正螺旋面 r ( u , v ) = ( u cos v , u sin v , b v ) \mathbf{r}(u,v)=(u\cos v,u\sin v,bv) r(u,v)=(ucosv,usinv,bv)是极小曲面
证明：直接计算，有
r u = ( cos v , sin v , 0 ) , r v = ( − u cos v , u sin v , b ) , n = 1 u 2 + b 2 ( b sin v , b cos v , u ) . \mathbf{r}_u=(\cos v,\sin v,0),\:\mathbf{r}_v=(-u\cos v,u\sin v,b),\:\mathbf{n}=\frac{1}{\sqrt{u^2+b^2}}(b\sin v,b\cos v,u). ru=(cosv,sinv,0),rv=(−ucosv,usinv,b),n=u2+b21(bsinv,bcosv,u).
故
E = 1 , F = 0 , G = u 2 + b 2 . \begin{matrix}E=1,&F=0,&G=u^2+b^2.\end{matrix} E=1,F=0,G=u2+b2.
又
r u u = 0 , r u v = ( − sin v , cos v , 0 ) , r v v = ( − u cos v , − u sin v , 0 ) , \mathbf{r}_{uu}=\mathbf{0},\:\mathbf{r}_{uv}=(-\sin v,\cos v,0),\:\mathbf{r}_{vv}=(-u\cos v,-u\sin v,0), ruu=0,ruv=(−sinv,cosv,0),rvv=(−ucosv,−usinv,0),
故
L = 0 , M = − b u 2 + b 2 , N = 0. L=0,\quad M=-\frac{b}{\sqrt{u^2+b^2}},\quad N=0. L=0,M=−u2+b2b,N=0.
从而，平均曲率
H = L G − 2 M F + N E 2 ( E G − F 2 ) = 0. H=\frac{LG-2MF+NE}{2(EG-F^2)}=0. H=2(EG−F2)LG−2MF+NE=0.
即：正螺旋面是极小曲面.
