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Homework Help: Volume of a tetrahedron

  1. May 30, 2006 #1
    Find the volume of "A tetrahedron with three mutually perpendicular faces and three mutually perpendicular edges with lengths 3 cm, 4 cm, and 5cm."

    This is how I visualized it:
    http://img282.imageshack.us/img282/9466/calculus31re.th.jpg [Broken]

    The area of a triangle along the x-axis is:
    A(x) = .5 * x * (3/5x), 3/5x is from similar triangles.
    A(x) = 3/10 * x2
    [tex]V = \int^{5}_{0} A(x) dx = \frac{3}{10} \int^{5}_{0} x^2 dx
    = \frac{1}{10} [x^3]^{5}_{0}
    = 12.5 cm^3[/tex]
    But the answer is 10 cm3. Why doesn't my method work?
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. May 30, 2006 #2

    StatusX

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    Right now your dx is wrong. It corresponds to a change in the length of the short side of the triangluar slice, where as it should be the thickness of the slice, which is dy. You need to find dx/dy, the change in the length of this side for a change in height y.
     
  4. May 30, 2006 #3

    benorin

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    The area of a right triangle is A=b*h/2, where b=base, h=height
    In your diagram, slice through your tetrahedron with a plane passing through a given x value (the shape made by this is a right triangle whose right angle has its corner on the x-axis). The height of that triangle is
    h=4(1-x/5) since the line that connects the vetricies on the y-axis and the x-axis is y/4+x/5=1; the base of that right triangle is b=3(1-x/5), since the line that connects the vetricies on the x-axis and the z-axis is
    z/3+x/5=1; hence the cross-sectional area is A(x)=3(1-x/5)*4(1-x/5)/2
    simplified this gives A(x)=6(1-x/5)^2, so the volume is

    [tex]6\int_{0}^{5}\left( 1-\frac{x}{5}\right) ^2dx =10[/tex]
     
  5. May 30, 2006 #4
    Would it not be easier to do that using a triple scalar product ( [itex] V = \frac{1}{6}a \cdot (b \times c) [/itex] ), where a, b and c are the vectors of the sides?
     
  6. May 30, 2006 #5
    This works...
    If I used a triangle perpendicular to the y-axis (along the x and z axes), then it seems like I would have to find the area in terms of y, right? But why doesn't finding the area in terms of x work?
     
  7. May 30, 2006 #6

    benorin

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    x is never the height nor base of any right triangle form by slicing that tetra hedron by a plane parallell to one of the coordinate planes.
     
  8. Jun 1, 2006 #7

    Tide

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    You can check your answer by noting that the volume of a pyramid is 1/3 the area of its base times its height.
     
  9. Mar 25, 2008 #8
    Sorry to resurrect this old thread, but I can't figure out why the line y/4+x/5=1 or the line z/3+x/5=1 are equal to 1?
     
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