To solve this problem, we need to find the volume of the solid generated by revolving the triangle with vertices at $(4,0), (4,4), (5,4)$ about the y-axis.

Steps:

1. Find the area of the triangular region: The triangle has vertices at $(4,0), (4,4), (5,4)$. To calculate the area, we can use the formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$:

   $\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

   Plugging in the points $(4,0), (4,4), (5,4)$, we get:

   = \frac{1}{2} \left| 0 + 16 - 20 \right| = \frac{1}{2} \times 4 = 2$$ So, the area of the triangle is 2 square units.

2. Set up the integral for the volume using the method of cylindrical shells. When revolving around the y-axis, the volume of the solid can be found by:

   $V = 2\pi \int_{x_1}^{x_2} x \cdot f(x) \, dx$ Here, $f(x)$ represents the height of the triangle at each point along the x-axis. The equation of the line connecting $(4,4)$ to $(5,4)$ is a horizontal line, and from $(4,0)$ to $(4,4)$ is vertical.

   • From $x = 4$ to $x = 5$, the height remains 4.

   Therefore, the volume is:

   = 2\pi \left( 2(25) - 2(16) \right) = 2\pi (50 - 32) = 2\pi \times 18 = 36\pi$$

Final Answer:

The volume of the solid is $\boxed{36\pi}$ cubic units.

Would you like any more details or clarifications on this problem? Here are some related questions:

1. What are the steps involved in the method of cylindrical shells?
2. How would the formula change if we revolved the region around the x-axis instead of the y-axis?
3. Could we use the disk method for this problem, and if so, how?
4. How can we derive the equation of the lines forming the sides of the triangle?
5. What happens to the volume if we move one of the triangle's vertices?

Tip: When solving solids of revolution, the method of cylindrical shells is especially useful when the axis of revolution is parallel to the function.