# How to parameterize this surface: $x_{3}^{2}+x_{4}^{2}=x_{1}^{2}+x_{2}^{2}$ s.t. $0<x_{1}^{2}+x_{2}^{2}<R^{2}$?

The following equation represents a surface in $$\mathbb{R}^{4}$$, that is a 3-dimensional manifold: $$x_{3}^{2}+x_{4}^{2}=x_{1}^{2}+x_{2}^{2}\qquad\text{s.t.}\qquad 0 We can see that by defining $$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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$$x_1,x_2$$ 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.
$$x_1 = \rho \cos \theta\\ x_2 = \rho \sin\theta\\ x_3 = \rho \cos \phi\\ x_4 = \rho \sin \phi$$
$$0<\rho < R\\ 0\le \theta < 2\pi\\ 0\le \phi < 2\pi$$