Answer
$$
\left( f(x)=3x-1, \quad y =0, \quad x=2 \right)
$$
If $f(x)=3x-1 $ is nonnegative and $R$ is the region between $f(x) $ , the x-axis and $x=2 $ , from the x-intercepts we can find lower bound $a$. If $y=0 $ then $x=\frac{1}{3} $.
So, the volume of the solid formed by rotating $R$ about the x-axis is given by
$$
\begin{aligned} V &=\int_{a}^{b} \pi [f(x)]^{2}dx \\
&=\pi \int_{1 / 3}^{2}(3 x-1)^{2} d x \\
&=\pi \int_{1 / 3}^{2}\left(9 x^{2}-6 x+1\right) d x \\ &=\left.\pi\left(3 x^{3}-3 x^{2}+x\right)\right|_{1 / 3} ^{2} \\ &=\pi\left[(24-12+2)-\left(\frac{1}{9}-\frac{1}{3}+\frac{1}{3}\right)\right] \\
&=\pi\left(14-\frac{1}{9}\right)\\
&=\frac{125 \pi}{9} \\
&\approx 43.63 \end{aligned}
$$
Work Step by Step
$$
\left( f(x)=3x-1, \quad y =0, \quad x=2 \right)
$$
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)=3x-1 $ is nonnegative and $R$ is the region between $f(x) $ , the x-axis and $x=2 $ , from the x-intercepts we can find lower bound $a$. If $y=0 $ then $x=\frac{1}{3} $.
So, the volume of the solid formed by rotating $R$ about the x-axis is given by
$$
\begin{aligned} V &=\pi \int_{1 / 3}^{2}(3 x-1)^{2} d x \\
&=\pi \int_{1 / 3}^{2}\left(9 x^{2}-6 x+1\right) d x \\ &=\left.\pi\left(3 x^{3}-3 x^{2}+x\right)\right|_{1 / 3} ^{2} \\ &=\pi\left[(24-12+2)-\left(\frac{1}{9}-\frac{1}{3}+\frac{1}{3}\right)\right] \\
&=\pi\left(14-\frac{1}{9}\right)\\
&=\frac{125 \pi}{9} \\
&\approx 43.63 \end{aligned}
$$
