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Volumes, Strang

  1. May 29, 2008 #1
    I have a simple question on the set-up of a triangular prism.

    Strang, Ch. 8: Applications of the Integral

    pdf page 4 (bottom) and correlates with example 5 on pdf page 5

    How is he getting the length of [tex]1-\frac x h[/tex]
  2. jcsd
  3. May 29, 2008 #2


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    Look at the last figure on 8.3. 6(1-x/h)^2 is the area of a cross section at x. The line going through the height of each triangle is H = 3(1 - x/h). Using similar triangles the base of each triangle is B = 4(1-x/h). Thus, area = A(x) = 1/2*B*H = 6(1-x/h)^2. The volume is then obtained by adding together all the areas of each cross section which is the integral of A(x) from x = 0 to x =h.
  4. May 29, 2008 #3


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    The bottom leg of the triangular section has a length of 4(1-x/h) and the side leg is 3(1-x/h).
  5. May 29, 2008 #4
    That is a length, he says so in the next page. I don't get how he's setting it up by using similar triangles.

    My main confusion comes from the 1 minus part, and the x/h part makes sense to me by just looking at the figures and what happens when x=0 & x=h.
  6. May 29, 2008 #5


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    The slope of the line is -3/h and it intersects the z axis at z = 3. Thus z = -3/h*x + 3 = 3(1 - x/h). Let y be the base. 3/y*(1-x/h) = 3/4 by similar triangles. Solving for y, y= 4(1-x/h).
  7. May 29, 2008 #6
    Oh that's easy! Lol, I didn't even think about that. Thanks Vid :)
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