Volume of the Solid formed by revolving the region bounded b

[Lime]

Find the volume of the solid formed by revolving the region bounded by y = x^3, y = 8, and x = 0 about the line x = -1.

I'm sorry but I don't know what the anti-derivative symbol is in internet lingo. I'll use 8~0 with 8 being the upper integer and 0 being the lower.

int[8~0] {pi [(y^1/3 + 1)^2 - 1)]} dy

. . .= pi int[8~0] [(y^2/3 + 2y^1/3)] dy

I understand all that. But I don't understand the next step:

. . .= pi [3/5y^5/3 + 3/2y^4/3]8~0

Help please.

 

[galactus]

A cylindrical shell may be the way to go here.

\(\displaystyle \L\\2{\pi}\int_{0}^{2}(8-x^{3})(x+1)dx\)

Here's your solid revolved about x=-1:

 

[Lime]

I still don't get how you get from

= pi int[8~0] [(y^2/3 + 2y^1/3)] dy

to

= pi [3/5y^5/3 + 3/2y^4/3]8~0