I am trying the compute the surface of a given hyperboloid: $x^2+y^2-z^2=1$ between the planes $z=-4$ and $z=2$.
- I have found that the surface element is $$d\vec{S}=(\sqrt{1+z^2}cos\theta,\sqrt{1+z^2}sin\theta,-z),$$ so in order to compute the surface I should use: $$dS=||d\vec{S}||=\sqrt{1+z^2+z^2}=\mathbf{\sqrt{1+2z^2}}d\theta dz.$$ So my doubt it is now:
- Why if I use the cilindrical surface element $dS=\rho d\theta dz$, where $\rho=\sqrt{1+z^2}$, giving $$dS=\mathbf{\sqrt{1+z^2}}d\theta dz,$$ I don't get the same answer? It is that $2$ in front of the $z$ that bothers me.
Thanks in advance.