Find the volume of the solid generated by revolving the $$ \begin{array}{l}y=\frac{1}{\sqrt{4 x+9}} \ y=0 \ x=0 \ x=6\end{array} $$

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To find the volume of the solid generated by revolving the given region about the x-axis, we can use the method of disks.

The volume of the solid generated by revolving the region about the x-axis is given by the formula:

$$ V = \pi \int_{a}^{b} [f(x)]^2 dx $$

where:
- $$ V $$ is the volume of the solid,
- $$ f(x) $$ is the function that defines the region,
- $$ a $$ and $$ b $$ are the limits of integration.

In this case, the function that defines the region is $$ y = \frac{1}{\sqrt{4x+9}} $$, and the limits of integration are from $$ x = 0 $$ to $$ x = 6 $$.

Therefore, the volume of the solid generated by revolving the given region about the x-axis is:

$$ V = \pi \int_{0}^{6} \left(\frac{1}{\sqrt{4x+9}}\right)^2 dx $$

Now, we can calculate the volume by evaluating the integral.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\pi \times \int_{0}^{6} \left(\frac{1}{\sqrt{4x+9}}\right)^{2} dx$$
- step1: Evaluate the power:
  $$\pi \times \int_{0}^{6} \left(4x+9\right)^{-1} dx$$
- step2: Evaluate the integral:
  $$\pi \times \frac{\ln{\left(\frac{11}{3}\right)}}{4}$$
- step3: Multiply:
  $$\frac{\pi \ln{\left(\frac{11}{3}\right)}}{4}$$
The volume of the solid generated by revolving the given region about the x-axis is $$ \frac{\pi \ln\left(\frac{11}{3}\right)}{4} $$ or $$ \frac{\pi \ln\left(3\frac{2}{3}\right)}{4} $$.
The volume of the solid generated by revolving the given region about the x-axis is $$ \frac{\pi \ln\left(\frac{11}{3}\right)}{4} $$ or $$ \frac{\pi \ln\left(3\frac{2}{3}\right)}{4} $$.

Find the volume of the solid generated by revolving the

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