calculate the surface of the manifold in $\Bbb{R}^4$

Question

How to calculate the surface area of the following manifold :

$$ x_1^2 + x_2^2 = x_3^2 + x_4^2, 0 \le x_1^2+x_2^2 \le a^2$$

I know I should first describe this manifold as a map or a graph of a function.

but I dont know how to describe this manifold.

Answer 1

Imagine $x_4$ as time coordinate and the others as space coordinates. At time $x_4=0$ you have $x_1^2+x_2^2-x_3^2=0$, which are dual cones with height $a$ (according to the side condition). At times $0<x_4<a$ you have one-sheeted hyperboloids with $x_4$ as smallest diameter at $x_3=0$ and $2(a-x_4)$ as overall height. Finally, at time $x_4=a$ you get a circular disk of radius $a$ at $x_3=0$.

So by eploiting the symmetry regarding the $x_3$ plane, you should be able to set up the integral.