Differential Geometry and Its Application - Exercise 2.1.11

Problem:

For a surface of revolution:

$x(u, v) = (g(u), h(u)\cos v, h(u) \sin v)$

Check that $x_u \times x_v = h(\frac{dh}{du}, -\frac{dg}{du}\cos v, -\frac{dg}{du}\sin v)$.

Why is $x_u \times x_v \neq 0$ for all $u$, $v$?

Solution:

$x_u = (g'(u), h'(u) \cos v, h'(u) \sin v)$
$x_v = (0, -h(u) \sin v, h(u) \cos v)$

$\begin{eqnarray*} & & x_u \times x_v \\ &=& \left|\begin{array}{ccc}i & j & k \\ g'(u) & h'(u) \cos v & h'(u) \sin v \\ 0 & -h(u) \sin v & h(u) \cos v\end{array}\right| \\ &=& ((h'(u)\cos v) (h(u) \cos v) - (h'(u)\sin v)(-h(u)\sin v))i + ((h'(u)\sin v) (0) - (g'(u))(h(u)\cos v)) j + ((g'(u)) (-h(u)\sin v) - (h'(u)\cos v)(0)) k \\ &=& h(u)(h'(u), -\cos(v), -\sin(v)) \end{eqnarray*}$

If $h(u) = 0$, all bets are off because in fact the normal is $0$. Assuming $h(u) \neq 0$, then the normal vector is always non zero because $\cos v$ and $\sin v$ can never be 0 at the same $v$.