# Volume of solids of revolutions

• April 13th 2010, 12:50 PM
jimss2
Volume of solids of revolutions
I am trying to solve a question in my integration topic using volumes of solids of revolutions the attatchment shows my workings and the part I would like some guidance with
• April 13th 2010, 01:06 PM
Quote:

Originally Posted by jimss2
I am trying to solve a question in my integration topic using volumes of solids of revolutions the attatchment shows my workings and the part I would like some guidance with

Hi jimss2,

$y=2x^3$

$\frac{y}{2}=x^3$

$x=\left(\frac{y
}{2}\right)^{\frac{1}{3}}$

as x and y are positive in the region of interest.

Both methods you mentioned are equivalent.
• April 13th 2010, 04:52 PM
jimss2
Thankyou for your reply it was very helpful I am still struggling with integrating x=(y/2)^1/3 as I do not know what to do when bringing the power down as the 2 is already there each method I try give me the wrong answer
• April 14th 2010, 01:53 AM
For part 2,

$0.5\int_{y=0}^{54}{\pi}x^2dy=\frac{{\pi}}{2}\int_{ 0}^{54}\left(\frac{y}{2}\right)^{\frac{2}{3}}dy$

integrating this, we bring out the denominator under y first, then raise the power of y by 1
and divide by the new power

$\frac{{\pi}}{2}\int_{0}^{54}\left(\frac{1}{2}\righ t)^{\frac{2}{3}}y^{\frac{2}{3}}dy=\frac{{\pi}}{2}\ left(\frac{1}{4}\right)^{\frac{1}{3}}y^{\frac{2}{3 }+\frac{3}{3}}\frac{1}{\frac{2}{3}+\frac{3}{3}}$

evaluated from y=0 to y=54

$\frac{{\pi}}{2}\left(\frac{1}{4}\right)^{\frac{1}{ 3}}y^{\frac{5}{3}}\left(\frac{3}{5}\right)$

from 0 to 54

$=\frac{3{\pi}}{10}\left(\frac{1}{4}\right)^{\frac{ 1}{3}}54^{\frac{3}{5}}-0$
• April 14th 2010, 02:53 AM
jimss2
thankyou this has been very helpful