Answer
$$
\left( f(x)=\sqrt {x-4}, \quad y =0, \quad x=13 \right)
$$
If $f(x)=\sqrt {x-4} $ is nonnegative and $R$ is the region between $f(x) $ , the x-axis and $x=13 $ , from the x-intercepts we can find lower bound $a$. If $y=0 $ then $x=4 $.
So, the volume of the solid formed by rotating $R$ about the x-axis is given by
$$
\begin{aligned} V &=\int_{a}^{b} \pi [f(x)]^{2}dx \\
&=\pi \int_{4}^{13}(\sqrt{x-4})^{2} d x \\
&=\frac{81}{2} \pi \\
&\approx 127.2 \end{aligned}
$$
Work Step by Step
$$
\left( f(x)=\sqrt {x-4}, \quad y =0, \quad x=13 \right)
$$
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)=\sqrt {x-4} $ is nonnegative and $R$ is the region between $f(x) $ , the x-axis and $x=13 $ , from the x-intercepts we can find lower bound $a$. If $y=0 $ then $x=4 $.
So, the volume of the solid formed by rotating $R$ about the x-axis is given by
$$
\begin{aligned} V &=\pi \int_{4}^{13}(\sqrt{x-4})^{2} d x \\
&=\pi \int_{4}^{13}(x-4) d x \\
&=\left.\pi\left(\frac{x^{2}}{2}-4 x\right)\right|_{4} ^{13} \\ &=\pi\left[\left(\frac{169}{2}-52\right)-(8-16) |\right.\\ &=\pi\left(\frac{65}{2}+8\right)=\\
&=\frac{81}{2} \pi \\
&\approx 127.2 \end{aligned}
$$