How to set up an integral to find the volume of this solid of revolution?

Question

I am asked to find the area between the curves $y=x^3$,$y=x$ that rotate around the axis $x=2$. The integral that I set up was $$V=\int_{1}^{2}\pi\left(y-y^{1/3}\right)^2dy\space + \int_{2}^{8}\pi\left(2-y^{1/3})^2\right)dy\approx3.23$$

This is incorrect because the answer should be $2\pi$ what did I do wrong and how do I find this volume ?

Answer 1

It is easier if you use cylindrical shell method. $$ V=\int_{-1}^{0}2\pi(2-x)(x^3-x)dx + \int _0^1 2\pi (2-x)(x-x^3)dx = 2\pi $$

Answer 2

The intersections of the functions are $(-1,-1),(0,0),(1,1)$. Setup your problem like so $$\int_{-1}^{0}2\pi(2-x)(-x+x^3)dx \quad+\quad \int_{0}^{1}2\pi(2-x)(-x+x^3)dx \\ =\frac{19\pi}{15}+\frac{11\pi}{15} \longrightarrow\frac{30\pi}{15}\longrightarrow 2\pi$$