How do you find the volume bounded by #y=(x + 1)^.5# and the line x=3 revolved about the x-axis?

To find the volume bounded by (y = \sqrt{x + 1}) and the line (x = 3) revolved about the x-axis, you can use the method of cylindrical shells. Here are the steps:

1. Determine the limits of integration. In this case, the limits are from (x = 0) to (x = 2) because the intersection point of (y = \sqrt{x + 1}) and (x = 3) is (x = 2).

2. Set up the integral for the volume using the formula for cylindrical shells: [V = \int_{a}^{b} 2\pi x \cdot h(x) , dx] where (h(x)) is the height of the shell at (x), and in this case, (h(x) = \sqrt{x + 1}).

3. Substitute (h(x) = \sqrt{x + 1}) into the formula: [V = \int_{0}^{2} 2\pi x \cdot \sqrt{x + 1} , dx]

4. Evaluate the integral to find the volume.

After integrating, you'll find the volume of the solid of revolution bounded by (y = \sqrt{x + 1}) and the line (x = 3) revolved about the x-axis.