If you're using the shell method, the setup would be
[tex]\displaystyle2\pi\int_1^2x(x^3-1)\,\mathrm dx[/tex]
Reasoning: consider just one shell (see the image below). The height of the shell is determined by the vertical distance between the curve [tex]y=x^3[/tex] and the line [tex]y=1[/tex]. The radius of each shell is determined by the value of [tex]x[/tex], which can only range from 1 to 2 within the region of interest, and the height is determined exactly by [tex]x^3-1[/tex] at this [tex]x[/tex]. Multiply by [tex]2\pi[/tex] to get the surface area of this infinitesimally thin cylindrical shell, then integrate along the interval [1, 2] to get the volume.
We have
[tex]\displaystyle2\pi\int_1^2x(x^3-1)\,\mathrm dx=2\pi\int_1^2(x^4-x)\,\mathrm dx=2\pi\left(\dfrac{x^5}5-\dfrac{x^2}2\right)\bigg|_1^2=\dfrac{47\pi}5[/tex]
