Subsections

• Background
• Exercises

Cylindrical and Spherical Coordinates

Background

Defining surfaces with rectangular coordinates often times becomes more complicated than necessary. A change in coordinates can simplify things. The easiest examples are a sphere and a cylinder.

> with(plots):
> f1:=x^2+y^2+z^2=49;
> g1:=rho=7;
> implicitplot3d(f1,x=-7..7,y=-7..7,z=-7..7,axes=boxed,scaling=comstrained);
> implicitplot3d(g1,rho=0..7.5,theta=0..2*Pi,phi=0..Pi,coords=spherical,
numpoints=5000,axes=boxed);
> f2:=x^2+y^2=49;
> h2:=r=7;
> implicitplot3d(f2,x=-7..7,y=-7..7,z=-8..8,axes=boxed);
> implicitplot3d(h2,r=0..7.5,theta=0..2*Pi,z=-8..8,coords=cylindrical,
numpoints=3000,axes=boxed);

To change to cylindrical coordinates from rectangular coordinates use the conversion:

$x=r\cos(\theta)$

$y=r\sin(\theta)$

$z=z$

Where $r$ is the radius in the x-y plane and $\theta$ is the angle in the x-y plane. To change to spherical coordinates from rectangular coordinates use the conversion:

$x=\rho\sin(\phi)\cos(\theta)$

$y=\rho\sin(\phi)\sin(\theta)$

$z=\rho\cos(\phi)$

Where $\theta$ is the angle in the x-y plane; $\rho$ is the radius from the origin in any direction; and $\phi$ is the angle in the x-z plane. As an example, the equation of an ellipsoid in rectangular coordinates is

$$\frac{x^2}{23}+\frac{y^2}{23}+\frac{z^2}{122}=1$$

> f3:=x^2/23+y^2/23+z^2/122=1;
>implicitplot3d(f3,x=-5..5,y=-5..5,z=-12..12,scaling=constrained,axes=boxed);

Changing to sherical coordinates:

> g3:=simplify(subs({x=rho*sin(phi)*cos(theta),y=rho*sin(phi)*sin(theta),
z=rho*cos(phi)},f3));
> implicitplot3d(g3,rho=0..12,theta=0..2*Pi,phi=0..Pi,coords=spherical,axes=boxed,
scaling=constrained,numpoints=2000);

Exercises

1. Given the rectangular equation for a hyperboloid of one sheet:
   
   $$2(x^2+y^2)-z^2=2$$
   
   

   A)

   Graph the equation using the domain values of $-8 \leq x \leq 8$, $-8 \leq y \leq 8$ and the range values $-8 \leq z \leq 8$.

   B)

   Write the equation in spherical coordinates and then graph the equation.

   C)

   Write the equation in cylindrical coordinates and graph it.

   D)

   Looking at the three equations, which coordinates appears to give the simplest equation?

2. Given the equation of a torus (a.k.a. donut):
   
   $$36(x^2+y^2)=(x^2+y^2+z^2+5)^2$$
   
   

   A)

   Graph the equation using the domain values $-6 \leq x \leq 6$, $-6 \leq y \leq 6$ and the range values $-2 \leq z \leq 2$.

   B)

   Write the equation in spherical coordinates and graph it.

   C)

   Write the equation in cylindrical coordinates (hint: use the factor command outside the simplify command to simplify even more). Then graph your equation.

   D)

   Looking at the three equations, which coordinates appear to give the simplest equation?

3. A sphere centered at the origin (as seen in the background) is a very simple equation in spherical coordinates. Now let's look at a sphere moved away from the origin.
   
   $$(x-3)^2+(y-3)^2+z^2=9$$
   
   

   A)

   Graph the equation using the domain values $0 \leq x \leq 6$, $0 \leq y \leq 6$ and the range values $-3 \leq z \leq 3$.

   B)

   Write the equation in spherical coordinates and graph it. (You may want to use axes=normal instead of boxed)

   C)

   Write the equation in cylindrical coordinates and graph it. (You may want to use axes=normal instead of boxed)

   D)

   Looking at the three equations, which coordinates appear to give the simplest equation?

   E)

   Looking at the rectangular domain values, a hemisphere can be graphed two ways. Changing the $x$ values to $0..3$ (or $3..6$). The second way would be to change the y values to $0..3$. Graph a hemisphere four ways:

   i.

   In spherical coordinates change the $\theta$ values.

   ii.

   In spherical coordinates change the $\phi$ values.

   iii.

   In cylindrical coordinates change the $\theta$ values.

   iv.

   In cylindrical coordinates change the $z$ values.