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Math Help - Integration of a volume

  1. #1
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    Integration of a volume

    Hi, here are the parameters:

    26-3-3
    Find volume of the equation
    y = 2 - x
    rotated about X axis using disk method, boundaries are y = 0 and x = 0
    Formula: V = pi (int) y^2 dx

    Attempt:
    since y = 2 - (2) = 0, 2 is an x-coordinate boundary. So the upper and lower limits of this integral are 2 and 0. X axis intercepts. Correct?

    from there:
    = pi (int) (2-x)^2
    (2-x)^2 = 4 - 3x

    = pi (int) 4 - 3x dx
    = pi/3 (int) 4 - x dx

    Integration:

    = pi/3 4x - 1/2(x^2)

    Plugging in 2 = 6.3776

    However, the answer is 8.776. Any idea where I went wrong? thanks.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Integration of a volume

    One problem is where you expanded the integrand... (2-x)^2\ne4-3x.
    Thanks from topsquark
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  3. #3
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    Re: Integration of a volume

    Do you really need Calculus to do this? y= 2- x is a straight line. When x= 0, y= 2. When y= 0, x= 2. Rotating around the x-axis gives a cone with height 2 and radius 2. The volume of a cone of height h and radius r is given by V= (1/3)\pi r^2 h.
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  4. #4
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    Re: Integration of a volume

    No don't need calculus, however it is a calculus question and they will get harder.

    It was my understanding that everything to the right of the integral symbol you simplify algebraically and then integrate but obviously not
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