Solution
= Solution
For the <surface of revolution>,
$$
\mathbf r_u=(f'\cos v,f'\sin v,1),\qquad
\mathbf r_v=(-f\sin v,f\cos v,0).
$$
The mixed <inner product> vanishes, giving
$$
\boxed{I=(1+f'(u)^2)du^2+f(u)^2dv^2.}
$$
Taking $f(u)=\cosh u$ makes both coefficients $\cosh^2u$, as used in the parent calculation.