2.7 Cylindrical and spherical coordinates  (Page 4/11)

By convention, the origin is represented as $\left(0,0,0\right)$ in spherical coordinates.

The formulas to convert from spherical coordinates to rectangular coordinates may seem complex, but they are straightforward applications of trigonometry. Looking at [link] , it is easy to see that $r=\rho \text{sin}\phi .$ Then, looking at the triangle in the xy -plane with $r$ as its hypotenuse, we have $x=r\text{cos}\theta =\rho \text{sin}\phi \text{cos}\theta .$ The derivation of the formula for $y$ is similar. [link] also shows that ${\rho }^2=r^2+z^2=x^2+y^2+z^2$ and $z=\rho \text{cos}\phi .$ Solving this last equation for $\phi$ and then substituting $\rho =\sqrt{r^2+z^2}$ (from the first equation) yields $\phi =\text{arccos}\left(\frac{z}{\sqrt{r^2+z^2}}\right).$ Also, note that, as before, we must be careful when using the formula $\text{tan}\theta =\frac{y}{x}$ to choose the correct value of $\theta .$

As we did with cylindrical coordinates, let’s consider the surfaces that are generated when each of the coordinates is held constant. Let $c$ be a constant, and consider surfaces of the form $\rho =c.$ Points on these surfaces are at a fixed distance from the origin and form a sphere. The coordinate $\theta$ in the spherical coordinate system is the same as in the cylindrical coordinate system, so surfaces of the form $\theta =c$ are half-planes, as before. Last, consider surfaces of the form $\phi =0.$ The points on these surfaces are at a fixed angle from the z -axis and form a half-cone ( [link] ).

Converting from spherical coordinates

Plot the point with spherical coordinates $\left(8,\frac{\pi }{3},\frac{\pi }{6}\right)$ and express its location in both rectangular and cylindrical coordinates.

Use the equations in [link] to translate between spherical and cylindrical coordinates ( [link] ):

$\begin{array}{}\\ \\ x=\rho \text{sin}\phi \text{cos}\theta =8\text{sin}\left(\frac{\pi }{6}\right)\text{cos}\left(\frac{\pi }{3}\right)=8\left(\frac{1}{2}\right)\frac{1}{2}=2\hfill \\ y=\rho \text{sin}\phi \text{sin}\theta =8\text{sin}\left(\frac{\pi }{6}\right)\text{sin}\left(\frac{\pi }{3}\right)=8\left(\frac{1}{2}\right)\frac{\sqrt{3}}{2}=2\sqrt{3}\hfill \\ z=\rho \text{cos}\phi =8\text{cos}\left(\frac{\pi }{6}\right)=8\left(\frac{\sqrt{3}}{2}\right)=4\sqrt{3}.\hfill \end{array}$

The point with spherical coordinates $\left(8,\frac{\pi }{3},\frac{\pi }{6}\right)$ has rectangular coordinates $\left(2,2\sqrt{3},4\sqrt{3}\right).$

Finding the values in cylindrical coordinates is equally straightforward:

$\begin{array}{}\\ \hfill r& =\hfill & \rho \text{sin}\phi =8\text{sin}\frac{\pi }{6}=4\hfill \\ \hfill \theta & =\hfill & \theta \hfill \\ \hfill z& =\hfill & \rho \text{cos}\phi =8\text{cos}\frac{\pi }{6}=4\sqrt{3}.\hfill \end{array}$

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

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