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Calculate the Volume of a Lemonsqueezer

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data

    [itex]f(x)=\frac{1}{81}*x^4-\frac{5}{9}*x^2+4[/itex]
    The tangent in Point P(6|0) when rotated around the y-Axis gives the Shape of the Squeezer. The bottom is at y=-5, the top at y=0

    3. The attempt at a solution

    First I calculated the tangent and got

    [itex]t: y=4x-24[/itex]
    Then I converted that to [itex]x^2[/itex] since I need that for the rotation around the y-Axis
    [itex]x^2=\frac{y^2}{16}+3y+36[/itex]

    Then I rotated it:

    [itex]V=\pi\int\frac{y^2}{16}+3y+36dy = \pi(\frac{y^3}{48}+\frac{3y^2}{2}+36y)(0 to -5)[/itex]
    [itex] = \pi(0+0-(-\frac{125}{48}+\frac{75}{2}-180) = \pi(\frac{125}{48}-\frac{1800}{48}+\frac{8640}{48})[/itex]
    [itex] = \frac{6965\pi}{48} = 455.86[/itex]

    Since this is in cL, which would equal around 4.6L I find the result a bit excessive. Have I made a mistake somewhere? Can someone please check for me? Thank you in advance :)
     
    Last edited by a moderator: Apr 2, 2014
  2. jcsd
  3. Apr 2, 2014 #2

    LCKurtz

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    No mistake. That is correct.
     
    Last edited by a moderator: Apr 2, 2014
  4. Apr 2, 2014 #3
    4.6L Lemonsqueezer? Okay thank you^^
     
  5. Apr 2, 2014 #4

    LCKurtz

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    The problem states nothing about units. You can tell the 455.8 is reasonably close since the volume would be a bit less than a disk of radius ##6## and thickness ##5##$$
    5\pi 6^2 = 565.48$$
     
  6. Apr 2, 2014 #5
    Yeah it does. It states 1 unit = 1cm and the Volume in cl
    Thank you though :)
     
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