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Calculus Volume 3

2.7 Cylindrical and Spherical Coordinates

Calculus Volume 32.7 Cylindrical and Spherical Coordinates
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  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 2.7.1. Convert from cylindrical to rectangular coordinates.
  • 2.7.2. Convert from rectangular to cylindrical coordinates.
  • 2.7.3. Convert from spherical to rectangular coordinates.
  • 2.7.4. Convert from rectangular to spherical coordinates.

The Cartesian coordinate system provides a straightforward way to describe the location of points in space. Some surfaces, however, can be difficult to model with equations based on the Cartesian system. This is a familiar problem; recall that in two dimensions, polar coordinates often provide a useful alternative system for describing the location of a point in the plane, particularly in cases involving circles. In this section, we look at two different ways of describing the location of points in space, both of them based on extensions of polar coordinates. As the name suggests, cylindrical coordinates are useful for dealing with problems involving cylinders, such as calculating the volume of a round water tank or the amount of oil flowing through a pipe. Similarly, spherical coordinates are useful for dealing with problems involving spheres, such as finding the volume of domed structures.

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 2.89) is represented by the ordered triple (r,θ,z),(r,θ,z), where

  • (r,θ)(r,θ) are the polar coordinates of the point’s projection in the xy-plane
  • zz is the usual z-coordinatez-coordinate in the Cartesian coordinate system
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 2.89 The right triangle lies in the xy-plane. The length of the hypotenuse is rr and θθ is the measure of the angle formed by the positive x-axis and the hypotenuse. The
z-coordinate describes the location of the point above or below the xy-plane.

In the xy-plane, the right triangle shown in Figure 2.89 provides the key to transformation between cylindrical and Cartesian, or rectangular, coordinates.

Theorem 2.15

Conversion between Cylindrical and Cartesian Coordinates

The rectangular coordinates (x,y,z)(x,y,z) and the cylindrical coordinates (r,θ,z)(r,θ,z) of a point are related as follows:

x=rcosθThese equations are used to convert fromy=rsinθcylindrical coordinates to rectangularz=zcoordinates.andr2=x2+y2These equations are used to convert fromtanθ=yxrectangular coordinates to cylindricalz=zcoordinates.x=rcosθThese equations are used to convert fromy=rsinθcylindrical coordinates to rectangularz=zcoordinates.andr2=x2+y2These equations are used to convert fromtanθ=yxrectangular coordinates to cylindricalz=zcoordinates.

As when we discussed conversion from rectangular coordinates to polar coordinates in two dimensions, it should be noted that the equation tanθ=yxtanθ=yx has an infinite number of solutions. However, if we restrict θθ to values between 00 and 2π,2π, then we can find a unique solution based on the quadrant of the xy-plane in which original point (x,y,z)(x,y,z) is located. Note that if x=0,x=0, then the value of θθ is either π2,3π2,π2,3π2, or 0,0, depending on the value of y.y.

Notice that these equations are derived from properties of right triangles. To make this easy to see, consider point PP in the xy-plane with rectangular coordinates (x,y,0)(x,y,0) and with cylindrical coordinates (r,θ,0),(r,θ,0), 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 2.90 The Pythagorean theorem provides equation r2=x2+y2.r2=x2+y2. Right-triangle relationships tell us that x=rcosθ,x=rcosθ, y=rsinθ,y=rsinθ, and tanθ=y/x.tanθ=y/x.

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 cc is a constant, then in rectangular coordinates, surfaces of the form x=c,x=c, y=c,y=c, or z=cz=c are all planes. Planes of these forms are parallel to the yz-plane, the xz-plane, and the xy-plane, respectively. When we convert to cylindrical coordinates, the z-coordinate does not change. Therefore, in cylindrical coordinates, surfaces of the form z=cz=c are planes parallel to the xy-plane. Now, let’s think about surfaces of the form r=c.r=c. The points on these surfaces are at a fixed distance from the z-axis. In other words, these surfaces are vertical circular cylinders. Last, what about θ=c?θ=c? The points on a surface of the form θ=cθ=c are at a fixed angle from the x-axis, which gives us a half-plane that starts at the z-axis (Figure 2.91 and Figure 2.92).

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 2.91 In rectangular coordinates, (a) surfaces of the form x=cx=c are planes parallel to the yz-plane, (b) surfaces of the form y=cy=c are planes parallel to the xz-plane, and (c) surfaces of the form z=cz=c are planes parallel to the xy-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 2.92 In cylindrical coordinates, (a) surfaces of the form r=cr=c are vertical cylinders of radius r,r, (b) surfaces of the form θ=cθ=c are half-planes at angle θθ from the x-axis, and (c) surfaces of the form z=cz=c are planes parallel to the xy-plane.

Example 2.60

Converting from Cylindrical to Rectangular Coordinates

Plot the point with cylindrical coordinates (4,2π3,−2)(4,2π3,−2) and express its location in rectangular coordinates.

Solution

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

x=rcosθ=4cos2π3=−2y=rsinθ=4sin2π3=23z=−2.x=rcosθ=4cos2π3=−2y=rsinθ=4sin2π3=23z=−2.

The point with cylindrical coordinates (4,2π3,−2)(4,2π3,−2) has rectangular coordinates (−2,23,−2)(−2,23,−2) (see the following figure).

This figure is the 3-dimensional coordinate system. It has a point where r = 4, z = -2 and theta = 2 pi /3.
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 2π32π3 with the positive x-axis. The point lies 22 units below the xy-plane.
Checkpoint 2.55

Point RR has cylindrical coordinates (5,π6,4)(5,π6,4). Plot RR and describe its location in space using rectangular, or Cartesian, coordinates.

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 2.61

Converting from Rectangular to Cylindrical Coordinates

Convert the rectangular coordinates (1,−3,5)(1,−3,5) to cylindrical coordinates.

Solution

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

r2=x2+y2r=±12+(−3)2=±10.r2=x2+y2r=±12+(−3)2=±10.

We choose the positive square root, so r=10.r=10. Now, we apply the formula to find θ.θ. In this case, yy is negative and xx is positive, which means we must select the value of θθ between 3π23π2 and 2π:2π:

tanθ=yx=−31θ=arctan(−3)5.03rad.tanθ=yx=−31θ=arctan(−3)5.03rad.

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

z=5.z=5.

The point with rectangular coordinates (1,−3,5)(1,−3,5) has cylindrical coordinates approximately equal to (10,5.03,5).(10,5.03,5).

Checkpoint 2.56

Convert point (−8,8,−7)(−8,8,−7) from Cartesian coordinates to cylindrical coordinates.

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 x2+y2=25x2+y2=25 in the Cartesian system can be represented by cylindrical equation r=5.r=5.

Example 2.62

Identifying Surfaces in the Cylindrical Coordinate System

Describe the surfaces with the given cylindrical equations.

  1. θ=π4θ=π4
  2. r2+z2=9r2+z2=9
  3. z=rz=r

Solution

  1. When the angle θθ is held constant while rr and zz 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 2.94 In polar coordinates, the equation θ=π/4θ=π/4 describes the ray extending diagonally through the first quadrant. In three dimensions, this same equation describes a half-plane.
  2. Substitute r2=x2+y2r2=x2+y2 into equation r2+z2=9r2+z2=9 to express the rectangular form of the equation: x2+y2+z2=9.x2+y2+z2=9. This equation describes a sphere centered at the origin with radius 33 (see the following figure).
    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 2.95 The sphere centered at the origin with radius 33 can be described by the cylindrical equation r2+z2=9.r2+z2=9.
  3. To describe the surface defined by equation z=r,z=r, is it useful to examine traces parallel to the xy-plane. For example, the trace in plane z=1z=1 is circle r=1,r=1, the trace in plane z=3z=3 is circle r=3,r=3, and so on. Each trace is a circle. As the value of zz increases, the radius of the circle also increases. The resulting surface is a cone (see the following figure).
    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 2.96 The traces in planes parallel to the xy-plane are circles. The radius of the circles increases as zz increases.
Checkpoint 2.57

Describe the surface with cylindrical equation r=6.r=6.

Spherical Coordinates

In the Cartesian coordinate system, the location of a point in space is described using an ordered triple in which each coordinate represents a distance. In the cylindrical coordinate system, location of a point in space is described using two distances (randz)(randz) and an angle measure (θ).(θ). In the spherical coordinate system, we again use an ordered triple to describe the location of a point in space. In this case, the triple describes one distance and two angles. Spherical coordinates make it simple to describe a sphere, just as cylindrical coordinates make it easy to describe a cylinder. Grid lines for spherical coordinates are based on angle measures, like those for polar coordinates.

Definition

In the spherical coordinate system, a point PP in space (Figure 2.97) is represented by the ordered triple (ρ,θ,φ)(ρ,θ,φ) where

  • ρρ (the Greek letter rho) is the distance between PP and the origin (ρ0);(ρ0);
  • θθ is the same angle used to describe the location in cylindrical coordinates;
  • φφ (the Greek letter phi) is the angle formed by the positive z-axis and line segment OP,OP, where OO is the origin and 0φπ.0φπ.
This figure is the first quadrant of the 3-dimensional coordinate system. It has a point labeled “(x, y, z) = (rho, theta, phi).” There is a line segment from the origin to the point. It is labeled “rho.” The angle between this line segment and the z-axis is phi. There is a line segment in the x y-plane from the origin to the shadow of the point. This segment is labeled “r.” The angle between the x-axis and r is theta.
Figure 2.97 The relationship among spherical, rectangular, and cylindrical coordinates.

By convention, the origin is represented as (0,0,0)(0,0,0) in spherical coordinates.

Theorem 2.16

Converting among Spherical, Cylindrical, and Rectangular Coordinates

Rectangular coordinates (x,y,z)(x,y,z) and spherical coordinates (ρ,θ,φ)(ρ,θ,φ) of a point are related as follows:

x=ρsinφcosθThese equations are used to convert fromy=ρsinφsinθspherical coordinates to rectangularz=ρcosφcoordinates.andρ2=x2+y2+z2These equations are used to convert fromtanθ=yxrectangular coordinates to sphericalφ=arccos(zx2+y2+z2).coordinates.x=ρsinφcosθThese equations are used to convert fromy=ρsinφsinθspherical coordinates to rectangularz=ρcosφcoordinates.andρ2=x2+y2+z2These equations are used to convert fromtanθ=yxrectangular coordinates to sphericalφ=arccos(zx2+y2+z2).coordinates.

If a point has cylindrical coordinates (r,θ,z),(r,θ,z), then these equations define the relationship between cylindrical and spherical coordinates.

r=ρsinφThese equations are used to convert fromθ=θspherical coordinates to cylindricalz=ρcosφcoordinates.andρ=r2+z2These equations are used to convert fromθ=θcylindrical coordinates to sphericalφ=arccos(zr2+z2)coordinates.r=ρsinφThese equations are used to convert fromθ=θspherical coordinates to cylindricalz=ρcosφcoordinates.andρ=r2+z2These equations are used to convert fromθ=θcylindrical coordinates to sphericalφ=arccos(zr2+z2)coordinates.

The formulas to convert from spherical coordinates to rectangular coordinates may seem complex, but they are straightforward applications of trigonometry. Looking at Figure 2.98, it is easy to see that r=ρsinφ.r=ρsinφ. Then, looking at the triangle in the xy-plane with rr as its hypotenuse, we have x=rcosθ=ρsinφcosθ.x=rcosθ=ρsinφcosθ. The derivation of the formula for yy is similar. Figure 2.96 also shows that ρ2=r2+z2=x2+y2+z2ρ2=r2+z2=x2+y2+z2 and z=ρcosφ.z=ρcosφ. Solving this last equation for φφ and then substituting ρ=r2+z2ρ=r2+z2 (from the first equation) yields φ=arccos(zr2+z2).φ=arccos(zr2+z2). Also, note that, as before, we must be careful when using the formula tanθ=yxtanθ=yx to choose the correct value of θ.θ.

This figure is the first quadrant of the 3-dimensional coordinate system. It has a point labeled “(x, y, z) = (r, theta, z) = (rho, theta, phi).” There is a line segment from the origin to the point. It is labeled “rho.” The angle between this line segment and the z-axis is phi. There is a line segment in the x y-plane from the origin to the shadow of the point. This segment is labeled “r.” The angle between the x-axis and r is theta.The distance from r to the point is labeled “z.”
Figure 2.98 The equations that convert from one system to another are derived from right-triangle relationships.

As we did with cylindrical coordinates, let’s consider the surfaces that are generated when each of the coordinates is held constant. Let cc be a constant, and consider surfaces of the form ρ=c.ρ=c. Points on these surfaces are at a fixed distance from the origin and form a sphere. The coordinate θθ in the spherical coordinate system is the same as in the cylindrical coordinate system, so surfaces of the form θ=cθ=c are half-planes, as before. Last, consider surfaces of the form φ=c.φ=c. The points on these surfaces are at a fixed angle from the z-axis and form a half-cone (Figure 2.99).

This figure has three images. The first image is a sphere centered in the 3-dimensional coordinate system. The second figure is a vertical plane with an edge on the z-axis in the 3-dimensional coordinate system. The third image is an elliptical cone with the center at the origin of the 3-dimensional coordinate system.
Figure 2.99 In spherical coordinates, surfaces of the form ρ=cρ=c are spheres of radius ρρ (a), surfaces of the form θ=cθ=c are half-planes at an angle θθ from the x-axis (b), and surfaces of the form ϕ=cϕ=c are half-cones at an angle ϕϕ from the z-axis (c).

Example 2.63

Converting from Spherical Coordinates

Plot the point with spherical coordinates (8,π3,π6)(8,π3,π6) and express its location in both rectangular and cylindrical coordinates.

Solution

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

x=ρsinφcosθ=8sin(π6)cos(π3)=8(12)12=2y=ρsinφsinθ=8sin(π6)sin(π3)=8(12)32=23z=ρcosφ=8cos(π6)=8(32)=43.x=ρsinφcosθ=8sin(π6)cos(π3)=8(12)12=2y=ρsinφsinθ=8sin(π6)sin(π3)=8(12)32=23z=ρcosφ=8cos(π6)=8(32)=43.
This figure is the first quadrant of the 3-dimensional coordinate system. It has a point labeled “(8, pi/3, pi/6).” There is a line segment from the origin to the point. It is labeled “rho = 8.” The angle between this line segment and the z-axis is labeled “phi = pi/6.” There is a line segment in the x y-plane from the origin to the shadow of the point. The angle between the x-axis and r is labeled “theta = pi/3.”
Figure 2.100 The projection of the point in the xy-plane is 44 units from the origin. The line from the origin to the point’s projection forms an angle of π/3π/3 with the positive x-axis. The point lies 4343 units above the xy-plane.

The point with spherical coordinates (8,π3,π6)(8,π3,π6) has rectangular coordinates (2,23,43).(2,23,43).

Finding the values in cylindrical coordinates is equally straightforward:

r=ρsinφ=8sinπ6=4θ=θz=ρcosφ=8cosπ6=43.r=ρsinφ=8sinπ6=4θ=θz=ρcosφ=8cosπ6=43.

Thus, cylindrical coordinates for the point are (4,π3,43).(4,π3,43).

Checkpoint 2.58

Plot the point with spherical coordinates (2,5π6,π6)(2,5π6,π6) and describe its location in both rectangular and cylindrical coordinates.

Example 2.64

Converting from Rectangular Coordinates

Convert the rectangular coordinates (−1,1,6)(−1,1,6) to both spherical and cylindrical coordinates.

Solution

Start by converting from rectangular to spherical coordinates:

ρ2=x2+y2+z2=(−1)2+12+(6)2=8ρ=22tanθ=1−1θ=arctan(−1)=3π4.ρ2=x2+y2+z2=(−1)2+12+(6)2=8ρ=22tanθ=1−1θ=arctan(−1)=3π4.

Because (x,y)=(−1,1),(x,y)=(−1,1), then the correct choice for θθ is 3π4.3π4.

There are actually two ways to identify φ.φ. We can use the equation φ=arccos(zx2+y2+z2).φ=arccos(zx2+y2+z2). A more simple approach, however, is to use equation z=ρcosφ.z=ρcosφ. We know that z=6z=6 and ρ=22,ρ=22, so

6=22cosφ,socosφ=622=326=22cosφ,socosφ=622=32

and therefore φ=π6.φ=π6. The spherical coordinates of the point are (22,3π4,π6).(22,3π4,π6).

To find the cylindrical coordinates for the point, we need only find r:r:

r=ρsinφ=22sin(π6)=2.r=ρsinφ=22sin(π6)=2.

The cylindrical coordinates for the point are (2,3π4,6).(2,3π4,6).

Example 2.65

Identifying Surfaces in the Spherical Coordinate System

Describe the surfaces with the given spherical equations.

  1. θ=π3θ=π3
  2. φ=5π6φ=5π6
  3. ρ=6ρ=6
  4. ρ=sinθsinφρ=sinθsinφ

Solution

  1. The variable θθ represents the measure of the same angle in both the cylindrical and spherical coordinate systems. Points with coordinates (ρ,π3,φ)(ρ,π3,φ) lie on the plane that forms angle θ=π3θ=π3 with the positive x-axis. Because ρ>0,ρ>0, the surface described by equation θ=π3θ=π3 is the half-plane shown in Figure 2.101.
    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 theta = pi/3.
    Figure 2.101 The surface described by equation θ=π3θ=π3 is a half-plane.
  2. Equation φ=5π6φ=5π6 describes all points in the spherical coordinate system that lie on a line from the origin forming an angle measuring 5π65π6 rad with the positive z-axis. These points form a half-cone (Figure 2.102). Because there is only one value for φφ that is measured from the positive z-axis, we do not get the full cone (with two pieces).
    This figure is the upper part of an elliptical cone. The bottom point of the cone is at the origin of the 3-dimensional coordinate system.
    Figure 2.102 The equation φ=5π6φ=5π6 describes a cone.

    To find the equation in rectangular coordinates, use equation φ=arccos(zx2+y2+z2).φ=arccos(zx2+y2+z2).
    5π6=arccos(zx2+y2+z2)cos5π6=zx2+y2+z232=zx2+y2+z234=z2x2+y2+z23x24+3y24+3z24=z23x24+3y24z24=0.5π6=arccos(zx2+y2+z2)cos5π6=zx2+y2+z232=zx2+y2+z234=z2x2+y2+z23x24+3y24+3z24=z23x24+3y24z24=0.

    This is the equation of a cone centered on the z-axis.
  3. Equation ρ=6ρ=6 describes the set of all points 66 units away from the origin—a sphere with radius 66 (Figure 2.103).
    This figure is a sphere. The z-axis is vertically through the center and intersects the sphere at (0, 0, 6). The y-axis is horizontally through the center and intersects the sphere at (0, 6, 0).
    Figure 2.103 Equation ρ=6ρ=6 describes a sphere with radius 6.6.
  4. To identify this surface, convert the equation from spherical to rectangular coordinates, using equations y=ρsinφsinθy=ρsinφsinθ and ρ2=x2+y2+z2:ρ2=x2+y2+z2:
    ρ=sinθsinφρ2=ρsinθsinφMultiply both sides of the equation byρ.x2+y2+z2=ySubstitute rectangular variables using the equations above.x2+y2y+z2=0Subtractyfrom both sides of the equation.x2+y2y+14+z2=14Complete the square.x2+(y12)2+z2=14.Rewrite the middle terms as a perfect square.ρ=sinθsinφρ2=ρsinθsinφMultiply both sides of the equation byρ.x2+y2+z2=ySubstitute rectangular variables using the equations above.x2+y2y+z2=0Subtractyfrom both sides of the equation.x2+y2y+14+z2=14Complete the square.x2+(y12)2+z2=14.Rewrite the middle terms as a perfect square.

    The equation describes a sphere centered at point (0,12,0)(0,12,0) with radius 12.12.
Checkpoint 2.59

Describe the surfaces defined by the following equations.

  1. ρ=13ρ=13
  2. θ=2π3θ=2π3
  3. φ=π4φ=π4

Spherical coordinates are useful in analyzing systems that have some degree of symmetry about a point, such as the volume of the space inside a domed stadium or wind speeds in a planet’s atmosphere. A sphere that has Cartesian equation x2+y2+z2=c2x2+y2+z2=c2 has the simple equation ρ=cρ=c in spherical coordinates.

In geography, latitude and longitude are used to describe locations on Earth’s surface, as shown in Figure 2.104. Although the shape of Earth is not a perfect sphere, we use spherical coordinates to communicate the locations of points on Earth. Let’s assume Earth has the shape of a sphere with radius 40004000 mi. We express angle measures in degrees rather than radians because latitude and longitude are measured in degrees.

This figure is an image of the Earth. It has the prime meridian labeled, which is a circle on the surface circumnavigating the Earth vertically through the poles. The equator is also labeled which is a horizontal circle circumnavigating the Earth. Three vectors extend out from the center of Earth. Two of them extend to the equator and indicate a measurement of longitude. Two of them extend to a vertical polar circle and indicate a measurement of latitude.
Figure 2.104 In the latitude–longitude system, angles describe the location of a point on Earth relative to the equator and the prime meridian.

Let the center of Earth be the center of the sphere, with the ray from the center through the North Pole representing the positive z-axis. The prime meridian represents the trace of the surface as it intersects the xz-plane. The equator is the trace of the sphere intersecting the xy-plane.

Example 2.66

Converting Latitude and Longitude to Spherical Coordinates

The latitude of Columbus, Ohio, is 40°40° N and the longitude is 83°83° W, which means that Columbus is 40°40° north of the equator. Imagine a ray from the center of Earth through Columbus and a ray from the center of Earth through the equator directly south of Columbus. The measure of the angle formed by the rays is 40°.40°. In the same way, measuring from the prime meridian, Columbus lies 83°83° to the west. Express the location of Columbus in spherical coordinates.

Solution

The radius of Earth is 40004000 mi, so ρ=4000.ρ=4000. The intersection of the prime meridian and the equator lies on the positive x-axis. Movement to the west is then described with negative angle measures, which shows that θ=−83°,θ=−83°, Because Columbus lies 40°40° north of the equator, it lies 50°50° south of the North Pole, so φ=50°.φ=50°. In spherical coordinates, Columbus lies at point (4000,−83°,50°).(4000,−83°,50°).

Checkpoint 2.60

Sydney, Australia is at 34°S34°S and 151°E.151°E. Express Sydney’s location in spherical coordinates.

Cylindrical and spherical coordinates give us the flexibility to select a coordinate system appropriate to the problem at hand. A thoughtful choice of coordinate system can make a problem much easier to solve, whereas a poor choice can lead to unnecessarily complex calculations. In the following example, we examine several different problems and discuss how to select the best coordinate system for each one.

Example 2.67

Choosing the Best Coordinate System

In each of the following situations, we determine which coordinate system is most appropriate and describe how we would orient the coordinate axes. There could be more than one right answer for how the axes should be oriented, but we select an orientation that makes sense in the context of the problem. Note: There is not enough information to set up or solve these problems; we simply select the coordinate system (Figure 2.105).

  1. Find the center of gravity of a bowling ball.
  2. Determine the velocity of a submarine subjected to an ocean current.
  3. Calculate the pressure in a conical water tank.
  4. Find the volume of oil flowing through a pipeline.
  5. Determine the amount of leather required to make a football.
    This figure has 5 images. The first image shows bowling balls. The second image is a submarine traveling on an ocean surface. The third image is a traffic cone. The fourth image is a pipeline across some barren land. The fifth image is a football.
    Figure 2.105 (credit: (a) modification of work by scl hua, Wikimedia, (b) modification of work by DVIDSHUB, Flickr, (c) modification of work by Michael Malak, Wikimedia, (d) modification of work by Sean Mack, Wikimedia, (e) modification of work by Elvert Barnes, Flickr)

Solution

  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,z=kr, where kk is a constant. In spherical coordinates, we have seen that surfaces of the form φ=cφ=c are half-cones. Last, in rectangular coordinates, elliptic cones are quadric surfaces and can be represented by equations of the form z2=x2a2+y2b2.z2=x2a2+y2b2. 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.
Checkpoint 2.61

Which coordinate system is most appropriate for creating a star map, as viewed from Earth (see the following figure)?

This figure is a circle with a star chart in the middle.

How should we orient the coordinate axes?

Section 2.7 Exercises

Use the following figure as an aid in identifying the relationship between the rectangular, cylindrical, and spherical coordinate systems.

This figure is the first octant of the 3-dimensional coordinate system. There is a line segment from the origin upwards. It is labeled “rho.” The angle between this line segment and the z-axis is labeled “phi.” There is also a broken line from the origin to the shadow of the point. This line segment is in the x y-plane and is labeled “r.” The angle between r and the x-axis is labeled “theta.”

For the following exercises, the cylindrical coordinates (r,θ,z)(r,θ,z) of a point are given. Find the rectangular coordinates (x,y,z)(x,y,z) of the point.

363.

(4,π6,3)(4,π6,3)

364.

(3,π3,5)(3,π3,5)

365.

(4,7π6,3)(4,7π6,3)

366.

(2,π,−4)(2,π,−4)

For the following exercises, the rectangular coordinates (x,y,z)(x,y,z) of a point are given. Find the cylindrical coordinates (r,θ,z)(r,θ,z) of the point.

367.

(1,3,2)(1,3,2)

368.

(1,1,5)(1,1,5)

369.

(3,−3,7)(3,−3,7)

370.

(−22,22,4)(−22,22,4)

For the following exercises, the equation of a surface in cylindrical coordinates is given.

Find the equation of the surface in rectangular coordinates. Identify and graph the surface.

371.

[T] r=4r=4

372.

[T] z=r2cos2θz=r2cos2θ

373.

[T] r2cos(2θ)+z2+1=0r2cos(2θ)+z2+1=0

374.

[T] r=3sinθr=3sinθ

375.

[T] r=2cosθr=2cosθ

376.

[T] r2+z2=5r2+z2=5

377.

[T] r=2secθr=2secθ

378.

[T] r=3cscθr=3cscθ

For the following exercises, the equation of a surface in rectangular coordinates is given. Find the equation of the surface in cylindrical coordinates.

379.

z=3z=3

380.

x=6x=6

381.

x2+y2+z2=9x2+y2+z2=9

382.

y=2x2y=2x2

383.

x2+y216x=0x2+y216x=0

384.

x2+y23x2+y2+2=0x2+y23x2+y2+2=0

For the following exercises, the spherical coordinates (ρ,θ,φ)(ρ,θ,φ) of a point are given. Find the rectangular coordinates (x,y,z)(x,y,z) of the point.

385.

(3,0,π)(3,0,π)

386.

(1,π6,π6)(1,π6,π6)

387.

(12,π4,π4)(12,π4,π4)

388.

(3,π4,π6)(3,π4,π6)

For the following exercises, the rectangular coordinates (x,y,z)(x,y,z) of a point are given. Find the spherical coordinates (ρ,θ,φ)(ρ,θ,φ) of the point. Express the measure of the angles in degrees rounded to the nearest integer.

389.

(4,0,0)(4,0,0)

390.

(−1,2,1)(−1,2,1)

391.

(0,3,0)(0,3,0)

392.

(−2,23,4)(−2,23,4)

For the following exercises, the equation of a surface in spherical coordinates is given. Find the equation of the surface in rectangular coordinates. Identify and graph the surface.

393.

[T] ρ=3ρ=3

394.

[T] φ=π3φ=π3

395.

[T] ρ=2cosφρ=2cosφ

396.

[T] ρ=4cscφρ=4cscφ

397.

[T] φ=π2φ=π2

398.

[T] ρ=6cscφsecθρ=6cscφsecθ

For the following exercises, the equation of a surface in rectangular coordinates is given. Find the equation of the surface in spherical coordinates. Identify the surface.

399.

x2+y23z2=0,x2+y23z2=0, z0z0

400.

x2+y2+z24z=0x2+y2+z24z=0

401.

z=6z=6

402.

x2+y2=9x2+y2=9

For the following exercises, the cylindrical coordinates of a point are given. Find its associated spherical coordinates, with the measure of the angle φφ in radians rounded to four decimal places.

403.

[T] (1,π4,3)(1,π4,3)

404.

[T] (5,π,12)(5,π,12)

405.

(3,π2,3)(3,π2,3)

406.

(3,π6,3)(3,π6,3)

For the following exercises, the spherical coordinates of a point are given. Find its associated cylindrical coordinates.

407.

(2,π4,π2)(2,π4,π2)

408.

(4,π4,π6)(4,π4,π6)

409.

(8,π3,π2)(8,π3,π2)

410.

(9,π6,π3)(9,π6,π3)

For the following exercises, find the most suitable system of coordinates to describe the solids.

411.

The solid situated in the first octant with a vertex at the origin and enclosed by a cube of edge length a,a, where a>0a>0

412.

A spherical shell determined by the region between two concentric spheres centered at the origin, of radii of aa and b,b, respectively, where b>a>0b>a>0

413.

A solid inside sphere x2+y2+z2=9x2+y2+z2=9 and outside cylinder (x32)2+y2=94(x32)2+y2=94

414.

A cylindrical shell of height 1010 determined by the region between two cylinders with the same center, parallel rulings, and radii of 22 and 5,5, respectively

415.

[T] Use a CAS to graph in cylindrical coordinates the region between elliptic paraboloid z=x2+y2z=x2+y2 and cone x2+y2z2=0.x2+y2z2=0.

416.

[T] Use a CAS to graph in spherical coordinates the “ice cream-cone region” situated above the xy-plane between sphere x2+y2+z2=4x2+y2+z2=4 and elliptical cone x2+y2z2=0.x2+y2z2=0.

417.

Washington, DC, is located at 39°39° N and 77°77° W (see the following figure). Assume the radius of Earth is 40004000 mi. Express the location of Washington, DC, in spherical coordinates.

This figure is an image of a globe. On the globe there is a point labeled where Washington, DC is located. It is labeled with 39 degrees north latitude and 77 degrees west longitude.
418.

San Francisco is located at 37.78°N37.78°N and 122.42°W.122.42°W. Assume the radius of Earth is 40004000 mi. Express the location of San Francisco in spherical coordinates.

419.

Find the latitude and longitude of Rio de Janeiro if its spherical coordinates are (4000,43.17°,102.91°).(4000,43.17°,102.91°).

420.

Find the latitude and longitude of Berlin if its spherical coordinates are (4000,13.38°,37.48°).(4000,13.38°,37.48°).

421.

[T] Consider the torus of equation (x2+y2+z2+R2r2)2=4R2(x2+y2),(x2+y2+z2+R2r2)2=4R2(x2+y2), where Rr>0.Rr>0.

  1. Write the equation of the torus in spherical coordinates.
  2. If R=r,R=r, the surface is called a horn torus. Show that the equation of a horn torus in spherical coordinates is ρ=2Rsinφ.ρ=2Rsinφ.
  3. Use a CAS to graph the horn torus with R=r=2R=r=2 in spherical coordinates.
422.

[T] The “bumpy sphere” with an equation in spherical coordinates is ρ=a+bcos(mθ)sin(nφ),ρ=a+bcos(mθ)sin(nφ), with θ[0,2π]θ[0,2π] and φ[0,π],φ[0,π], where aa and bb are positive numbers and mm and nn are positive integers, may be used in applied mathematics to model tumor growth.

  1. Show that the “bumpy sphere” is contained inside a sphere of equation ρ=a+b.ρ=a+b. Find the values of θθ and φφ at which the two surfaces intersect.
  2. Use a CAS to graph the surface for a=14,a=14, b=2,b=2, m=4,m=4, and n=6n=6 along with sphere ρ=a+b.ρ=a+b.
  3. Find the equation of the intersection curve of the surface at b. with the cone φ=π12.φ=π12. Graph the intersection curve in the plane of intersection.
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