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Magnitude of a Base

  1. Feb 4, 2004 #1
    The base of a solid is a circle of radius a, and its vertical cross sections are equilateral triangles. Find the radius of the circle if the volume of the solid is 10 cubic meters.

    Eq. Triangle: A = [sqrt(3)/4]s^2
    V = [sqrt(3)/4]*Integral{-a to a} 4(a^2 - x^2) dx

    V = 2* [sqrt(3)*Integral{0 to a} (a^2 - x^2) ]dx

    What next?
    (Do I use x^2+y^2=a^2 somewhere?)
  2. jcsd
  3. Feb 5, 2004 #2


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    I would use the fact that the base is a circle then use disks, integrating over the height to get the volume.

    [tex] V= \int \pi r^2 dh =10[/tex]

    Since the cross section is an equilateral it is easy to come up with the relationship between h and r, the limits of integration would be from 0 to x. Solve for x to get the height of the cone. Use the relationship between h and r to get the radius.
  4. Feb 5, 2004 #3


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    I think that Roary DID exactly what Integral is suggesting: that was how he got the integral as he did.

    To answer Roary's question: "(Do I use x^2+y^2=a^2 somewhere?)"
    Actually, you already have when you wrote the square of the base of the triangle as a^2- x^2. What you HAVEN'T used is the fact that the volume is 10 cubic centimeters.

    What do you do next? You have V = 2* [sqrt(3)*Integral{0 to a} (a^2 - x^2) ]dx so go ahead and evaluate that, then set it equal to 10 and solve for a.
    Last edited: Feb 6, 2004
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