Find a parametrization of the surface $S$ defined by the cone $$z=\sqrt{3x^{2}+3y^{2}}\qquad x^{2}+y^{2}\leq 16$$

Solution (a) The equation is given in rectangular coordinates and is written explicitly as a function $z=f(x,y) .$ So, we define $x=u,$ $y=v,$ $z=\sqrt{3u^{2}+3v^{2}}.$ Then a parametrization of the cone is $$\mathbf{r}(u,v) =u\mathbf{i}+v\mathbf{j}+\sqrt{3u^{2}+3v^{2}}\, \mathbf{k}\qquad u^{2}+v^{2}\leq 16$$

(b) In cylindrical coordinates, the cone is given by $$z=\sqrt{3x^{2}+3y^{2}}=\sqrt{3}\sqrt{( r\cos \theta ) ^{2}+(r\sin \theta ) ^{2}}=\sqrt{3}\sqrt{r^{2}}=\sqrt{3}\,r$$

where $z$ is expressed explicitly in terms of $r$ and $\theta .$ So if we use $r$ and $\theta$ as parameters, then a parametrization of the cone is $$\begin{eqnarray*} \mathbf{r}( r,\theta ) =r\cos \theta \,\mathbf{i}+ \underset{\underset{\color{#0066A7}{x=r\cos \theta ,y= r\sin \theta , z=\sqrt{3}r}}{\color{#0066A7}{\uparrow}}}{r\sin \theta \, \mathbf{j}}+\sqrt{3}r\,\mathbf{k}\qquad 0\leq r\leq 4\quad 0\leq \theta \leq 2\pi \end{eqnarray*}$$

Figure 45

(c) In spherical coordinates, the equation of a half cone, where $z\geq 0,$ is $\phi =a,$ $0< a <\dfrac{\pi }{2}.$ We use Figure 45 to find $a.$ $$\begin{eqnarray*} \tan \phi &=&\dfrac{r}{z}=\dfrac{r}{\sqrt{3}r}=\dfrac{1}{\sqrt{3}}\qquad \color{#0066A7}{{\hbox{\(z=\sqrt{3}r\)}}}\\[6pt] \phi &=&\tan ^{-1}\dfrac{1}{\sqrt{3}}=\dfrac{\pi }{6}=a \end{eqnarray*}$$

Using $\rho$ and $\theta$ as parameters, parametric equations for the cone in spherical coordinates are $$\begin{eqnarray*} x\underset{\underset{\color{#0066A7}{\sin \tfrac{\pi }{6}=\tfrac{1}{2}}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{ \rho }{2}\cos \theta \qquad y\underset{\underset{\color{#0066A7}{\sin \tfrac{\pi }{6}=\tfrac{1}{2}}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{\rho }{2}\sin \theta \qquad z \underset{\underset{\color{#0066A7}{\cos \tfrac{\pi }{6}=\tfrac{\sqrt{3}}{2}}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{\sqrt{3}}{2}\rho \end{eqnarray*}$$

Now we find the parameter domain. Since $$z^{2}=3x^{2}+3y^{2} \ \hbox{and}\ \rho ^{2}=x^{2}+y^{2}+z^{2}=x^{2}+y^{2}+ (\underset{\color{#0066A7}{z^{2}}}{\underbrace{ 3x^{2}+3y^{2}}})=4x^{2}+4y^{2}$$

we have $$0\leq \rho ^{2}=4(x^{2}+y^{2}) \leq 4\,{\cdot}\, 16=64,\ \hbox{or equivalently,}\ 0\leq \rho \leq 8.$$

The parameter domain of the cone is $0\leq \rho \,\leq 8$ and $0\leq \theta \leq 2\pi .$ Then a parametrization of the cone is $$\mathbf{r}( \rho ,\theta ) =\dfrac{\rho }{2}\cos \theta \,\mathbf{ i}+\dfrac{\rho }{2}\sin \theta \,\mathbf{j}+\dfrac{\sqrt{3}}{2}\rho \, \mathbf{k}\qquad 0\leq \rho \leq 8\quad 0\leq \theta \leq 2\pi$$