We can use the cylindrical shell method, or the slicing (area of cross-section) method. In this case, shells are easier.
$1$. Take a thin vertical strip, of width "$dx$" with base running from $x$ to $x+dx$. This strip is at distance $2-x$, roughly, from the line $x=2$. When you rotate the strip, you get a "shell" of thickness $dx$, height $x^3$, and perimeter $2\pi(2-x)$. "Add up," $x=0$ to $x=1$. We get volume $\int_{x=0}^1 2\pi(2-x)x^3\,dx.$
$2$. Or use slicing, parallel to the $x$-axis, so we will be integrating with respect to $y$. Take a slice of thickness $dy$, at height $y$. The volume of the slice is $dy$ times the area of cross-section. Note the hole. The area of cross-section is $\pi(2-x)^2-\pi$. In terms of $y$ it is $\pi(2-y^{1/3})^2-\pi$. So the volume is $\int_{y=0}^1 \pi\left((2-y^{1/3})^2-1\right)\,dy.$ Or else you can forget about the hole for a while, integrate $\pi(2-y^{1/3})^2$, and remove the volume of the hole later. The hole is a cylinder, with volume easy to find without calculus.