= Solution
Set $C=\cos u$, $S=\sin u$, and $D=1+S^2=2-C^2$. At $v=0$, the first <partial derivatives> are
$$
\mathbf r_u=(-S,C,0),\qquad\mathbf r_v=(-S,1+C,1).
$$
Their <dot products> give $E=1$, $F=1+C$, $G=3+2C$. Hence
$$
\boxed{I=du^2+2(1+\cos u)\,du\,dv+(3+2\cos u)\,dv^2.}
$$
Since $EG-F^2=3+2C-(1+C)^2=D>0$, the <parametric surface> is regular everywhere. Its <cross product> and <unit normal> at $v=0$ are
$$
\mathbf r_u\times\mathbf r_v=(C,S,-S),\qquad N=\frac{(C,S,-S)}{\sqrt D}.
$$
The second <partial derivatives> at $v=0$ are
$$
\mathbf r_{uu}=(-C,-S,0),\qquad
\mathbf r_{uv}=(-C,-S,0),\qquad
\mathbf r_{vv}=(-1-C,-S,0).
$$
Their <dot products> with $N$ give $e=f=-1/\sqrt D$ and $g_2=-(1+C)/\sqrt D$. Thus
$$
\boxed{II=-\frac{du^2+2\,du\,dv+(1+\cos u)\,dv^2}{\sqrt{1+\sin^2u}}.}
$$
Part (b) transports these <first fundamental form> and <second fundamental form> coefficients to every $v$. Finally,
$$
\boxed{\kappa(u,v)=\frac{(1+C)/D-1/D}{D}=\frac{\cos u}{(1+\sin^2u)^2},\qquad dA=\sqrt{1+\sin^2u}\,du\,dv.}
$$
The last expression is the <surface area> element from $\sqrt{EG-F^2}$. In particular the <Gaussian curvature> changes sign where $\cos u$ does; it vanishes at $u=\pi/2$ modulo $\pi$.
Back to article page
