2.7 Cylindrical and spherical coordinates  (Page 5/11)

Start by converting from rectangular to spherical coordinates:

Because $\left(x,y\right)=\left(\mathrm{-1},1\right),$ then the correct choice for $\theta$ is $\frac{3\pi }{4}.$

There are actually two ways to identify $\phi .$ We can use the equation $\phi =\text{arccos}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right).$ A more simple approach, however, is to use equation $z=\rho \text{cos}\phi .$ We know that $z=\sqrt{6}$ and $\rho =2\sqrt{2},$ so

and therefore $\phi =\frac{\pi }{6}.$ The spherical coordinates of the point are $\left(2\sqrt{2},\frac{3\pi }{4},\frac{\pi }{6}\right).$

To find the cylindrical coordinates for the point, we need only find $r\text{:}$

The cylindrical coordinates for the point are $\left(\sqrt{2},\frac{3\pi }{4},\sqrt{6}\right).$

1. The variable $\theta$ represents the measure of the same angle in both the cylindrical and spherical coordinate systems. Points with coordinates $\left(\rho ,\frac{\pi }{3},\phi \right)$ lie on the plane that forms angle $\theta =\frac{\pi }{3}$ with the positive x -axis. Because $\rho >0,$ the surface described by equation $\theta =\frac{\pi }{3}$ is the half-plane shown in [link] .
   

   

2. Equation $\phi =\frac{5\pi }{6}$ describes all points in the spherical coordinate system that lie on a line from the origin forming an angle measuring $\frac{5\pi }{6}$ rad with the positive z -axis. These points form a half-cone ( [link] ). Because there is only one value for $\phi$ that is measured from the positive z -axis, we do not get the full cone (with two pieces).
   

   

   
   To find the equation in rectangular coordinates, use equation $\phi =\text{arccos}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right).$
   
   $\begin{array}{ccc}\hfill \frac{5\pi }{6}& =\hfill & \text{arccos}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)\hfill \\ \hfill \text{cos}\frac{5\pi }{6}& =\hfill & \frac{z}{\sqrt{x^2+y^2+z^2}}\hfill \\ \hfill -\frac{\sqrt{3}}{2}& =\hfill & \frac{z}{\sqrt{x^2+y^2+z^2}}\hfill \\ \hfill \frac{3}{4}& =\hfill & \frac{z^2}{x^2+y^2+z^2}\hfill \\ \hfill \frac{3x^2}{4}+\frac{3y^2}{4}+\frac{3z^2}{4}& =\hfill & z^2\hfill \\ \hfill \frac{3x^2}{4}+\frac{3y^2}{4}-\frac{z^2}{4}& =\hfill & 0.\hfill \end{array}$
   
   This is the equation of a cone centered on the z -axis.
3. Equation $\rho =6$ describes the set of all points $6$ units away from the origin—a sphere with radius $6$ ( [link] ).
   

   

4. To identify this surface, convert the equation from spherical to rectangular coordinates, using equations $y=\rho \text{sin}\phi \text{sin}\theta$ and ${\rho }^2=x^2+y^2+z^2\text{:}$
   
   $\begin{array}{cccccc}\hfill \rho & =\hfill & \text{sin}\theta \text{sin}\phi \hfill & & & \\ \hfill {\rho }^2& =\hfill & \rho \text{sin}\theta \text{sin}\phi \hfill & & & \text{Multiply both sides of the equation by}\rho .\hfill \\ \hfill x^2+y^2+z^2& =\hfill & y\hfill & & & \text{Substitute rectangular variables using the equations above.}\hfill \\ \hfill x^2+y^2-y+z^2& =\hfill & 0\hfill & & & \text{Subtract}y\text{from both sides of the equation.}\hfill \\ \hfill x^2+y^2-y+\frac{1}{4}+z^2& =\hfill & \frac{1}{4}\hfill & & & \text{Complete the square.}\hfill \\ \hfill x^2+{\left(y-\frac{1}{2}\right)}^2+z^2& =\hfill & \frac{1}{4}.\hfill & & & \text{Rewrite the middle terms as a perfect square.}\hfill \end{array}$
   
   The equation describes a sphere centered at point $\left(0,\frac{1}{2},0\right)$ with radius $\frac{1}{2}.$