Answer
$3.68$
Work Step by Step
Given
$$y= \sqrt{1+x^3} ,\ \ \ 0\leq x\leq 1 \ \ \text{about } \ y-\text{axis} $$
Since the volume given by
\begin{aligned}
V&= 2\pi \int_a^b r(x)h(x) dx
\end{aligned}
Here
$$r(x)= x,\ \ \ h(x)= \sqrt{1+x^3}$$
Then
\begin{aligned}
V&= 2\pi \int_a^b r(x)h(x) dx\\
&= 2\pi\int_0^1x\sqrt{1+x^3}dx\\
\end{aligned}
Here
\begin{aligned}
\Delta x&= \frac{b-a}{n}\\
&= \frac{1}{5}
\end{aligned}
Divide the interval $[0,1]$ into $n=5$ subintervals of the length $\Delta x=\frac{1}{5}$ with the following endpoints:
$$
a=0, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, 1=b
$$
It follows that
\begin{aligned}
\int_a^b f(x) d x \approx & \Delta x\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+\cdots+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_n}{2}\right)\right)\\
\int_0^1 f(x) d x & \approx \frac{1-0}{5}[f(0.1)+f(0.3)+f(0.5)+f(0.7)+f(0.9)] \\
& \approx 0.2(2.9290)
\end{aligned}
Hence
$$V=(2\pi)0.2(2.9290) \approx 3.68$$
