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Cylindrical coordinates

  1. May 4, 2009 #1
    1. The problem statement, all variables and given/known data
    z = 4y^2, x = 0, is rotated about the z axis. write the equation of the resulting surface in cylindrical coordinates

    2. Relevant equations



    3. The attempt at a solution

    not really sure what the x = 0 means so i ignored it

    i solved for y because that would be my radius ,y = (z/4)^1/2 = r

    x = (z/4)^1/2*cos
    y = (z/4)^1/2*sin
    z = z

    am i correct
     
  2. jcsd
  3. May 4, 2009 #2

    dx

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    No. First, draw the curve. x = 0 means the curve lies in the yz plane, and consists of the points that satisfy z = 4y². After you draw the curve, draw the surface of revolution. What does it look like?
     
  4. May 4, 2009 #3
    well i understand that it is initialy in the yx plane, but once you rotate it, it does go into the x plane doesnt it. like if u viwed it from above the z axis you sould see a circle.
     

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  5. May 5, 2009 #4

    Mark44

    Staff: Mentor

    No, the original curve is a parabola in the y-z plane. If x were not specified, the surface would be a parabolic cylinder (sort of a trough shape). Since x = 0 is specified, the curve is strictly in the y-z plane.

    There is no x plane, or y plane, or z plane. There is an x-axis, and a y-axis, and a z-axis.

    After you rotate the parabola z = 4y2 around the z-axis, you get a paraboloid, all of whose horizontal cross sections are circles. Some points that are on one of these cross sections are (0, y, 4y2) and (x, 0, 4x2). Can you figure out the radius and center of this typical circular cross section? That should go a long way in helping you get the equation of the surface in polar form.
     
  6. May 5, 2009 #5
    that all makes sense but i dont understand where i am incorect

    to find the radius of the circle would the same as finding the y value. so i solved for y. and as you mentioned all the circles, well the larger the z value, the larger the circle, just as my y accomidates. am i missing something here.
     
  7. May 11, 2009 #6

    dx

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    Re: 3) Cylindrical Coordinates

    No.

    Take a point P on the y axis with y = y, x = 0, z = 0. Now rotate it around the z axis by some angle θ. Where does it go to? What are the cylindrical coordinates of this point? (cylindrical coordinates are r, θ and z)

    What is the height of the surface above the point (r, θ, 0)?

    (Visualize the situation while you do this.)
     
    Last edited: May 11, 2009
  8. May 11, 2009 #7
    Re: 3) Cylindrical Coordinates

    Ok this problem is really confusing me.

    cylindrical coordinates are
    x = rcos(theta)
    y = r csin(theta)
    z = z

    Right

    So isnt the radius of the rotated surface = y value of z = 4y^2


    like if u drew it in just the yz plane you get a parabola of which the radius is always the y value.

    where am confused at
     
  9. May 11, 2009 #8

    dx

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    Re: 3) Cylindrical Coordinates

    x, y and z are Cartesian coordinates. r, θ and z are cylindrical coordinates. The equations you wrote, i.e.

    x = r cos θ
    y = r sin θ
    z = z

    tell you how these two systems of coordinates are related.

    When they ask you for the equation of the surface in in cylindrical coordinates, they want you to write down the equation that the points on the surface satisfy in terms of the r, θ and z coordinates.

    You're right when you say the radius of the rotated point is r = y. You're also right when you say the height of the surface above that point is 4y².

    This means that z = 4r², which is the answer.
     
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