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Math Help - Quick Surface of Revolution Question

  1. #1
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    Quick Surface of Revolution Question

    Hi I have a quick question about surface of revolution. I have the problem:

    Find the area of the surface obtained by rotating the curve about the x-axis.
    x = 1 + 2y^2
    2<=y<=3

    I know the formula for S = 2\pi y ds

    When I set up the integral can I use the bounds in the same way? I.e. from 2 to 3?
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  2. #2
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    Please define "quick" in this context. How long have you been working on it?

    To answer your question directly, maybe, if that's the right thing to do. A formula does you no good if it doesn't mean anything.

    Please define 'ds'. This definition, done properly, should lead you on your way.
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  3. #3
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    Sorry, I was just reffering to the bounds. Im used to stuff like 2<=x<=3. Do I treat the bounds the same way (i.e. from 2 to 3) when setting up the integral? In this case ds = \sqrt[]{1+(\frac{dy}{dx})^2}
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  4. #4
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    Ok if someone would be so kind as to check my work and show me how to simplify the following:

    Rotate x = 1 + 2y^2 about the x-axis from 2 <= y <= 3

    y = \sqrt{\frac{x-1}{2}}

    \frac{dy}{dx} = \frac{\sqrt{2}}{4\sqrt{x-1}}

    ds = \sqrt{1+ (\frac{dy}{dx})^2} =  \sqrt{1+ (\frac{1}{8(x-1)})}= \frac{\sqrt{8x-7}}{4}

    S = 2 \pi y ds = \int_2^3 \! 2\pi (\sqrt{\frac{x-1}{2}})(\frac{\sqrt{8x-7}}{4}) \ dx.

    ?
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