Answer
$2.36163$
Work Step by Step
a) Given
$$y=x, y=2 x /\left(1+x^3\right) ; \quad \text { about } x=-1$$
First, we find the intersection points
\begin{aligned}
x&= \frac{2x}{1+x^3}\\
x+ x^4 &= 2x\\
x^4-x&=0\\
x(x^3-1)&=0\\
x(x-1)(x^2+x+1)&=0
\end{aligned}
Then $x=0,\ \ x= 1 $.
Since the volume of the generated solid given by
\begin{aligned}
V&=2\pi \int_a^b r(x)h(x)dx
\end{aligned}
Since $\frac{2x}{1+x^3}\geq x,\ \ 0\leq x\leq 1$ , then
$$ h(x) =\frac{2x}{1+x^3}-x,\ \ r(x)= x+1$$
Then
\begin{aligned}
V&=2\pi \int_a^b r(x)h(x)dx\\
&= 2\pi \int_{0}^{1} \left(x+1\right)\left(\frac{2x}{1+x^3}-x \right)dx
\end{aligned}
b) Using the calculator, we get
\begin{aligned}
V &= 2\pi \int_{0}^{1} \left(x+1\right)\left(\frac{2x}{1+x^3}-x \right)dx \\
&=2\pi \left(\frac{-5}{6}+ \frac{2\sqrt{3}\pi}{9}\right)\\
&\approx 2.36163
\end{aligned}