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Math Help - Solids of Revolution using Integration

  1. #1
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    Volume of Solids of Revolution using Integration

    The region between the curve y=(6/5-2x), the x-axis and the lines x=0 and x=1 is rotated through 360 degrees about the x-axis. Show that the volume obtained is (12/5)pi

    Please emphasize on the integration of (5-2x)^-2 in some detail or i'll be grateful if your explain that particular procedure. Thanks

    Please Help
    Last edited by creatively12; October 12th 2009 at 04:35 PM.
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by creatively12 View Post
    The region between the curve y=(6/5-2x), the x-axis and the lines x=0 and x=1 is rotated through 360 degrees about the x-axis. Show that the volume obtained is (12/5)pi

    Please emphasize on the integration of (5-2x)^-2 in some detail or i'll be grateful if your explain that particular procedure. Thanks

    Please Help
    V = \pi \int\limits_0^1 {{{\left( {\frac{6}<br />
{{5 - 2x}}} \right)}^2}dx}  = 36\pi \int\limits_0^1 {{{\left( {5 - 2x} \right)}^{ - 2}}dx}  =  - 18\pi \int\limits_0^1 {{{\left( {5 - 2x} \right)}^{ - 2}}d\left( {5 - 2x} \right)}  =

    = \left. {\frac{{18\pi }}<br />
{{5 - 2x}}} \right|_0^1 = 18\pi \left( {\frac{1}<br />
{3} - \frac{1}<br />
{5}} \right) = 18\pi  \cdot \frac{2}<br />
{{15}} = \frac{{36\pi }}<br />
{{15}} = \frac{{12\pi }}<br />
{5}.
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  3. #3
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    thnx for the help but i couldn't understand the process after you canceled 36 to get 18, please if you could explain from there onwards, i'd be really ver grateful
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