Question

Find the volume of the solid obtained by revolving the loop of the curve $2a{y}^2=x{(x-a)}^2$ about x-axis. Here $a>0$.

Answer 1

We have to find the volume of the solid $2a{y}^2=x(x-a{)}^2$

The graph of the curve is shown below

Put $y=0$

$\Rightarrow x=0,x=a$

Therefore Required volume $=\pi {\int }_0^a\frac{x(x-a{)}^2}{2a}dx$

$=\frac{\pi }{2a}{\int }_0^{a}x(x^2-2ax+a^2)dx$

$=\frac{\pi }{2a}{\int }_0^a(x^3-2a{x}^2+a^{2}x)dx$

$=\frac{\pi }{2a}{[\frac{x^4}{4}-\frac{2a{x}^3}{3}+\frac{a^2{x}^2}{2}]}_0^a$

$=\frac{\pi a^4}{2a}\left[\frac{3-8+6}{12}\right]$

$=\frac{\pi a^3}{24}$

Hence the required volume is $\frac{\pi a^3}{24}cu.units$