Magnitude of a Base

  • Thread starter Roary
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  • #1
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Main Question or Discussion Point

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?)
 

Answers and Replies

  • #2
Integral
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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.
 
  • #3
HallsofIvy
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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.
 
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