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A simple volume problem

  1. Jan 20, 2009 #1
    Q: A solid has a base in the form of the ellipse: x^2/25 + y^2/16 = 1. Find the volume if every cross section perpendicular to the x-axis is an isosceles triangle whose altitude is 6 inches.
    I got 11.781 (or 3.75pi) but I just wanted to check my answer.
    Q: Use the same base and cross sections as #3, but change the axis to the y-axis.
    Here I got 7.54 (or 2.4pi)
     
  2. jcsd
  3. Jan 20, 2009 #2

    Dick

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    I think your answers are way too low. You'd better show your work.
     
  4. Jan 20, 2009 #3
    For the first one

    [tex]
    \frac{y^2}{16} = 1-\frac{x^2}{25}[/tex]

    [tex]
    y=\frac{\sqrt{25-x^2}}{20}[/tex]

    [tex]
    A=\frac{3\sqrt{25-x^2}}{10}[/tex]

    [tex]
    V=\int^5_{-5}Adx[/tex]

    [tex]
    V=\frac{3}{10}\left[\frac{1}{2}\sqrt{25-x^2}x + \frac{25}{2}\arcsin\left(\frac{x}{5}\right)\right]\left|^5_{-5}[/tex]

    [tex]
    \approx 11.781
    [/tex]
     
  5. Jan 20, 2009 #4

    Mentallic

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    Making y the subject of the ellipse equation I got:

    [tex]y=\pm \frac{4}{5}\sqrt{25-x^2}[/tex]

    I can't seem to follow where you went wrong there.
     
  6. Jan 20, 2009 #5

    Dick

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    y^2/16=1-x^2/25. y=4*sqrt(1-x^2/25). y=(4/5)*sqrt(25-x^2). A=(1/2)*6*(2*y). Stuff like that. You are doing something wrong.
     
  7. Jan 21, 2009 #6
    EDIT: I accidentally divided by 4 on the other side instead of multiplying.
     
    Last edited: Jan 21, 2009
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