Find the volume of a solid rotated around the x-axis $y=x^3, y=x, x \geq 0$
$$ \begin{align}
V &= \int_{0}^{1} A(x)dx = \int_{0}^{1} \pi (x-x^3)^2dx = \pi \int_{0}^{1}(x^2-2x^4 + x^6)dx \\
&= \pi \bigg[ \frac{1}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7\bigg]_{0}^{1}
\end{align}$$
The book answer says the answer is $\frac{4\pi}{21}$ but I am not getting anywhere close to that.
I think there are three main areas where I checked that I could have messed up, but I am not seeing anything.
1) setup
2) integration
3) getting the volume  
Here is a visual of what the graph looks like and the area I am trying to integrate


The correct setup is the difference of the squares $$\int_{0}^{1} \pi \left( x^2 - (x^3)^2 \right) dx$$
and not the square of the difference $(x-x^3)^2$.
The radius of the larger disk is $x$ and its area is $\pi x^2$. The radius of the small disk is $x^3$ thus area $\pi (x^3)^2$. The ring (annulus) is the large disk minus the small disk thus $\pi\left( x^2 - (x^3)^2 \right)$.
