Determining dimension of manifold

Question

I have a question concerning differential manifolds. I need to prove that $$M=\{z-x=\sqrt{x+y^2},0<x<z\}$$ is a $2$ dimensional manifold. I define the function $F(x,y,z)=z-x-\sqrt{x+y^2}=0$. Obviously it is $C^1$ and its differential has maximum rank so it's a differential manifold. Now I know it's a manifold how should I determine its dimension? I know it's a pretty basic question but I would love to get some help.

Answer 1

The regular value theorem implies that the zero set of $F : \Bbb R^3 \to \Bbb R$, where $F$ has surjective differential, is either empty or a submanifold of dimension $3-1 = 2$.