Olentangy Construction Math

March Puzzler

We all use math every day in many different ways. Each month a problem will show how math is used by architects, engineers and builders in the construction of our schools.

See if you can solve the problem!

We removed (“stripped”) topsoil and placed it in a mound (“stockpile”) in order to reshape land to build some field hockey fields, and we now have to place (“respread”) the topsoil on the new field areas. If the topsoil stockpile is:

  1. circular with a diameter of 100 feet
  2. symmetrical about its vertical center line with a height at its center of 25 feet
  3. defined in cross-section by the equation y = -0.01x2,

then how thick will the topsoil be after it is spread on the new field hockey fields area of 3 acres?

mound of topsoil

Hints:

  1. There are 43,560 square feet in one acre
  2. This problem can be solved using either integration or approximation using stacked cylinders
  3. Assume that the ground is the x-axis and the y-axis passes through the center of the pile (the vertex of the parabola)

How thick will the topsoil be after it is spread on the new field hockey fields area of 3 acres?

Answer:

The topsoil on the field hockey fields will be 0.75 feet (9 inches) thick.

Solution:

mound of topsoil
  1. The volume of the pile of dirt can be found by rotating about the y axis the region bounded by the line y = 0 (the x axis) and the parabola y = -0.01x2 + 25
  2. Each “slice” taken perpendicular to the y axis forms the circle of area A = πr2, where the radius, r, is the distance x to the edge of the dirt pile (the parabola)
  3. The volume of this pile is found by adding the areas of all the circles, of negligible thickness, that are inside the parabola. This can be done by taking the integral of the equation for the area of these circles
    A = πr2
    r = x, so A = πx2
    Solving the equation y = -0.01x2 + 25 for x2, we get x2 = -100y + 2500
    Using substitution: A = π (-100y + 2500)
  4. The integral for this problem is bounded by the parabola and the line y = 0. The vertex of the parabola is at the point (0,25), so the limits of integration would be 0 to 25
    V = ∫π(-100y + 2500) dy
    = ∫ -100πy + 2500π dy
    = -50πy2 + 2500πy from y = 0 to y = 25
    = (-50π (25)2 + 2500π(25)) – (-50π (0)2 + 2500π(0))
    = 98,174.77 cubic feet of topsoil in the mound
  5. Now that we know the volume of topsoil we have available, we will find the area of the field hockey fields that we need to cover with the soil
    Area of fields = 3 acres

    1 acre=3 acres
    43,560 ft2x ft2
    x = 130,680 ft2
  6. Assuming the topsoil is like a rectangular prism on top of the fields, we can set the volume of topsoil available equal to the area found in step 5 times the thickness (T) of the topsoil
    98,174.77 ft3 = (130,680 ft3) (T)
    T = 0.75 feet

The topsoil will be 0.75 feet (9 inches) thick