Describe the area of the following manifold: $M=\{(x,y,z)|x^2+y^2=f(z)^2, 0\lt z \lt 1\}$, as a one dimensional integral.
My attempt -
I define $F(x,y,z) = x^2+y^2-f(z)^2$ and therefore $M=\{(x,y,z)| F(x,y,z) = 0\}$.
The formula(as much as I know) for such defintion of manifolds over $\mathbb{R}^3$ is: $\int_M dA = \int F(x,y,z)\frac{\sqrt{|\nabla F|}}{|F_z|}dxdydz$.
In our case: $F_z = -2f(z)f'(z)$, and $|\nabla F|= |4x^2+4y^2+4f(z)^2f'(z)^2|$.
Putting it all together: $\int_M dA = \int_0^1\int_0^{f(z)^2}\int_0^{f(z)^2} (x^2+y^2-f(z)^2)\frac{\sqrt{|4x^2+4y^2+4f(z)^2f'(z)^2|}}{|-2f(z)f'(z)|}dxdydz$
But I'm really not sure how to continue from here... Any hints?
Or perhaps, is there any way doing it using some parameterization?
Switching to cylinder coordinates gives $r^2 = f(z)^2$, so $r = |f(z)|, z\in(0,1), \theta \in [0,2\pi]$. Your surface is then composed of horizontal strips (perpendicular to the $z$-axis) of thickness $dz$, radius $r(z) = |f(z)|$, and hence area $2\pi |f(z)| dz$. So the total area would be
$$ A = 2 \pi \int_0^1 |f(z)| dz $$