Rotaion Surfaces and Complex Numbers

Question

Consider a continuous invertible map $\varphi:\mathbb{R}^+ \longrightarrow \mathbb{R}$, and define the follwing surface $$ s:\mathbb{C} \longrightarrow \mathbb{R} \times \mathbb{C} $$ $$ \qquad xe^{i\theta} \longmapsto (x, \varphi(x)e^{i\theta}) $$ Can we say that $s$ defines a manifold which is isomorphic to the rotation surface generated by $\varphi$? In addition, consider a disc $$ \mu : [0, r)\times[0, 2\pi)\longrightarrow \mathbb{C} $$ $$ \qquad \qquad \quad \ (x, \theta) \longmapsto xe^{i\theta} $$ I would like to compute the integral $$ \int_{[0, r)\times [0, 2\pi)}s\circ \mu $$ In other words, I would like to find a formula to compute the "area" of a disc that "streches out" on the surface defined by $s$. I need such a formulation for practical purposes. Please forgive me if the notation is ambiguous, or if the question is obvious to theoreticians.

Answer 1

This is actually one of the standard formulas in a first-year calculus course for the surface area of a surface of revolution. You want to compute the integral $$\int_0^{2\pi}\int_0^r \phi(u)\sqrt{1+\phi'(u)^2}\,du\,d\theta.$$ (This is the surface obtained by rotating the graph $y=\phi(x)$ about the $x$-axis.)

Alternatively, you can derive this using the formula for a surface integral with a parametrization $\mathbf g(r,\theta)$, $$\iint \left\|\frac{\partial\mathbf g}{\partial r}\times\frac{\partial\mathbf g}{\partial\theta}\right\|\,dr\,d\theta,$$ which you'll find in the latter parts of any calculus book.