2.7 Cylindrical and Spherical Coordinates - Calculus Volume 3 | OpenStax (2024)

Conversion from cylindrical to rectangular coordinates requires a simple application of the equations listed in Conversion between Cylindrical and Cartesian Coordinates:

$$\begin{array}{ccc}\hfill x& =\hfill & r\text{cos}\theta =4\text{cos}\frac{2\pi }{3}=\mathrm{-2}\hfill \\ \hfill y& =\hfill & r\text{sin}\theta =4\text{sin}\frac{2\pi }{3}=2\sqrt{3}\hfill \\ \hfill z& =\hfill & \mathrm{-2.}\hfill \end{array}$$

The point with cylindrical coordinates $\left(4,\frac{2\pi }{3},\mathrm{-2}\right)$ has rectangular coordinates $\left(\mathrm{-2},2\sqrt{3},\mathrm{-2}\right)$ (see the following figure).

Figure 2.93 The projection of the point in the xy-plane is 4 units from the origin. The line from the origin to the point’s projection forms an angle of $\frac{2\pi }{3}$ with the positive x-axis. The point lies $2$ units below the xy-plane.

Use the second set of equations from Conversion between Cylindrical and Cartesian Coordinates to translate from rectangular to cylindrical coordinates:

$$\begin{array}{ccc}\hfill r^2& =\hfill & x^2+y^2\hfill \\ \hfill r& =\hfill & \text{±}\sqrt{1^2+{\left(\mathrm{-3}\right)}^2}=\text{±}\sqrt{10}.\hfill \end{array}$$

We choose the positive square root, so $r=\sqrt{10}.$ Now, we apply the formula to find $\theta .$ In this case, $y$ is negative and $x$ is positive, which means we must select the value of $\theta$ between $\frac{3\pi }{2}$ and $2\pi \text{:}$

$$\begin{array}{ccc}\hfill \text{tan}\theta & =\hfill & \frac{y}{x}=\frac{\mathrm{-3}}{1}\hfill \\ \hfill \theta & =\hfill & \text{arctan}\left(\mathrm{-3}\right)+2\pi \approx 5.03\text{rad}.\hfill \end{array}$$

In this case, the z-coordinates are the same in both rectangular and cylindrical coordinates:

$$z=5.$$

The point with rectangular coordinates $(1,\mathrm{-3},5)$ has cylindrical coordinates approximately equal to $\left(\sqrt{10},5.03,5\right).$

1. When the angle $\theta$ is held constant while $r$ and $z$ are allowed to vary, the result is a half-plane (see the following figure).
   

   

   Figure 2.94 In polar coordinates, the equation $\theta =\pi \text{/}4$ describes the ray extending diagonally through the first quadrant. In three dimensions, this same equation describes a half-plane.

2. Substitute $r^2=x^2+y^2$ into equation $r^2+z^2=9$ to express the rectangular form of the equation: $x^2+y^2+z^2=9.$ This equation describes a sphere centered at the origin with radius $3$ (see the following figure).
   

   

   Figure 2.95 The sphere centered at the origin with radius $3$ can be described by the cylindrical equation $r^2+z^2=9.$

3. To describe the surface defined by equation $z=r,$ is it useful to examine traces parallel to the xy-plane. For example, the trace in plane $z=1$ is circle $r=1,$ the trace in plane $z=3$ is circle $r=3,$ and so on. Each trace is a circle. As the value of $z$ increases, the radius of the circle also increases. The resulting surface is a cone (see the following figure).
   

   

   Figure 2.96 The traces in planes parallel to the xy-plane are circles. The radius of the circles increases as $z$ increases.

Use the equations in Converting among Spherical, Cylindrical, and Rectangular Coordinates to translate between spherical and cylindrical coordinates (Figure 2.100):

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

Figure 2.100 The projection of the point in the xy-plane is $4$ units from the origin. The line from the origin to the point’s projection forms an angle of $\pi \text{/}3$ with the positive x-axis. The point lies $4\sqrt{3}$ units above the xy-plane.

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 $\left(4,\frac{\pi }{3},4\sqrt{3}\right).$

Start by converting from rectangular to spherical coordinates:

$$\begin{array}{cccc}\begin{array}{ccc}\hfill {\rho }^2& =\hfill & x^2+y^2+z^2={\left(\mathrm{-1}\right)}^2+1^2+{\left(\sqrt{6}\right)}^2=8\hfill \\ \hfill \rho & =\hfill & 2\sqrt{2}\hfill \end{array}\hfill & & & \begin{array}{ccc}\hfill \text{tan}\theta & =\hfill & \frac{1}{\mathrm{-1}}\hfill \\ \hfill \theta & =\hfill & \text{arctan}\left(\mathrm{-1}\right)=\frac{3\pi }{4}.\hfill \end{array}\hfill \end{array}$$

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

$$\sqrt{6}=2\sqrt{2}\text{cos}\phi ,\text{so}\text{cos}\phi =\frac{\sqrt{6}}{2\sqrt{2}}=\frac{\sqrt{3}}{2}$$

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 $r\text{:}$

$$r=\rho \text{sin}\phi =2\sqrt{2}\text{sin}\left(\frac{\pi }{6}\right)=\sqrt{2}.$$

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 Figure 2.101.
   

   

   Figure 2.101 The surface described by equation $\theta =\frac{\pi }{3}$ is a half-plane.

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 (Figure 2.102). 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).
   

   

   Figure 2.102 The equation $\phi =\frac{5\pi }{6}$ describes a cone.

   
   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 $6$ (Figure 2.103).
   

   

   Figure 2.103 Equation $\rho =6$ describes a sphere with radius $6.$

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 $\frac{1}{2}.$

1. Clearly, a bowling ball is a sphere, so spherical coordinates would probably work best here. The origin should be located at the physical center of the ball. There is no obvious choice for how the x-, y- and z-axes should be oriented. Bowling balls normally have a weight block in the center. One possible choice is to align the z-axis with the axis of symmetry of the weight block.
2. A submarine generally moves in a straight line. There is no rotational or spherical symmetry that applies in this situation, so rectangular coordinates are a good choice. The z-axis should probably point upward. The x- and y-axes could be aligned to point east and north, respectively. The origin should be some convenient physical location, such as the starting position of the submarine or the location of a particular port.
3. A cone has several kinds of symmetry. In cylindrical coordinates, a cone can be represented by equation $z=kr,$ where $k$ is a constant. In spherical coordinates, we have seen that surfaces of the form $\phi =c$ are half-cones. Last, in rectangular coordinates, elliptic cones are quadric surfaces and can be represented by equations of the form $z^2=\frac{x^2}{a^2}+\frac{y^2}{b^2}.$ In this case, we could choose any of the three. However, the equation for the surface is more complicated in rectangular coordinates than in the other two systems, so we might want to avoid that choice. In addition, we are talking about a water tank, and the depth of the water might come into play at some point in our calculations, so it might be nice to have a component that represents height and depth directly. Based on this reasoning, cylindrical coordinates might be the best choice. Choose the z-axis to align with the axis of the cone. The orientation of the other two axes is arbitrary. The origin should be the bottom point of the cone.
4. A pipeline is a cylinder, so cylindrical coordinates would be best the best choice. In this case, however, we would likely choose to orient our z-axis with the center axis of the pipeline. The x-axis could be chosen to point straight downward or to some other logical direction. The origin should be chosen based on the problem statement. Note that this puts the z-axis in a horizontal orientation, which is a little different from what we usually do. It may make sense to choose an unusual orientation for the axes if it makes sense for the problem.
5. A football has rotational symmetry about a central axis, so cylindrical coordinates would work best. The z-axis should align with the axis of the ball. The origin could be the center of the ball or perhaps one of the ends. The position of the x-axis is arbitrary.