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Let $H_n$ be a $n$-dimensional hypersurface covered by a parametrization $\Phi=\Phi(u_1,\ldots, u_n)$, for $(u_1, \ldots, u_n) \in D$, and let $H_k$ be a k-dimensional hypersurface contained in $H_n$, where $1 \leq k \leq n$, given by $\Psi=\Phi(u_1(t_1, \ldots, t_k), \ldots, u_n(t_1, \ldots, t_k))$ for $(t_1, \ldots, t_k) \in \Delta$.

I want to know what is a formula for area $|H_k|$ of $H_k$.

I know only that

$$|H_n|=\int_{D} \sqrt{g} du_1 \ldots du_n ,$$


$$ |H_1|= \int_{\Delta} \sqrt{\sum_{i,j=1}^k g_{ij}(u_1(t_1), \ldots, u_n(t_1)) \frac{du_i}{dt_1} \frac{du_j}{dt_1}} \ dt_1 ,$$

where $g_{ij}= \langle \Phi|_i , \Phi|_j \rangle$ for $i,j=1, \ldots, n$, and $g=\det[g_{ij}]$.


P.S. I search for formulas which contain only coefficients $g_{ij}$ and derivatives $\frac{\partial u_i}{\partial t_j}$.

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up vote 2 down vote accepted

The metric $g$ on $H_n$ is the pull-back of the metric in the ambient space through $\Phi$, i.e. as you just have said: $g=g_{ij}du^i du^j=\langle\frac{\partial\Phi}{\partial u^i},\frac{\partial\Phi}{\partial u^j}\rangle du^i du^j$.

Now similarly the metric $\overline{g}$ on $H_k$ is the pull-back of the metric $g$ on $H_n$ through the map $t\mapsto u(t)$, i.e. $\overline{g}=\overline{g}_{hk}dt^h dt^k=g_{ij}\frac{\partial u^i}{\partial t^h}\frac{\partial u^j}{\partial t^k}dt^h dt^k$.

So $\overline{g}_{hk}=g_{ij}\frac{\partial u^i}{\partial t^h}\frac{\partial u^j}{\partial t^k}$, and $|H_k|=\int_{\Delta}\sqrt{\overline{g}}dt_1\dots dt_k$.

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Thanks for answer. – Richard Oct 17 '11 at 11:57

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