Cylindrical and Spherical Coordinates

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Plot the point with cylindrical coordinates \((4,\frac{2\pi }{3},-2)\) and express its location in rectangular coordinates.

   Жавобни кўрсатиш

   Conversion from cylindrical to rectangular coordinates requires a simple application of the equations listed in :

   \[\begin{array}{lll}x & = & r\ \text{cos}\ \theta =4\ \text{cos}\ \frac{2\pi }{3}=-2 \\ y & = & r\ \text{sin}\ \theta =4\ \text{sin}\ \frac{2\pi }{3}=2\sqrt{3} \\ z & = & -2.\end{array}\]

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

2. Point \(R\) has cylindrical coordinates \((5,\frac{\pi }{6},4)\). Plot \(R\) and describe its location in space using rectangular, or Cartesian, coordinates.

   Жавобни кўрсатиш

   The rectangular coordinates of the point are \((\frac{5\sqrt{3}}{2},\frac{5}{2},4).\)
   

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

   Жавобни кўрсатиш

   Use the second set of equations from to translate from rectangular to cylindrical coordinates:

   \[\begin{array}{lll}{r}^{2} & = & {x}^{2}+{y}^{2} \\ r & = & \text{\pm }\sqrt{{1}^{2}+{(-3)}^{2}}=\text{\pm }\sqrt{10}.\end{array}\]

   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 \(2\pi \text{:}\)

   \[\begin{array}{lll}\text{tan}\ \theta & = & \frac{y}{x}=\frac{-3}{1} \\ \theta & = & \text{arctan}(-3)+2\pi \approx 5.03\ \text{rad}.\end{array}\]

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

   \[z=5.\]

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

4. Convert point \((-8,8,-7)\) from Cartesian coordinates to cylindrical coordinates.

   Жавобни кўрсатиш

   \((8\sqrt{2},\frac{3\pi }{4},-7)\)

5. Describe the surfaces with the given cylindrical equations.

   1. \(\theta =\frac{\pi }{4}\)
   2. \({r}^{2}+{z}^{2}=9\)
   3. \(z=r\)
   Жавобни кўрсатиш

   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).
      
   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).
      
   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).
      

6. Describe the surface with cylindrical equation \(r=6.\)

   Жавобни кўрсатиш

   This surface is a cylinder with radius \(6.\)
   

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

   Жавобни кўрсатиш

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

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

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

   Finding the values in cylindrical coordinates is equally straightforward:

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

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

8. Plot the point with spherical coordinates \((2,-\frac{5\pi }{6},\frac{\pi }{6})\) and describe its location in both rectangular and cylindrical coordinates.

   Жавобни кўрсатиш

   
   
   Cartesian: \((-\frac{\sqrt{3}}{2},-\frac{1}{2},\sqrt{3}),\) cylindrical: \((1,-\frac{5\pi }{6},\sqrt{3})\)

9. Convert the rectangular coordinates \((-1,1,\sqrt{6})\) to both spherical and cylindrical coordinates.

   Жавобни кўрсатиш

   Start by converting from rectangular to spherical coordinates:

   \[\begin{array}{llllllllllll}\begin{array}{lll}{\rho }^{2} & = & {x}^{2}+{y}^{2}+{z}^{2}={(-1)}^{2}+{1}^{2}+{(\sqrt{6})}^{2}=8 \\ \rho & = & 2\sqrt{2}\end{array} & & & \begin{array}{lll}\text{tan}\ \theta & = & \frac{1}{-1} \\ \theta & = & \text{arctan}(-1)=\frac{3\pi }{4}.\end{array}\end{array}\]

   Because \((x,y)=(-1,1),\) then the correct choice for \(\theta\) is \(\frac{3\pi }{4}.\)

   There are actually two ways to identify \(\phi .\) We can use the equation \(\phi =\text{arccos}(\frac{z}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}).\) 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 \((2\sqrt{2},\frac{3\pi }{4},\frac{\pi }{6}).\)

   To find the cylindrical coordinates for the point, we need only find \(r\text{:}\)

   \[r=\rho \ \text{sin}\ \phi =2\sqrt{2}\ \text{sin}(\frac{\pi }{6})=\sqrt{2}.\]

   The cylindrical coordinates for the point are \((\sqrt{2},\frac{3\pi }{4},\sqrt{6}).\)

10. Describe the surfaces with the given spherical equations.

    1. \(\theta =\frac{\pi }{3}\)
    2. \(\phi =\frac{5\pi }{6}\)
    3. \(\rho =6\)
    4. \(\rho =\text{sin}\ \theta \ \text{sin}\ \phi\)
    Жавобни кўрсатиш

    1. The variable \(\theta\) represents the measure of the same angle in both the cylindrical and spherical coordinate systems. Points with coordinates \((\rho ,\frac{\pi }{3},\phi )\) 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 .
       
    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 (). 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).
       
       To find the equation in rectangular coordinates, use equation \(\phi =\text{arccos}(\frac{z}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}).\)
       \[\begin{array}{lll}\frac{5\pi }{6} & = & \text{arccos}(\frac{z}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}) \\ \text{cos}\ \frac{5\pi }{6} & = & \frac{z}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}} \\ -\frac{\sqrt{3}}{2} & = & \frac{z}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}} \\ \frac{3}{4} & = & \frac{{z}^{2}}{{x}^{2}+{y}^{2}+{z}^{2}} \\ \frac{3{x}^{2}}{4}+\frac{3{y}^{2}}{4}+\frac{3{z}^{2}}{4} & = & {z}^{2} \\ \frac{3{x}^{2}}{4}+\frac{3{y}^{2}}{4}-\frac{{z}^{2}}{4} & = & 0.\end{array}\]
       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 \(6\) ().
       
    4. To identify this surface, convert the equation from spherical to rectangular coordinates, using equations \(y=\rho \ \text{sin}\ \phi \ \text{sin}\ \theta\) and \({\rho }^{2}={x}^{2}+{y}^{2}+{z}^{2}\text{:}\)
       \[\begin{array}{llllll}\rho & = & \text{sin}\ \theta \ \text{sin}\ \phi & & & \\ {\rho }^{2} & = & \rho \ \text{sin}\ \theta \ \text{sin}\ \phi & & & \text{Multiply both sides of the equation by}\ \rho . \\ {x}^{2}+{y}^{2}+{z}^{2} & = & y & & & \text{Substitute rectangular variables using the equations above.} \\ {x}^{2}+{y}^{2}-y+{z}^{2} & = & 0 & & & \text{Subtract}\ y\ \text{from both sides of the equation.} \\ {x}^{2}+{y}^{2}-y+\frac{1}{4}+{z}^{2} & = & \frac{1}{4} & & & \text{Complete the square.} \\ {x}^{2}+{(y-\frac{1}{2})}^{2}+{z}^{2} & = & \frac{1}{4}. & & & \text{Rewrite the middle terms as a perfect square.}\end{array}\]
       The equation describes a sphere centered at point \((0,\frac{1}{2},0)\) with radius \(\frac{1}{2}.\)

11. Describe the surfaces defined by the following equations.

    1. \(\rho =13\)
    2. \(\theta =\frac{2\pi }{3}\)
    3. \(\phi =\frac{\pi }{4}\)
    Жавобни кўрсатиш

    a. This is the set of all points \(13\) units from the origin. This set forms a sphere with radius \(13.\) b. This set of points forms a half plane. The angle between the half plane and the positive x-axis is \(\theta =\frac{2\pi }{3}.\) c. Let \(P\) be a point on this surface. The position vector of this point forms an angle of \(\phi =\frac{\pi }{4}\) with the positive z-axis, which means that points closer to the origin are closer to the axis. These points form a half-cone.

12. The latitude of Columbus, Ohio, is \(40\text{^{\circ}}\) N and the longitude is \(83\text{^{\circ}}\) W, which means that Columbus is \(40\text{^{\circ}}\) 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 \(40\text{^{\circ}}.\) In the same way, measuring from the prime meridian, Columbus lies \(83\text{^{\circ}}\) to the west. Express the location of Columbus in spherical coordinates.

    Жавобни кўрсатиш

    The radius of Earth is \(4000\) mi, so \(\rho =4000.\) 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 \(\theta =-83\text{^{\circ}},\) Because Columbus lies \(40\text{^{\circ}}\) north of the equator, it lies \(50\text{^{\circ}}\) south of the North Pole, so \(\phi =50\text{^{\circ}}.\) In spherical coordinates, Columbus lies at point \((4000,-83\text{^{\circ}},50\text{^{\circ}}).\)

13. Sydney, Australia is at \(34\text{^{\circ}}\text{S}\) and \(151\text{^{\circ}}\text{E}.\) Express Sydney’s location in spherical coordinates.

    Жавобни кўрсатиш

    \((4000,151\text{^{\circ}},124\text{^{\circ}})\)

14. 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 ().

    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.
       
    Жавобни кўрсатиш

    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.

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

    How should we orient the coordinate axes?

    Жавобни кўрсатиш

    Spherical coordinates with the origin located at the center of the earth, the z-axis aligned with the North Pole, and the x-axis aligned with the prime meridian

16. \((4,\frac{\pi }{6},3)\)

    Жавобни кўрсатиш

    \((2\sqrt{3},2,3)\)

17. \((3,\frac{\pi }{3},5)\)

18. \((4,\frac{7\pi }{6},3)\)

    Жавобни кўрсатиш

    \((-2\sqrt{3},-2,3)\)

19. \((2,\pi ,-4)\)

20. \((1,\sqrt{3},2)\)

    Жавобни кўрсатиш

    \((2,\frac{\pi }{3},2)\)

21. \((3,-3,7)\)

    Жавобни кўрсатиш

    \((3\sqrt{2},-\frac{\pi }{4},7)\)

22. \((-2\sqrt{2},2\sqrt{2},4)\)

23. [T] \(z={r}^{2}{\text{cos}}^{2}\theta\)

24. [T] \({r}^{2}\text{cos}(2\theta )+{z}^{2}+1=0\)

    Жавобни кўрсатиш

    Hyperboloid of two sheets of equation \(\text{-}{x}^{2}+{y}^{2}-{z}^{2}=1,\) with the y-axis as the axis of symmetry,
    

25. [T] \(r=3\ \text{sin}\ \theta\)

26. [T] \(r=2\ \text{cos}\ \theta\)

    Жавобни кўрсатиш

    Cylinder of equation \({x}^{2}-2x+{y}^{2}=0,\) with a center at \((1,0,0)\) and radius \(1,\) with rulings parallel to the z-axis,
    

27. [T] \({r}^{2}+{z}^{2}=5\)

28. [T] \(r=2\ \text{sec}\ \theta\)

    Жавобни кўрсатиш

    Plane of equation \(x=2,\)
    

29. [T] \(r=3\ \text{csc}\ \theta\)

30. \({x}^{2}+{y}^{2}+{z}^{2}=9\)

    Жавобни кўрсатиш

    \({r}^{2}+{z}^{2}=9\)

31. \(y=2{x}^{2}\)

32. \({x}^{2}+{y}^{2}-16x=0\)

    Жавобни кўрсатиш

    \(r=16\ \text{cos}\ \theta ,r=0\)

33. \({x}^{2}+{y}^{2}-3\sqrt{{x}^{2}+{y}^{2}}+2=0\)

34. \((3,0,\pi )\)

    Жавобни кўрсатиш

    \((0,0,-3)\)

35. \((1,\frac{\pi }{6},\frac{\pi }{6})\)

36. \((12,-\frac{\pi }{4},\frac{\pi }{4})\)

    Жавобни кўрсатиш

    \((6,-6,6\sqrt{2})\)

37. \((3,\frac{\pi }{4},\frac{\pi }{6})\)

38. \((-1,2,1)\)

39. \((-2,2\sqrt{3},4)\)

40. [T] \(\rho =3\)

    Жавобни кўрсатиш

    Sphere of equation \({x}^{2}+{y}^{2}+{z}^{2}=9\) centered at the origin with radius \(3,\)
    

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