Answer to Question #143635 in Differential Geometry | Topology for anjali

Question

Calculate the principal, Gaussian, and mean curvatures of the surface of revolution
σ(u, v) = (f(u) cos v, f(u) sin v, g(u)), where f, g are smooth real-valued functions
with f > 0.

Answer 1

$\sigma_u=(f' cos \ v, f' sin \ v, g' ),$ $\sigma_v=(-f sin \ v, f cos \ v, 0 ),$ $\sigma_{uu}=(f'' cos \ v, f'' sin \ v, g'' ),$ $\sigma_{vv}=(-f cos \ v, f sin \ v, 0 ).$ We note $\sigma_u\cdot\sigma_v=0.$ $\hat{n}=\frac{\sigma_u\times\sigma_v}{||\sigma_u\times\sigma_v||}=$ $\hat{n}=\frac{(fg'cos \ v , -fg' sin \ v ,ff')}{\sqrt{g'^2 f^2+f'^2f^2}}.$ Hence $\sigma_{uu}\cdot \hat{n}=-\frac{f}{|f|}\frac{g''f'-g'f''}{\sqrt{g'^2+f'^2}}$ , $\sigma_{vv}\cdot \hat{n}= -\frac{|f|g'}{\sqrt{g'^2+f'^2}}.$ Hence, we get, that the principal curvatures are $\frac{\sigma_{uu}\cdot \hat{n}}{\sigma_u\cdot\sigma_u}=-\frac{f}{|f|}\frac{g''f'-g'f''}{(g'^2+f'^2)^{3/2}}$ and $\frac{\sigma_{vv}\cdot \hat{n}}{\sigma_v\cdot\sigma_v}=-\frac{g'}{|f|\sqrt{g'^2+f'^2}}$ .

Hence the Gaussian curvature is product of the two= $\frac{1}{f}\frac{g'(g''f'-g'f'')}{(g'^2+f'^2)^{2}}$ and mean curvature = mean of the two= $\frac{f(-g''f'+g'f'')-g'(g'^2+f'^2)}{2|f|(g'^2+f'^2)^{3/2}}$