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Homework Help: Volume of the frustum of a pyramid

  1. Jun 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the frustum of a pyramid with square base of side b,
    square top of side a, and height h. What happens if a = b ? If a = 0 ?

    2. Relevant equations


    3. The attempt at a solution

    I know that i have to integrate from 0 to H. I make a generic square with sides X. The area of the square would be x*x or X^2 and the height would be dx. I have to integrate that, but how can i find what exactly X is? Since the base is a larger square, i know the x should be decreasing as it moves from base b towards base a. My problem is that i cannot figure out how to get the value of x since there are no numbers given for the problem.

    Also, can someone confirm this if its right or not?
    if a=b then the bottom square and the top square become the same length so the shape would become a rectangular box right?
    If a = 0 then the shape would just become a regular pyramid rather than a frustum of a pyramid?
  2. jcsd
  3. Jun 17, 2010 #2


    Staff: Mentor

    No, the height or thickness of your volume element would be dy.

    If you haven't already done so, draw a picture of the vertical cross-section of your frustum of a pyramid, with the base of the pyramid along the x-axis. The endpoints of the base should be at (-b/2, 0) and (b/2, 0). The endpoints of the top of the frustum should be at (-a/2, H) and (a/2, H). You will need equations for the slanted sides.
  4. Jun 18, 2010 #3
    once i find the equations of the slanted sides what exactly am i supposed to do?
  5. Jun 19, 2010 #4


    Staff: Mentor

    Then you can get an equation for x in terms of y. You'll be integrating with respect to y, so you need x in terms of y.
  6. Jun 19, 2010 #5
    maybe im not understanding your answer correctly. This is what i have for y so far. y = 2(b/2 +(a/2-b/2)*y/h). I put the two outside so whatever the x value comes out to be it will be multiplied by 2 that way i get the whole length and not just half of it. If h = 0 then y = 0 , thus we are left with 2b/2 or just b, if y = h then its just 2(b/2+(a/2-b/2). Which comes out to be b/2 + a-b/2 which results in 2a/2 or a. Since i have that in terms of y could i integrate that from 0 to h? The integral would be o to h of (2(b/2+(1/2-b/2)) ^2. Since its a square in order to get the area i just squared the side. Is it right so far?
  7. Jun 19, 2010 #6


    Staff: Mentor

    This equation makes no sense at all, since it does not include x. Any line that is neither vertical nor horizontal will have an equation that includes two variables, usually x and y.

    Since you are attempting to work an calculus problem, you should know how to find the equation of a line, given two points on the line. In this case, the two points are (b/2, 0) and (a/2, H).

    When you get the equation of the line, it will probably be y = <an expression involving x>. Solve that equation so that you get x as a function of (in terms of) y.
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