# 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 $$S^{n-1}$$.

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.

• Many thanks! Indeed the determinant formula is the key to the problem. I missed that. – guest61 May 26 '19 at 18:14