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Math Help - Cylindrical Coordinate Problem

  1. #1
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    Cylindrical Coordinate Problem

    Im trying to work out the answer to the following question

    Describe the set in cylindrical coordinates.
    x2 + y2 + z2 ≤ 1

    you need to find the limit for r, theta and z.

    so far i have found that 0 ≤ r ≤ 1 and pi/4 ≤ θ

    i havent a clue how to find the rest, can anyone help me?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    x2 + y2 + z2 ≤ 1

    or r^2 +z^2 = 1 which is a sphere

    x2 + y2 + z2 ≤ 1

    is the solid ball 0 < r < 1 0 < theta < 2pi -1 < z < 1
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  3. #3
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    yeah but its not...

    right now an accepted answer is -7pi/4 < theta < pi/2
    and -1 < z < 1 is not the right answer... which doesnt make ANY sense, but still its not right

    i mean if y = 0 and x = 0, then that HAS to be true

    this stuff is driving me nuts
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  4. #4
    MHF Contributor Calculus26's Avatar
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    How is it not a sphere?

    does (0,0,-1) satisfy the inequality?

    does (0,0,1) satisfy the inequality ?

    does (0,0,z) for any z st -1 < z < 1 satisfy the inequality?
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  5. #5
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    Quote Originally Posted by Calculus26 View Post
    How is it not a sphere?

    does (0,0,-1) satisfy the inequality?

    does (0,0,1) satisfy the inequality ?

    does (0,0,z) for any z st -1 < z < 1 satisfy the inequality?

    Yeah your telling me...
    it is going into cylindrical polar coordinates, but it should be converted into a sphere non the less

    I hate this software....
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  6. #6
    MHF Contributor Calculus26's Avatar
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    You want

    0 < r < 1

    0 < theta < 2pi

    Then at each (r,theta) pair

    -sqrt(1-r^2) < z < sqrt(1-r^2)
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  7. #7
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    Quote Originally Posted by Calculus26 View Post
    You want

    0 < r < 1

    0 < theta < 2pi

    Then at each (r,theta) pair

    -sqrt(1-r^2) < z < sqrt(1-r^2)
    god i kinda hope its not... I was thinking about trying it but it didnt seem right for some reason

    i dont think it matters at this point, its due online in abou 20 mins n the freakin website is down due to high traffic

    load of @#$@ right there if u ask me...
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