Answer
$$
\left( f(x)=\frac{x^{2}}{4}, \quad y =0, \quad x=4 \right)
$$
If $f(x)=\frac{x^{2}}{4} $ is nonnegative and $R$ is the region between $f(x) $ , the x-axis and $x=4 $ , from the x-intercepts we can find lower bound $a$. If $y=0 $ then $x=0 $.
So, the volume of the solid formed by rotating $R$ about the x-axis is given by
$$
\begin{aligned} V &=\int_{a}^{b} \pi [f(x)]^{2}dx\\
&=\pi \int_{0}^{4}\left(\frac{x^{2}}{4}\right)^{2} d x\\
&=\frac{64 \pi}{5} \\
&\approx 40.21 \end{aligned}
$$
Work Step by Step
$$
\left( f(x)=\frac{x^{2}}{4}, \quad y =0, \quad x=4 \right)
$$
If $f(x) $ is nonnegative and $R$ is the region between $f(x) $ , the x-axis and from $x=a$ to $x=b$, the volume of the solid formed by rotating $R$ about the x-axis is given by
$$
\int_{a}^{b} \pi [f(x)]^{2}dx
$$
If $f(x)=\frac{x^{2}}{4} $ is nonnegative and $R$ is the region between $f(x) $ , the x-axis and $x=4 $ , from the x-intercepts we can find lower bound $a$. If $y=0 $ then $x=0 $.
So, the volume of the solid formed by rotating $R$ about the x-axis is given by
$$
\begin{aligned} V &=\int_{a}^{b} \pi [f(x)]^{2}dx\\
&=\pi \int_{0}^{4}\left(\frac{x^{2}}{4}\right)^{2} d x\\
&=\pi \int_{0}^{4} \frac{x^{4}}{16} \\
&=\left.\frac{\pi}{16}\left(\frac{x^{5}}{5}\right)\right|_{0} ^{4}\\
&=\frac{\pi}{16}\left(\frac{1024}{5}\right) \\
&=\frac{64 \pi}{5} \\
&\approx 40.21 \end{aligned}
$$
