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Volume of a double-lobed cam

  1. Mar 30, 2014 #1
    1. The problem statement, all variables and given/known data
    The surface of a double lobed cam are modeled by the inequalities:

    [itex]\frac{1}{4}[/itex][itex]\leq[/itex]r[itex]\leq[/itex][itex]\frac{1}{2}[/itex](1+cos2θ)

    and

    -9/(4(x2+y2+9)) ≤ z ≤ 9/(4(x2+y2+9))

    Find the volume of the steel in the cam.


    2. Relevant equations



    3. The attempt at a solution
    I know I need to use a double or triple integral to solve this. I was thinking since I was given r I could change to polar coordinates and solve that way.

    Please give me some guidance.
     
  2. jcsd
  3. Mar 30, 2014 #2

    LCKurtz

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    Hint: General formula for volumes:
    $$\iint_R z_{upper}-z_{lower}~dA$$
     
  4. Mar 30, 2014 #3
    Ok, that makes sense. Wasn't sure if I could do that.

    R would be the r given. Theta is from 0 to 2*pi.

    Therefore I can convert the bounds to polar coordinates.

    My z_upper - z_lower is taken from the z given in the inequality.

    x^2 + y^2 become r^2 and I can integrate completely from there.

    Right?
     
  5. Mar 30, 2014 #4

    LCKurtz

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    If you mean what I think you mean, yes.
     
  6. Mar 30, 2014 #5
    Ha ha ok. Thanks LCKurtz.

    When I evaluate the integral (using a calculator) I get 0.79993.

    This seems awfully low to be a volume of an shape like this.

    - I checked it twice for errors, I think it's accurate.
     
  7. Mar 30, 2014 #6

    LCKurtz

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    That looks like it might be about right. Here's a picture (I had a little time to waste):
    attachment.php?attachmentid=68156&stc=1&d=1396220575.jpg
     

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    • cam.jpg
      cam.jpg
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  8. Mar 30, 2014 #7
    Wow, thanks for wasting your time for me! ;-)

    I guess the cam is pretty small so the volume is more reasonable than I thought.
     
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