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The following equation represents a surface in $\mathbb{R}^{4}$, that is a 3-dimensional manifold: $$ x_{3}^{2}+x_{4}^{2}=x_{1}^{2}+x_{2}^{2}\qquad\text{s.t.}\qquad 0<x_{1}^{2}+x_{2}^{2}<R^{2} $$ We can see that by defining $F\colon\mathbb{R}^{4}\to\mathbb{R}$ to be $$F\left(x_{1},x_{2},x_{3},x_{4}\right)=x_{1}^{2}+x_{2}^{2}-x_{3}^{2}-x_{4}^{2}$$ so that $\nabla F\left(x\right)=2\left(x_{1},x_{2},-x_{3},-x_{4}\right)\neq0$, and the manifold is $M=F^{-1}\left(0\right)$.

My question is how can we parameterize this surface? I'm looking for a $V\subset\mathbb{R}^{3}$ and a function $r\colon V\to\mathbb{R}^{4}$ such that $M=r\left(V\right)$. That is in order to compute the surface area of $M$.

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$0<x_1^2+x_2^2 < R^2$

$x_1,x_2$ lie on a disk.

$x_3^2 + x_4^2 = x_1^2+x_2^2$ puts $x_3, x_4$ on a corresponding disk.

$x_1 = \rho \cos \theta\\ x_2 = \rho \sin\theta\\ x_3 = \rho \cos \phi\\ x_4 = \rho \sin \phi$

$0<\rho < R\\ 0\le \theta < 2\pi\\ 0\le \phi < 2\pi$

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