Finding volume of a frustum of a pyramidI need to find the volume of a frustum of a pyramid with square base of side b, square top of side a, and height h(using integrals). I have no idea how to do questions like these, I only know to use the disc/washer/cylindrical shells methods and rotate a region around any line and find its volume. For these type of question, I find myself at a loss as to where to even start. Any hints on general about starting these kind of problems are also appreciated!

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Allyson Vance · Accepted Answer

Step 1Ron Gordon is correct, but I do not know why his final answer was stated as such. It could be simplified further.  V  =      ∫          0              h        (  b  −            x      (      b      −      a      )        h        )    2    d  xHe got here because the line from the tip of the pyramid at origin to the end of the pyramid at H is of the form   m  x  +  b where the slope   M  =            (      b      −      a      )        h   and the area of a square is equal to   f  (  x      )    2  .Step 2From here you can multiply it out to be  V  =      ∫          0              h            b    2    −            2      b      x      (      b      −      a      )        h    +                    x        2            (      b      −      a              )        2                    h      2        d  xThen you can integrate it to be  V  =      b    2    h  −  b  h  (  b  −  a  )  +            h      (      b      −      a              )        2              3  From there it simplifies to  V  =      h    3    (      b    2    +  a  b  +      a    2    )This was verified as a correct answer in webassign and Wolfram

lamontalbanav6 · Accepted Answer

Step 1A volume element is   d  V  =  A  (  y  )  d  y where the sides in the square cross-sectional area A(y) behaves linearly with height:  A  (  y  )  =            [      b      −              y        h            (      b      −      a      )      ]        2    =      b    2    −  2  b  (  b  −  a  )      y    h    +            (      b      −      a              )        2                    h      2            y    2  Step 2So the integral is  V  =      ∫    0    h    d  y    A  (  y  )  =      b    2    h  −  b  (  b  −  a  )  h  +      1    3    (  b  −  a      )    2    hor, simplifying,   V  =      1    3                      b        3            −              a        3                    b      −      a        h

Find the volume of a frustum of a pyramid with square base of side b, square top of side a, and height h(using integrals).

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