Solids of Revolution using Integration

**creatively12** · October 12th 2009, 03:58 PM

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

**DeMath** · October 12th 2009, 04:35 PM

Originally Posted by creatively12

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

$V = \pi \int\limits_0^1 {{{\left( {\frac{6} {{5 - 2x}}} \right)}^2}dx} = 36\pi \int\limits_0^1 {{{\left( {5 - 2x} \right)}^{ - 2}}dx} = - 18\pi \int\limits_0^1 {{{\left( {5 - 2x} \right)}^{ - 2}}d\left( {5 - 2x} \right)} =$

$= \left. {\frac{{18\pi }} {{5 - 2x}}} \right|_0^1 = 18\pi \left( {\frac{1} {3} - \frac{1} {5}} \right) = 18\pi \cdot \frac{2} {{15}} = \frac{{36\pi }} {{15}} = \frac{{12\pi }} {5}.$

**creatively12** · October 12th 2009, 04:55 PM

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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