Answer
$$
y= \frac{1}{5}x^{5}, \quad \quad 0 \leq x \leq 5
$$
The area of the surface obtained by rotating the curve about the x-axis is
$$
\begin{aligned}
S&=\int 2 \pi y d s \\
&=\int_{0}^{5} \frac{2}{5}\pi x^{5}\sqrt{1+x^{8} } dx\\
& \approx 1,230,507 \quad\quad\left[\text {by using Simpson’s Rule } \right] \\
&\approx 1,227,192. \quad\quad\left[\text {by using a calculator } \right] \\
\end{aligned}
$$
Work Step by Step
$$
y= \frac{1}{5}x^{5}, \quad \quad 0 \leq x \leq 5
$$
$\Rightarrow$
$$
y^{\prime} =\frac{dy}{dx}= x^{4}
$$
$\Rightarrow$
$$
\begin{aligned} ds &= \sqrt {1+\left(y^{\prime}\right)^{2}} dx \\
&=\sqrt{1+\left(x^{4} \right)^{2}} dx\\
&=\sqrt{1+\left(x^{8} \right)} dx\\
\end{aligned}
$$
So, the area of the surface obtained by rotating the curve about the x-axis
$$
\begin{aligned}
S&=\int 2 \pi y d s \\
&=\int_{0}^{5} 2 \pi ( \frac{1}{5}x^{5})(\sqrt{1+\left(x^{8} \right)} )dx\\
&=\int_{0}^{5} \frac{2}{5}\pi x^{5}\sqrt{1+x^{8} } dx\\
\end{aligned}
$$
Simpson’s Rule to approximate the given integral with the
specified value of $ n=10$ , with $ n =10 , a = 0$, and $b = 5$ we have
$$
\Delta x=\frac{b-a}{n}=\frac{5-0}{10}=\frac{1}{2}
$$
Let
$$
f(x)=\frac{2}{5}\pi x^{5}\sqrt{1+x^{8} }
$$
then
$$
\begin{aligned} S &= \int_{0}^{5} \frac{2}{5}\pi x^{5}\sqrt{1+x^{8} } dx\\
& = \frac{\Delta x}{3}\left[f\left(x_{0}\right)\right. +4 f\left(x_{1}\right)+2 f\left(x_{2}\right)+4 f\left(x_{3}\right)+\cdots \\
& \quad \left.+2 f\left(x_{n-2}\right)+4 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right] \\
&\approx S_{10}\\
& \approx \frac{1 / 2}{3}[f(0)+4 f(0.5)+2 f(1)+\\
& \quad\quad+4 f(1.5)+2 f(2)+4 f(2.5)+2 f(3)+\\
& \quad\quad +4 f(3.5)+2 f(4)+4 f(4.5)+f(5)] \\
&\approx 1,230,507
\end{aligned}
$$
But the surface area by using the a calculator is approximately $ 1,227,192.$
