Solution
= Solution
The <cylinder> is a <level set> of $h=x^2+y^2-8$, so its outward <normal vector> is parallel to $\nabla h=(2x,2y,0)$. At the point this is $(4,4,0)$, with norm $4\sqrt2$. Choosing the outward <orientation> gives
$$
\boxed{\hat{\mathbf n}=\frac{(1,1,0)}{\sqrt2}}.
$$
Its negative is the equally valid inward <unit normal vector>; the next <directional derivative> then changes sign.