Using the shell method, the volume is
[tex]\displaystyle 2\pi \int_0^1 (2-x) \cdot 8x^3 \, dx = 16\pi \int_0^1 (2x^3 - x^4) \, dx[/tex]
Each cylindrical shell has radius [tex]2-x[/tex] (the horizontal distance from the axis of revolution to the curve [tex]y=8x^3[/tex]); has height [tex]8x^3[/tex] (the vertical distance between a point on the [tex]x[/tex]-axis in [tex]0\le x\le1[/tex] and the curve [tex]y=8x^3[/tex]).
Compute the integral.
[tex]\displaystyle 16 \pi \int_0^1 (2x^3 - x^4) \, dx = 16\pi \left(\frac{x^4}2 - \frac{x^5}5\right) \bigg|_{x=0}^{x=1} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 16\pi \left(\frac12 - \frac15\right) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac{24}5\pi = \boxed{4.8\pi}[/tex]
