# Thread: Volume of solids of revolutions

1. ## 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

2. 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,

$\displaystyle y=2x^3$

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

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

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

Both methods you mentioned are equivalent.

3. 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

4. For part 2,

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

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

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

evaluated from y=0 to y=54

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

from 0 to 54

$\displaystyle =\frac{3{\pi}}{10}\left(\frac{1}{4}\right)^{\frac{ 1}{3}}54^{\frac{3}{5}}-0$

5. thankyou this has been very helpful