I know it might seem like a trivial question but I honestly couldn't find this formula anywhere. If I have $$\Gamma(t)=(x(t),y(t),z(t))\qquad t\in[a,b]$$ I want to calculate the area of the rotation surface generated by a parametric curve around the $Z$ axis.
My process
The rotation surface has this equation: $$\Gamma(t)=\begin{pmatrix}x(t)\\y(t)\\z(t)\end{pmatrix}\mapsto \Gamma_R(u,v)= \begin{pmatrix}x(u)\cos(v)-y(u)\sin(v)\\ x(u)\sin(v)+y(u)\cos(v)\\ z(u)\end{pmatrix}\qquad u\in[a,b], v\in[0,2\pi]$$ $$A=\int_0^{2\pi}\int_a^b\left\Vert\frac{\partial\Gamma_R}{\partial u}\times\frac{\partial\Gamma_R}{\partial v}\right\Vert\mathrm{d}u\mathrm{d}v$$ (Various calculations.... for brevity $x:=x(u), y:=y(u)$ and $z:=z(u)$) $${A=\int_{0}^{2\pi}\int_{a}^{b}\sqrt{\left(xx'-xy'\sin(2v)-yy'\cos(2v)\right)^{2}+z'^{2}\left(x^{2}+y^{2}\right)}\mathrm{d}u\mathrm{d}v}$$ Is it possible to simplify this formula further? I would like to remove the dependence on $v$ and highlight a $\pi$ in front of the integral since these formulas generally have that form.
Let us define a curve $\gamma(t) = (x, y, z)(t)$. The area resulting surface around z-axis is given by formula $$ \int\limits_{z_0}^{z_1} 2 \pi r \,ds \overset{\substack{ds = \sqrt{r'(t)^2 + z'(t)^2}\,dt \\ r^2 = x^2 + y^2}}{=} 2 \pi \int\limits_{t_0}^{t_1} \sqrt{(x(t)^2 + y(t)^2)(r'(t)^2 + z'(t)^2)} \, dt. $$
The radius derivative $r'(t)$ corresponds to formula $\frac{x'(t) x(t) + y(t) y'(t)}{\sqrt{x(t)^2 + y(t)^2}}$.