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Math Help - Solid of revolution

  1. #1
    Member dokrbb's Avatar
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    Solid of revolution

    I tried to understand the solution to this problem but one thing I don't get:

    the problem states: find the volume of the solid obtained by rotating the region bounded by the curves y= 1 + sec(x), y = 3, about y = 1;

    so, the inner radius is (1 + sec(x)) - 1 = sex(x); the outer radius is 3-1 = 2; and we have the area A(x) = pi(4 - sec^2(x))

    and it is integrated from -pi/3 to pi/3; how does they get these x-axis coordinates of pi/3?

    thank you
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  2. #2
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    Re: Solid of revolution

    Set the two equations equal. You get 1+\sec(x) = 3 or \sec(x) = 2. That happens when \cos(x) = \dfrac{1}{2}. Since \cos\left(\pm \dfrac{\pi}{3} \right) = \dfrac{1}{2}, those are your x-coordinates. Now, since \sec(x) is periodic, there are actually an infinite number of those regions bounded by that curve and the line, so I must assume they only want you to look in a specific range, such as [-\pi,\pi].
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