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Math Help - Integration Application question

  1. #1
    Newbie br1tt204's Avatar
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    Integration Application question

    This may be hard to visualize without a graph, so I apologize in advance, but here is my question:

    There is a circle with radius 'b' that is 'a' units away from the y-axis. if the circle is rotated about the y-axis, how do you find the volume of the doughnut that is formed? (assume that b<a)

    I would even appreciate a jump start! Anything, really! Thank you in advanced!
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  2. #2
    Super Member
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    Quote Originally Posted by br1tt204 View Post
    This may be hard to visualize without a graph, so I apologize in advance, but here is my question:

    There is a circle with radius 'b' that is 'a' units away from the y-axis. if the circle is rotated about the y-axis, how do you find the volume of the doughnut that is formed? (assume that b<a)

    I would even appreciate a jump start! Anything, really! Thank you in advanced!

    For convinent calculation , I locate the center of the circle at

     (a,0) and thus the equation of the circle is

     (x-a)^2 + y^2 = b^2

    Now , let's go to the integration ,

    We know the volume of the revolution about the y-axis is

     \pi \int_p^q [ f(y)^2 - g(y)^2 ] ~dy , | f(y) |\geq |g(y) | , y \in [p,q]

    since  f(y) =  the distance from the y -axis (x)

    we should find x in terms of y , that is

     (x-a)^2 + y^2 = b^2  \implies x = a + \sqrt{b^2- y^2}

    and also  x = a - \sqrt{b^2- y^2}

    But we know  a + \sqrt{b^2- y^2}  \geq  a - \sqrt{b^2- y^2}

    For the interval , i choose  (-b,b)

    Now , we are able to see what the integral looks like !

    It should be

     \pi \int_{-b}^b \left ( [ a + \sqrt{b^2- y^2} ]^2 - [ a - \sqrt{b^2- y^2}]^2 \right ) ~dy

     = \pi \int_{-b}^b 4a\sqrt{b^2-y^2}~dy

     = 4a\pi \int_{-b}^b \sqrt{b^2 - y^2} ~dy

     = 4a \pi \times ( ~area~of ~a~semicircle ~with~radius~b ~)

     = 4a \pi \cdot (\frac{\pi b^2 }{2}  ) = 2ab^2\pi^2
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