Q. The region enclosed by the given curves is rotated about the specified line. Find the volume of the resulting solid.

y=x^3 and y=sqrt.of(x); about x = 1.

What I did: Rotating about a vertical line along the y-axis.

x^3=sqrt.of(x); x= 0 and 1

x=y^(1/3) and x=y^2.

A(y)=pi(outer radius)^2-pi(inner radius)^2

=pi(1+y^(1/3))^2-pi(1+y^2)^2

V= pi*Integral from 0 to 1 of [(1+(y^1/3))^2-(1+y^2)^2]dy

V= pi*Integral from 0 to 1 of [(1+y^(2/3)+2y^(1/3)-(1+y^4+2y^2)]dy

V= pi[(3y^(2/3))/2 + (6y^(4/3))/4 - (y^4)/4 - (2y^3)/3] from 0 to 1

But after plugging in for 0 and 1, I got 25pi/12. But the ans. is 17pi/30...8(