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Homework Help: Volume of Tetrahedron

  1. Mar 24, 2010 #1
    Volume of Tetrahedron[Solved]

    20fdn29.jpg

    My text book opts to integrate with respect to y before x(dydx vs dxdy), so I assumed that it would not affect the outcome.

    I set the upper and lower bounds of y, respectively, as y = 24 - 7x/4 (from 7x+4y=96) to y = x/4 (from x = 4y). For x I set it from upper bound x = 12 (using x/4 = 24 - 7x/4) to lower bound x = 0 (given).

    Integrating with respect to y, I get [tex]\int(96y - 7xy - 2y^2)|dx[/tex] which after inputing the bounds become [tex]\int(1152-144x+8x^2)dx[/tex]. After integrating this I get [tex](1152x - 72x^2 + 8x^3/3)|[/tex].
    Computing this I get 1152(12) - 72(144) + 8(1728)/3 = 13824-10368+4608 = 8064.

    I've been at this for a few hours now, but I can't seem to find my error.
    Any help would be great,
    Thanks in advance
     
    Last edited: Mar 25, 2010
  2. jcsd
  3. Mar 24, 2010 #2

    LCKurtz

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    It would be good to learn a little tex to make it easier to read. Just click on the equation to see it:

    [tex]\int_0^{12}\int_{\frac x 4}^{24 - \frac {7x}{4} }dydx[/tex]

    If I read your post correctly, this is how you set up the integral and it is correct. You just need to chase down the arithmetic error in your calculation, which, unfortunately, I don't have time right now to help you with. Good luck.
     
  4. Mar 24, 2010 #3
    I get 8x^2 - 192x + 1152 after integrating w/ respect to y.
     
  5. Mar 25, 2010 #4
    Thank you both! I'll definitely work on my latex once things settle down so that I don't cause so much confusion. And yes, in fact I add 24 to -168 instead of subtracting.
     
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