Answer
$$
\left( f(x)=\frac{1}{\sqrt {x-1}}, \quad y =0, \quad x=2, \quad x=4 \right)
$$
The volume of the solid formed by revolving the function, $f(x)=\frac{1}{\sqrt {x-1}} $ about the x-axis from $x=2$ to $x=4$, is
$$
\begin{aligned} V &=\int_{a}^{b} \pi [f(x)]^{2}dx \\
&=\pi \int_{2}^{4}\left(\frac{1}{\sqrt{x-1}}\right)^{2} d x \\
&=\pi \ln 3 \\
&\approx 3.451 \end{aligned}
$$
Work Step by Step
$$
\left( f(x)=\frac{1}{\sqrt {x-1}}, \quad y =0, \quad x=2, \quad x=4 \right)
$$
If $f(x) $ is nonnegative and $R$ is the region between $f(x) $ and the x-axis from $x=a$ to $x=b$, the volume of the solid formed by rotating $R$ about the x-axis is given by
$$
\int_{a}^{b} \pi [f(x)]^{2}dx
$$
So,the volume of the solid formed by revolving the function, $f(x)=\frac{1}{\sqrt {x-1}} $ about the x-axis from $x=2$ to $x=4$, is
$$
\begin{aligned} V &=\int_{a}^{b} \pi [f(x)]^{2}dx \\
&=\pi \int_{2}^{4}\left(\frac{1}{\sqrt{x-1}}\right)^{2} d x \\
&=\pi \int_{2}^{4} \frac{d x}{x-1} \\
&=\left.\pi(\ln |x-1|)\right|_{2} ^{4} \\
&=\pi \ln 3-(0) \\
&=\pi \ln 3 \\
&\approx 3.451 \end{aligned}
$$
