posted by jump on .
determine the volume of the solid of revolution generated by revolving the region bounded by y=x^3-x^5, y=0, x=0 and x=1 about the line x=3
check to be sure that the graph lies all on one side of the axis for the interval. It does.
So, volume is
v = ∫[0,1] pi (R^2-r^2) dx
where R = 3, r = 3-y
= pi∫[0,1] 9 - (3-(x^3-x^5))^2 dx
= pi∫[0,1] -x^10 + 2x^8 - x^6 - 6x^5 + 6x^3 dx
= pi(-1/11 x^11 + 2/9 x^9 - 1/7 x^7 - x^6 + 3/2 x^4) [0,1]