Express a surface integral of an axisymmetric function on an axisymmetric surface? An axisymmetric surface is represented by continuous and smooth function $z=F(r)$, where $0\le r \le R$, in a cylindrical coordinate $(r,z)$.
Now, I need to calculate the surface integral of an axisymmetric function $g(r)$, which is defined on the surface $z=F(r)$, with the follow expression
$$G=2 \pi \int_0^R r (1+F_r^2)^{1/2}g(r) \mathrm{d} r.$$
Is it true? Thank you very much.
 A: Let's describe the surface by a vector function
$${\bf x}(r, \phi) = r\cos\phi\ {\bf e}_x + r\sin\phi\ {\bf e}_y + F(r) {\bf e}_z.$$
At each point $(r, \phi)$ tangent vectors corresponding to each of these two coordinates are
$${\bf T}_r = \frac{\partial}{\partial r}{\bf x}(r, \phi) = \cos \phi \ {\bf e}_x + \sin \phi\ {\bf e}_y + F'(r)\ {\bf e}_z,\\
{\bf T}_\phi = \frac{\partial}{\partial \phi}{\bf x}(r, \phi) = -r \sin \phi \ {\bf e}_x + r \cos \phi\ {\bf e}_y.$$
The surface integral can be expressed through these tangent vectors as
$$\int\limits_S g({\bf x})\ dS = \iint\limits_S g(r) \left|{\bf T}_r \times {\bf T}_\phi\right|\ dr \ d\phi. $$
Calculating the cross product in ${\bf e}_x, {\bf e}_y, {\bf e}_z$ basis and taking its norm it is easy to obtain that
$${\bf T}_r \times {\bf T}_\phi = \left|
\begin{matrix}
{\bf e}_x & {\bf e}_y & {\bf e}_z \\
\cos \phi & \sin \phi\ & F'(r) \\
-r\sin \phi  &  r\cos \phi\ & 0
\end{matrix}
\right| = r\ F'(r)\cos\phi\ {\bf e}_x - r\ F'(r)\sin\phi\ {\bf e}_y + r\
 {\bf e}_z$$
and $$\left|{\bf T}_r \times {\bf T}_\phi\right| = r\sqrt{1 + \left(F'(r)\right)^2},$$
so it is possible to substitute this expression to the integral and integrate by $\phi$ from $0$ to $2\pi:$
$$\iint\limits_S g(r) \left|{\bf T}_r \times {\bf T}_\phi\right|\ dr \ d\phi  =
\int\limits_{0}^{2\pi} d\phi \int\limits_{0}^{R} dr\  g(r)\ r \sqrt{1 + \left(F'(r)\right)^2} 
= 2\pi \int\limits_{0}^R g(r)\ r \sqrt{1 + \left(F'(r)\right)^2}\ dr.$$
