How to find the surface element for the cylinder $x^2 + y^2 = r^2$? So if given a surface (cylindrical) which has radius r and equation $x^2 + y^2 = r^2$, I want to work out the line element for it.
How do I get it?
I know the final answer has to be $dS^2 = r^2dϕ^2 + dz^2$ as that is what the textbook gave but I don't know how to get to that. I'm sure it has something to do with cylindrical coordinates but not sure how to apply it
Any help would be appreciated~
 A: Lengths of curves, ares, angles ans so on are all based on a choice of an inner product on the tangent space at each point. In other words, given a parametrized curve $\alpha:(-\epsilon,\epsilon)\to S$on our surface, we want to measure its "velocity" - we want $\dot{\alpha}$ to have a norm. 
The above cylinder is parametrized locally by  $\varphi(\theta,z)=(r\cos\theta,r\sin\theta,z)$, where $\varphi$ is defined on some open rectangle $A=(\theta_0,\theta_1)\times(z_0,z_1)$. Given a path $\alpha:(-\epsilon,\epsilon)\to A,t\mapsto(\theta(t),z(t))$, the induced path $\varphi\circ\alpha$ is given by $t\mapsto(r\cos\theta(t),r\sin(\theta(t),z(t))$. The velocity is
$\frac{d}{dt}\varphi\circ\alpha=(-r\sin\theta\dot{\theta},r\cos\theta\dot{\theta},\dot{z})$,
and the squared norm of the velocity turns out to be
$\langle\frac{d}{dt}\varphi\circ\alpha,\frac{d}{dt}\varphi\circ\alpha\rangle=r^2\sin^2\theta\dot{\theta}^2+r^2\cos^2\theta\dot{\theta}^2+\dot{z}^2=r^2\dot{\theta}^2+\dot{z}^2$.
The induced inner product is thus given by $\|\partial_\theta\|^2=r^2,\|\partial_z\|^2=1,\langle\partial_\theta,\partial_r\rangle=0$.
