= Solution
For the <helicoid>,
$$
r_u=(-v\sin u,v\cos u,1),\qquad r_v=(\cos u,\sin u,0),\qquad n=\frac{(-\sin u,\cos u,-v)}{\sqrt{1+v^2}}.
$$
Hence $E=1+v^2$, $F=0$, $G=1$. Since $r_{uu}=(-v\cos u,-v\sin u,0)$ and $r_{vv}=0$, both $L$ and $N$ vanish. The mean-curvature formula therefore gives
$$
\boxed{H=0.}
$$
For comparison, $M=1/\sqrt{1+v^2}$, so $K=-1/(1+v^2)^2$; zero <mean curvature> does not mean zero <Gaussian curvature>.
\Image[/past-exam-of-the-geometry-course-of-the-university-of-oxford/2008-helicoid.png]
{title=The helicoid is ruled by straight lines and has zero mean curvature}
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