Given :
A be the bounded region enclosed by the graphs of [tex]f(x)=x[/tex] and [tex]g(x)=x^3[/tex] .
To Find :
The volume of the solid obtained by rotating the region A about the line x+2=0 .
Solution :
Point of intersection of f(x) and g(x) is 0 and 1 .
Volume of solid is given by :
[tex]V=2\pi\int\limits^a_b {(x+p)(f(x)-g(x))} \, dx[/tex]
Here , a and b is point of intersection and a = 1 and b = 0 .
Putting all given values in above equation , we get :
[tex]V=2\pi\int\limits^1_0 {(x+2)(x-x^3)} \, dx\\\\V=2\pi\int\limits^1_0(x^2-x^4+2x-2x^3)dx\\\\V=2\pi(\dfrac{x^3}{3}-\dfrac{x^5}{5}+2(\dfrac{x^2}{2})-2(\dfrac{x^4}{4}))_0^1\\\\V=2\pi[(\dfrac{1^3}{3}-\dfrac{1^5}{5}+2(\dfrac{1^2}{2})-2(\dfrac{1^4}{4}))-(\dfrac{0^3}{3}-\dfrac{0^5}{5}+2(\dfrac{0^2}{2})-2(\dfrac{0^4}{4}))]\\\\V=2\times \pi \times({\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{2})\\V=3.96\ units^2[/tex]
Therefore , the volume of the solid obtained is [tex]3.96 \ units^2[/tex] .
Hence , this is the required solution .
