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Math Help - Volume of a solid of revolution

  1. #1
    HMV
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    Volume of a solid of revolution

    Find the volume of the solid whose base is bounded by the circle x^2+y^2=4 with Equilateral triangle cross sections taken perpendicular to the x-axis.
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  2. #2
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    area of an equilateral triangle of side length " s" is A = \frac{\sqrt{3}}{4} s^2

    for your problem, the side of an equilateral cross-section is s = 2y.

    equation for the volume of the described solid is

    V = \int_{-2}^{2} A(x) \, dx

    get the cross-sectional area as a function of x and integrate.
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  3. #3
    HMV
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    I was able to work it out and got V=8. For semicircles, did i work it correctly?
    y=(4-x^2)^(1/2)
    A(x)=(pi r^2)/2
    r=2(4-x^2)^(1/2)

    V= Integral pi((4-x^2)^2)dx Limits [-2,2]= 32pi/3
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  4. #4
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    Quote Originally Posted by HMV View Post
    I was able to work it out and got V=8.

    not correct. how did you get V = 8?

    For semicircles, did i work it correctly?
    y=(4-x^2)^(1/2)
    A(x)=(pi r^2)/2
    r=2(4-x^2)^(1/2)

    V= Integral pi((4-x^2)^2)dx Limits [-2,2]= 32pi/3

    you got double the correct volume ... r = (4-x^2)^(1/2)

    note that it's much easier if you realized that it's just the volume of a hemisphere of radius 2 ...

    V = (2/3)pi*2^3 = 16pi/3

    ok?
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