Answer
A parametrization of Viviani's curve (Exercise 27) with $\theta$ as the parameter is
${\bf{r}}\left( \theta \right) = \left( {\sin \theta \cos \theta ,{{\sin }^2}\theta ,\sin \theta } \right)$, ${\ \ }$ for ${\ \ }$ $0 \le \theta \le 2\pi $
Work Step by Step
(a) Suppose there exist some $\theta$ such that $x = z\cos \theta $ and $y = z\sin \theta $. Thus,
${x^2} + {y^2} = {z^2}{\cos ^2}\theta + {z^2}{\sin ^2}\theta $
${x^2} + {y^2} = {z^2}$
Therefore, we may write $x = z\cos \theta $, $y = z\sin \theta $, and $z=z$, such that ${x^2} + {y^2} = {z^2}$. Hence, any point on ${x^2} + {y^2} = {z^2}$ can be written in the form $\left( {z\cos \theta ,z\sin \theta ,z} \right)$ for some $\theta$.
(b) Recall from Exercise 27, the Viviani's Curve $C$ is the intersection of the surfaces (Figure 13) ${x^2} + {y^2} = {z^2}$ and $y = {z^2}$.
From part (a), the surface ${x^2} + {y^2} = {z^2}$ can be parametrized by
${\bf{r}}\left( \theta \right) = \left( {z\cos \theta ,z\sin \theta ,z} \right)$,
where $x = z\cos \theta $ and $y = z\sin \theta $, for some $\theta$ in the interval $0 \le \theta \le 2\pi $.
Since $y = {z^2}$, so
$y = z\sin \theta = {z^2}$
It follows that $z = \sin \theta $.
Substituting $x$, $y$ and $z$ in ${\bf{r}}\left( \theta \right)$ gives
${\bf{r}}\left( \theta \right) = \left( {\sin \theta \cos \theta ,{{\sin }^2}\theta ,\sin \theta } \right)$
Hence, a parametrization of Viviani's curve (Exercise 27) with $\theta$ as the parameter is
${\bf{r}}\left( \theta \right) = \left( {\sin \theta \cos \theta ,{{\sin }^2}\theta ,\sin \theta } \right)$, ${\ \ }$ for ${\ \ }$ $0 \le \theta \le 2\pi $
