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Parametrize a cylinder

  1. Dec 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Parametrize the part of the cylinder 4y^2 + z^2 = 36 between the planes x= -3 and x=7

    3. The attempt at a solution
    Parametric equations:
    y=4 + 6cos(theta)

    in vector form
    [tex]\widehat{}r[/tex]= x[tex]\widehat{}i[/tex] + (4 + 6cos(theta))[tex]\widehat{}j[/tex] + 6sin(theta)[tex]\widehat{}k[/tex]

    -3 [tex]\leq[/tex] x [tex]\leq[/tex] 7

    I really don't know if I'm completing correctly, any direction would be appreciated. Thanks!
  2. jcsd
  3. Dec 5, 2008 #2
    These y,z do NOT satisfy 4y^2+z^2=36. Moreover, you cannot speak of a radius here, as the crosssection of the cylinder is not a circle.
  4. Dec 5, 2008 #3
    the cross sections of the cylinder perpendicular to the x axis are circles y^2 + z^2=6^2 correct?
    then this would be given parametrically as y=6cos(theta) and z=6sin(theta) right?
  5. Dec 5, 2008 #4
    This is correct, but in your original problem the y^2 is multiplied by a factor 4.
    If you let y'=2y, then you have

    y'^2+z^2=36 and so

    y'=6 cos(theta)
    z =6 sin(theta)

    Now use y=y'/2 and you're done.
  6. Dec 5, 2008 #5
    Thanks I did not know what to do with factor of 4

    so would this be correct???

    in vector form
    [tex]\widehat{}r[/tex]= x[tex]\widehat{}i[/tex] + 3cos(theta)[tex]\widehat{}j[/tex] + 6sin(theta)[tex]\widehat{}k[/tex]

    -3 [tex]\leq[/tex] x [tex]\leq[/tex] 7

    do i need to say anything about theta going from 0 to 2pi?
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