Find the volume of the solid obtained by rotating the region enclosed by the graphs x=y^(3) and x= y^(1/3) about the y-axis over the interval [0,1].

Answer 1

⌠pi[(y^1/3)^2 - (y^3)^2] dy
⌡

= pi INT [y^2/3 - y^6)dy

= pi[(3/5)y^5/3 - (1/7)y^7]│ from 0 to 1

=pi (3/5 - 1/7)
= 16pi/7