\[S\]
is smooth if the unit normal \[\mathbf{n}\]
exists and varies continuous over the surface.Take the curve defined by
\[f(x,y,z)=x^2 +y^2 +z^2 -9=0\]
\[\mathbf{\nabla}f= 2x \mathbf{i}+ 2y\mathbf{j} + 2z \mathbf{k}\]
and \[|\mathbf{\nabla} f | = 2 \sqrt{x^2 +y^2 +z^2} = 2 \times 3 =6\]
.\[\mathbf{n} = \frac{\mathbf{\nabla} f}{|\mathbf{\nabla} f|} = \frac{x}{3} \mathbf{i} + \frac{
y}{3} \mathbf{j} + \frac{z}{3} \mathbf{k}\]
.Obviously this is smooth. A cube is not a smooth surf face, The normal does not vary continuously at the edges, moving from face to face, or at the corners.