} dS_{R}=R^{n2}\sqrt{R^2+\vert\nabla R(\xi)\vert^2}\:dS_{\xi}\:? \end{eqnarray*} Notation: $dS_R$ denotes the area element on $\{R(\xi)\xi:\xi\in S^{n1}\}$ and $dS_{\xi}$ denotes the area element on $S^{n1}$.
1 Answer
Other than the chain rule, I think the only ingredient needed for this is the following formula for the determinant of a rank 1 perturbation of an invertible matrix; for $d\in \mathbb{N}$, $A\in \mathrm{GL}(d)$ and $v\in \mathbb{R}^d$,
$$ \mathrm{det}(A+vv^T) = \mathrm{det}(A)(1 + v^TA^{1}v). $$
Writing $\hat x$ for the embedding $S^{n1}\to \mathbb{R}^n$, the chain rule gives you the local expression
$$ \dfrac{\partial x}{\partial u^i} = R\dfrac{\partial \hat x}{\partial u^i} + \dfrac{\partial r}{\partial u^i}\hat x $$
in any chart $(U,(u^i)_{i=1}^{n1})$ on $S^{n1}$. Letting $g$ be the metric tensor for your hypersurface, $g$ is related to the metric tensor $\hat g$ on the sphere via
$$ g_{ij} = R^2 \hat g_{ij} + \dfrac{\partial r}{\partial u^i}\dfrac{\partial r}{\partial u^j}. $$
NB: this uses the fact that $\hat x \cdot \hat x = 1$ and $\hat x \cdot (\partial \hat x/\partial u^i)=0$.
If you write this as $g = R^2 \hat g + vv^T$ with $v$ the vector with components $\partial r/\partial u^i$, the formula above gives you
$$ \begin{array}{lll} \mathrm{det}(g) &=& \mathrm{det}(R^2 \hat g)(1 + v^T(R^2\hat g)^{1}v) \\ &=& R^{2n2}\left(R^2+ g^{ij}\dfrac{\partial r}{\partial u^i}\dfrac{\partial r}{\partial u^j} \right)\mathrm{det}(\hat g) \\ &=& R^{2n2}(R^2 + \nabla R^2)\mathrm{det}(\hat g), \end{array} $$
which is what you want when combined with the usual expression for the volume form on a Riemannian manifold.

$\begingroup$ Many thanks! Indeed the determinant formula is the key to the problem. I missed that. $\endgroup$– guest61May 26, 2019 at 18:14