# Calculate the integral over a surface.

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.

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.

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

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

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

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

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

So:

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

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

• Thank you very much for the elaborated and clear solution! :) – Tamir Shalev Apr 12 at 18:43
• You're welcome :D – Presage Apr 12 at 19:19