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Math Help - Area of a Surface of Revolution

  1. #1
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    Area of a Surface of Revolution

    x = 1 + 2y^2, 1 <= y <= 2

    so I figure out dy/dx = 4y

    2PI integral[ (1+2y^2)*sqrt(1+16y^2)]

    At this point, I've tried so many different integration techniques but I can't find the antiderivative.
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  2. #2
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    Quote Originally Posted by s3n4te View Post
    x = 1 + 2y^2, 1 <= y <= 2

    so I figure out dy/dx = 4y

    2PI integral[ (1+2y^2)*sqrt(1+16y^2)]

    At this point, I've tried so many different integration techniques but I can't find the antiderivative.
    I assume the rotation is around the y-axis ....?

    Try the substitution y = \frac{1}{4} \sinh u.
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  3. #3
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    sorry, it's around the x-axis, i thought it was y until i re-read the question. LOL, it's really simple when it's around the x-axis doh.
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