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Volume of a triangular based solid

  1. Apr 6, 2012 #1
    1. The problem statement, all variables and given/known data
    the base of a solid is the triangular region bounded by the line x=0 (on the y-axis), the line y = 0 (the x-axis), and the line y = −2x + 2. Cross-sections perpendicular to the
    y-axis are squares. Find the volume of S.

    2. Relevant equations



    3. The attempt at a solution
    I was not sure where to start with this question. Is it something where is should be using disk shell method and rotating about an axis in order to find a volume? or is it done by finding the area of the triangular base (which I found to be 1) and multiplying by a length? I would really appreciate some help on how I could picture this question and go about it.
     
  2. jcsd
  3. Apr 6, 2012 #2

    Mark44

    Staff: Mentor

    There's no rotation involved. Sketch the solid first to get an idea of what it looks like. Use the fact that cross sections perpendicular to the y-axis are squares to get the incremental volume, and then sum those volumes to get the total volume.
     
  4. Apr 6, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, this is not rotated about an axis nor is it a volume. What you can do is argue that, slicing perpendicular to the y-axis, the volume of each such slice is its area times dy, then integrate with respect to dy.

    I chose the y- axis because we are told that the cross sections perpendicular to the y-axis are squares. All you need to do is determine what the side length of each such square is. And a side of each square is from the y-axis to the line y= -2x+ 2. What is the length of that line segment in terms of y?
     
  5. Apr 6, 2012 #4
    so if I solved y=-2x+2 for x which would be x=-1/2y+1 it would give me the length of one of the squares and then if I squared it and took the integral I could find the volume?
     
  6. Apr 6, 2012 #5

    Mark44

    Staff: Mentor

    Yes. And you integrate from y = 0 to y = 2.
     
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