Chegg write the equations in cylindrical coordinates

When we expanded the traditional Cartesian coordinate system from two dimensions to three, we simply added a new axis to model the third dimension. Starting with polar coordinates, we can follow this same process to create a new three-dimensional coordinate system, called the cylindrical coordinate system. In this way, cylindrical coordinates provide a natural extension of polar coordinates to three dimensions.

DEFINITION

In the cylindrical coordinate system , a point in space (Figure 1) is represented by the ordered triple [latex](r,\theta,z)[/latex], where

This figure is the first octant of the 3-dimensional coordinate system. There is a point labeled “(x, y, z) = (r, theta, z).” In the x y-plane, there is a line segment extending to underneath the point. This line segment is labeled “r.” The angle between the line segment and the x-axis is theta. There is a line segment perpendicular to the x-axis. Along with the line segment labeled r, this line segment and the x-axis form a right triangle.

Figure 1. The right triangle lies in the [latex]xy[/latex]-plane. The length of the hypotenuse is [latex]r[/latex] and [latex]\theta[/latex] is the measure of the angle formed by the positive [latex]x[/latex]-axis and the hypotenuse. The [latex]z[/latex]-coordinate describes the location of the point above or below the [latex]xy[/latex]-plane.

In the [latex]xy[/latex]-plane, the right triangle shown in Figure 1 provides the key to transformation between cylindrical and Cartesian, or rectangular, coordinates.

THEOREM: conversion between cylindrical and cartesian coordinates

The rectangular coordinates [latex](x, y, z)[/latex] and the cylindrical coordinates [latex](r,\theta,z)[/latex] of a point are related as follows:

As when we discussed conversion from rectangular coordinates to polar coordinates in two dimensions, it should be noted that the equation [latex]\tan\theta=\frac[/latex] has an infinite number of solutions. However, if we restrict [latex]\theta[/latex] to values between [latex]0[/latex] and [latex]2\pi[/latex], then we can find a unique solution based on the quadrant of the [latex]xy[/latex]-plane in which original point [latex](x, y, z)[/latex] is located. Note that if [latex]x=0[/latex], then the value of [latex]\theta[/latex] is either [latex]\frac<\pi>2[/latex], [latex]\frac<3\pi>2[/latex], or [latex]0[/latex] depending on the value of [latex]y[/latex].

Notice that these equations are derived from properties of right triangles. To make this easy to see, consider point [latex]P[/latex] in the [latex]xy[/latex]-plane with rectangular coordinates [latex](x, y, 0)[/latex] and with cylindrical coordinates [latex](r,\theta,0)[/latex], as shown in the following figure.

This figure is the first quadrant of the rectangular coordinate system. There is a point labeled “P = (x, y, 0) = (r, theta, 0).” There is a line segment from the origin to point P. This line segment is labeled “r.” The angle between the x-axis and the line segment r is labeled “theta.” There is also a vertical line segment labeled “y” from P to the x-axis. It forms a right triangle.

Figure 2. The Pythagorean theorem provides equation [latex]r^=x^+y^[/latex]. Right-triangle relationships tell us that [latex]x=r\cos[/latex], [latex]y=r\sin[/latex], and [latex]\tan=\frac[/latex].

Let’s consider the differences between rectangular and cylindrical coordinates by looking at the surfaces generated when each of the coordinates is held constant. If [latex]c[/latex] is a constant, then in rectangular coordinates, surfaces of the form [latex]x=c[/latex], [latex]y=c[/latex], or [latex]z=c[/latex] are all planes. Planes of these forms are parallel to the [latex]yz[/latex]-plane, the [latex]xz[/latex]-plane, and the [latex]xy[/latex]-plane, respectively. When we convert to cylindrical coordinates, the [latex]z[/latex]-coordinate does not change. Therefore, in cylindrical coordinates, surfaces of the form [latex]z=c[/latex] are planes parallel to the [latex]xy[/latex]-plane. Now, let’s think about surfaces of the form [latex]r=c[/latex]. The points on these surfaces are at a fixed distance from the [latex]z[/latex]-axis. In other words, these surfaces are vertical circular cylinders. Last, what about [latex]\theta=c[/latex]? The points on a surface of the form [latex]\theta=c[/latex] are at a fixed angle from the [latex]x[/latex]-axis, which gives us a half-plane that starts at the [latex]z[/latex]-axis (Figure 3 and Figure 4).

This figure has 3 images. The first image is a plane in the 3-dimensional coordinate system. It is parallel to the y z-plane where x = c. The second image is a plane in the 3-dimensional coordinate system. It is parallel to the x z-plane where y = c. the third image is a plane in the 3-dimensional coordinate system where z = c.

Figure 3. In rectangular coordinates, (a) surfaces of the form [latex]x=c[/latex] are planes parallel to the [latex]yz[/latex]-plane, (b) surfaces of the form [latex]y=c[/latex] are planes parallel to the [latex]xz[/latex]-plane, and (c) surfaces of the form [latex]z=c[/latex] are planes parallel to the [latex]xy[/latex]-plane.

This figure has 3 images. The first image is a right circular cylinder in the 3-dimensional coordinate system. It has the z-axis in the middle. The second image is a plane in the 3-dimensional coordinate system. It is vertical with the z-axis on one edge. The third image is a plane in the 3-dimensional coordinate system where z = c.

Figure 4. In cylindrical coordinates, (a) surfaces of the form [latex]r=c[/latex] are vertical cylinders of radius [latex]c[/latex], (b) surfaces of the form [latex]\theta=c[/latex] are half-planes at angle [latex]c[/latex] from the [latex]x[/latex]-axis, and (c) surfaces of the form [latex]z=c[/latex] are planes parallel to the [latex]xy[/latex]-plane.

Example: converting from cylindrical to rectangular coordinates

Plot the point with cylindrical coordinates [latex]\left(4,\frac<2\pi>3,-2\right)[/latex] and express its location in rectangular coordinates.

Show Solution

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

[latex]\begin x&=r\cos\theta=4\cos\frac<2\pi>3=-2 \\ y&=r\sin\theta=4\sin\frac<2\pi>3=2\sqrt3 \\ z&=-2 \end[/latex].

The point with cylindrical coordinates [latex]\left(4,\frac<2\pi>3,-2\right)[/latex] has rectangular coordinates [latex](-2,2\sqrt3,-2)[/latex] (see the following figure).

This figure is the 3-dimensional coordinate system. It has a <a href=point where r = 4, z = -2 and theta = 2 pi /3." width="485" height="349" />

Figure 5. The projection of the point in the [latex]xy[/latex]-plane is [latex]4[/latex] units from the origin. The line from the origin to the point’s projection forms an angle of [latex]\frac<2\pi>[/latex] with the positive [latex]x[/latex]-axis. The point lies [latex]2[/latex] units below the [latex]xy[/latex]-plane.

try it

Point [latex]R[/latex] has cylindrical coordinates [latex]\left(5,\frac<\pi>6,4\right)[/latex]. Plot [latex]R[/latex] and describe its location in space using rectangular, or Cartesian, coordinates.

Show Solution

The rectangular coordinates of the point are [latex]\left(\frac2,\frac52,4\right)[/latex].

This figure is the 3-dimensional coordinate system. There is a point labeled “(5, pi/6, 4).” The point is located above a line segment in the x y-plane labeled r = 5 that is pi/6 degrees from the x-axis. The distance from the x y-plane to the point is labeled “z = 4.”

Figure 6. The rectangular coordinates of the point are [latex]\left(\frac2,\frac52,4\right)[/latex]

Watch the following video to see the worked solution to the above Try IT.

Try It

If this process seems familiar, it is with good reason. This is exactly the same process that we followed in Introduction to Parametric Equations and Polar Coordinates to convert from polar coordinates to two-dimensional rectangular coordinates.

Example: converting from rectangular to cylindrical coordinates

Convert the rectangular coordinates [latex](1, -3, 5)[/latex] to cylindrical coordinates.

Show Solution

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

We choose the positive square root, so [latex]r=\sqrt[/latex]. Now, we apply the formula to find [latex]\theta[/latex]. In this case, [latex]y[/latex] is negative and [latex]x[/latex] is positive, which means we must select the value of [latex]\theta[/latex] between [latex]\frac<3\pi>2[/latex] and [latex]2\pi[/latex]:

[latex]\begin \tan\theta&=\fracx=\frac1 \\ \theta&=\arctan(-3)\approx5.03\text< rad>. \end[/latex]

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

The point with rectangular coordinates [latex](1, -3, 5)[/latex] has cylindrical coordinates approximately equal to [latex](\sqrt,5.03,5)[/latex].

try it

Convert point [latex](-8, 8, -7)[/latex] from Cartesian coordinates to cylindrical coordinates.

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The use of cylindrical coordinates is common in fields such as physics. Physicists studying electrical charges and the capacitors used to store these charges have discovered that these systems sometimes have a cylindrical symmetry. These systems have complicated modeling equations in the Cartesian coordinate system, which make them difficult to describe and analyze. The equations can often be expressed in more simple terms using cylindrical coordinates. For example, the cylinder described by equation [latex]x^+y^=25[/latex] in the Cartesian system can be represented by cylindrical equation [latex]r=5[/latex].

Example: identifying surfaces in the cylindrical coordinate system

Describe the surfaces with the given cylindrical equations.

  1. [latex]\theta=\frac<\pi>4[/latex]
  2. [latex]r^+z^=9[/latex]
  3. [latex]z=r[/latex]
Show Solution
  1. When the angle [latex]\theta[/latex] is held constant while [latex]r[/latex] and [latex]z[/latex] are allowed to vary, the result is a half-plane (see the following figure).

This figure is the first quadrant of the 3-dimensional coordinate system. There is a plane attached to the z-axis, dividing the x y-plane with a diagonal line. The angle between the x-axis and this plane is pi/4.

Figure 7. In polar coordinates, the equation [latex]\theta=\frac<\pi>4[/latex] describes the ray extending diagonally through the first quadrant. In three dimensions, this same equation describes a half-plane.

This figure is a sphere. It has the z-axis through the center vertically. The point of intersection with the z-axis and the sphere is (0, 0, 3). There is also the y-axis through the center of the sphere horizontally. The intersection of the sphere and the y-axis is the point (0, 3, 0).

Figure 8. The sphere centered at the origin with radius 3 can be described by the cylindrical equation [latex]x^+y^+z^=9[/latex].

This figure is the 3-dimensional coordinate system. It has an elliptic cone with the z-axis down the center. The two cones, one right side up, the other upside down, meet at the origin.

Figure 9. The traces in planes parallel to the [latex]xy[/latex]-plane are circles. The radius of the circles increases as [latex]z[/latex] increases.

try it

Describe the surface with cylindrical equation [latex]r=6[/latex].

Show Solution

Describe the surface with cylindrical equation [latex]r=6[/latex].

This figure is a right circular cylinder. It is upright with the z-axis through the center. It is on top of the x y plane.

Figure 10. The surface with cylindrical equation [latex]r=6[/latex].