Find the volume of the solid generated by revolving the region bounded by the lines and curves $y=\frac{3}{\sqrt{x}}, y=0, x=8$, and $x=20$ about the $x$-axis
The volume is
(Type an exact answer, using $\pi$ as needed.)

Step 1 ：We are given the function \(y = \frac{3}{\sqrt{x}}\), and we are asked to find the volume of the solid generated by revolving the region bounded by the lines and curves \(y=\frac{3}{\sqrt{x}}, y=0, x=8\), and \(x=20\) about the x-axis.

Step 2 ：We can use the disk method to find the volume of the solid of revolution. The formula for the disk method is \(V = \pi \int_{a}^{b} [f(x)]^2 dx\), where \(f(x)\) is the function describing the curve, and \(a\) and \(b\) are the limits of integration.

Step 3 ：In this case, \(f(x) = \frac{3}{\sqrt{x}}\), \(a = 8\), and \(b = 20\).

Step 4 ：Substituting these values into the formula, we get \(V = \pi \int_{8}^{20} \left(\frac{3}{\sqrt{x}}\right)^2 dx\).

Step 5 ：Evaluating this integral, we get \(V = \pi(-9\log(8) + 9\log(20))\).

Step 6 ：We can simplify this expression further by using properties of logarithms. The simplified expression for the volume is \(V = -\log\left(\left(\frac{2}{5}\right)^{9\pi}\right)\).

Step 7 ：Final Answer: The volume of the solid generated by revolving the region bounded by the lines and curves \(y=\frac{3}{\sqrt{x}}, y=0, x=8\), and \(x=20\) about the x-axis is \(\boxed{-\log\left(\left(\frac{2}{5}\right)^{9\pi}\right)}\).