Find the volume generated by revolving about the x-axis the region bounded by the following curves

y=sqrt/(4x+3),x=0,x=4, and y=0.

(Use "pi" for π).

To find the volume generated by revolving a region about the x-axis, we can use the method of cylindrical shells.

First, let's start by sketching the graph of the given curves and the region bounded between them.

The region is bounded by the curves \(y = \sqrt{4x + 3}\), the x-axis, and the vertical lines \(x = 0\) and \(x = 4\). Here's a rough sketch of the region:

```
| ⌠ x=4
| ⌡ _______________
y | ___/
^ | __/
| | __/
| | /
| | _/
|___|__________________________________ x
0 1 2 3 4
```

Now, we need to find the volume of this region when it is rotated about the x-axis.

The volume of a cylindrical shell is given by the formula:

V = 2π * ∫[a,b] (f(x) * h(x)) dx

Where:
- a and b are the x-coordinates of the points where the curves intersect.
- f(x) is the radius of the shell, which is the distance from the x-axis to the curve.
- h(x) is the height of the shell, which is the difference between the upper and lower y-coordinates of the region.

To find the volume, we need to integrate the product of the radius and height of each cylindrical shell from x = 0 to x = 4.

Let's start by finding the points of intersection between the curves:

\(y = \sqrt{4x + 3}\) and \(y = 0\):
0 = √(4x + 3)
4x + 3 = 0
4x = -3
x = -3/4

\(y = sqrt{4x + 3}\) and \(x = 4\):
y = √(4(4) + 3)
y = √(16 + 3)
y = √19

So, the points of intersection are (-3/4, 0) and (4, √19).

Now, let's set up the integral:

V = 2π * ∫[-3/4, 4] (f(x) * h(x)) dx

The radius f(x) is the distance from the x-axis to the curve, which is given by \(f(x) = \sqrt{4x + 3}\).

The height h(x) is the difference between the upper and lower y-coordinates of the region, which is given by \(h(x) = \sqrt{19} - 0 = \sqrt{19}\).

Substituting f(x) and h(x) into the integral:

V = 2π * ∫[-3/4, 4] (sqrt{4x + 3} * sqrt{19}) dx

Now, we can compute this integral to find the volume.