I'm currently studying how to integrate over manifolds. I want to calculate the surface area of $M=\{x\in \Bbb{R}^4 : x_4^2+x_3^2=x_1^2+x_2^2, 0\le x_1^2+x_2^2\le R^2\}$. I need to find a parametrization of $M$ and use it to calculate the integral (A single parametrization covering $M$ could possibly not be good since it is not defined on an open set but we could ignore a null set of surface area zero to calculate it anyways), but I don't know how to approach that.

Any help would be highly appreciated.


1 Answer 1


I'll would try something like:

$\psi:(0,R)\times (0,2\pi)\times(0,2\pi) \to R^4$

Let $C=(0,R)\times (0,2\pi)\times(0,2\pi)$

$\psi(r,\alpha,\beta) = (rcos(\alpha),rsin(\alpha),rcos(\beta),rsin(\beta)) = (\psi_1(r,\alpha,\beta),\psi_2(r,\alpha,\beta),\psi_3(r,\alpha,\beta),\psi_4(r,\alpha,\beta))$

Indeed then:

$\psi_1^2(r,\alpha,\beta) + \psi_2^2(r,\alpha,\beta) \in (0,R)$

$\psi_1^2(r,\alpha,\beta) + \psi_2^2(r,\alpha,\beta) = \psi_3^2(r,\alpha,\beta) + \psi_4^2(r,\alpha,\beta)$

It's easy to see that $\psi(C) = M$ up to the set of measure 0.

$\nabla\psi_1(r,\alpha,\beta) = [cos(\alpha),-rsin(\alpha),0] $

$\nabla\psi_2(r,\alpha,\beta) = [sin(\alpha),rcos(\alpha),0] $

$\nabla\psi_3(r,\alpha,\beta) = [cos(\beta),0,-rsin(\beta)] $

$\nabla\psi_4(r,\alpha,\beta) = [sin(\beta),0,rcos(\beta)] $

$|\det(D(\psi)(r,\alpha,\beta)^T \cdot D(\psi)(r,\alpha,\beta))|$ = $r^4sin^2(\beta) + r^4cos^2(\beta) + r^4sin^2(\alpha) + r^4cos^2(\alpha) = 2r^4$


$\sigma_3(M) = \sqrt2\int_C r^2 d\lambda_3(r,\alpha,\beta) = \sqrt2\cdot 2\pi \cdot 2\pi \cdot \int_0^R r^2 dr = \frac{4\pi^2\sqrt2}{3}R^3 $

I've used Cauchy-Binett theorem to compute det, and Fubinii theorem to compute the integral.

  • 1
    $\begingroup$ Thank you very much for the elaborated and clear solution! :) $\endgroup$ Apr 12, 2019 at 18:43
  • $\begingroup$ You're welcome :D $\endgroup$
    – Presage
    Apr 12, 2019 at 19:19

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