# Thread: Solids of Revolution using Integration

1. ## 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

2. 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

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

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

3. 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