Area formula for parametric surfaces Assume for $\xi\in S^{n-1}$ the parametrization of a closed hypersurface is given by $x(\xi)=R(\xi)\xi\in \mathbb R^n$. Here $R: S^{n-1}\to \mathbb R$ is a positive function. Is there a reference for a proof of the formula
\begin{eqnarray*}
dS_{R}=R^{n-2}\sqrt{R^2+\vert\nabla R(\xi)\vert^2}\:dS_{\xi}\:?
\end{eqnarray*}
Notation: $dS_R$ denotes the area element on $\{R(\xi)\xi:\xi\in S^{n-1}\}$ and $dS_{\xi}$ denotes the area element on $S^{n-1}$.
 A: 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 $S^{n-1}\to \mathbb{R}^n$, 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 $(U,(u^i)_{i=1}^{n-1})$ on $S^{n-1}$. 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
$$
\begin{array}{lll}
\mathrm{det}(g) &=& 
\mathrm{det}(R^2 \hat g)(1 + v^T(R^2\hat g)^{-1}v) 
\\
&=& R^{2n-2}\left(R^2+ g^{ij}\dfrac{\partial r}{\partial u^i}\dfrac{\partial r}{\partial u^j} \right)\mathrm{det}(\hat g)
\\ 
&=& R^{2n-2}(R^2 + |\nabla R|^2)\mathrm{det}(\hat g),
\end{array}
$$
which is what you want when combined with the usual expression for the volume form on a Riemannian manifold. 
