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Solid formed by rotating equilateral triangle.

  1. Mar 4, 2007 #1

    Mo

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    1. The problem statement, all variables and given/known data
    Determine the surface area and volume of a solid formed by rotating an equilateral triangle of side a about its base.


    2. Relevant equations



    3. The attempt at a solution

    ----------working out surface area--------
    Integral of 2.pi.y dx with limits of a and 0.
    I worked out; y = root3 . x
    therefore surface area = root3.pi.a^2.
    ------------------------------------------

    ---------volume of revolution---------------
    integral of pi.y^2. dx with limits of a and 0.
    y = root3.x
    therefore volume of solid formed = pi.a^3
    ------------------------------------------

    I hope someone can check my answer to the first part (working out surface area) and then tell me where i went wrong with the volume of revolution.

    The answer for the second part (volume of revolution) should be (pi.a^3)/4

    Thanks.
     
  2. jcsd
  3. Mar 4, 2007 #2
    Pretty close. The problem is with your equation y= sqrt(3) * x
    I was working out the problem myself and was going to use symmetry, integrating from 0 to a/2, and multiply the answer by 2. This is when I noticed that your equation for y only works for values of x from 0 to a/2. i.e. the angle remains 60 degrees, and x keeps increasing. However, after a/2, the value of y starts decreasing.
     
  4. Mar 4, 2007 #3

    Mo

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    Yes, thats true. So now i have worked out the value of surface area for half of the triangle and found that to be (pi.root3.a^2)/4

    For the whole triangle, this value ahs to be multiplied by 2 surely?. Therefore final answer = (pi.root3.a^2)/2

    The thing is, the answer i got given for this (first) part of the question was (pi.root3.a^2).

    What else am i misssing in the solution?

    Thanks for your help.
     
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