To find the volume of the solid formed by revolving the given triangular region about the line $y = 4$, we can use the method of cylindrical shells.

Step 1: Identify the bounds and shape of the region

The vertices of the triangle are at $(0, 0)$, $(5, 0)$, and $(5, 4)$, which means the region is bounded by:

• The $x$-axis (from $x = 0$ to $x = 5$),
• The vertical line at $x = 5$,
• The line connecting $(0, 0)$ to $(5, 4)$, which has the equation $y = \frac{4}{5}x$.

This region is a right triangle with a base of length 5 (along the $x$-axis) and a height of 4.

Step 2: Set up the volume integral using cylindrical shells

When using the method of cylindrical shells, the formula for the volume of a solid of revolution about a vertical line (like $y = 4$) is given by:

$V = 2 \pi \int_{a}^{b} \left( \text{radius} \right) \cdot \left( \text{height} \right) \, dx$

Here:

• The radius is the distance from the line $y = 4$ to a point on the curve. Since we're rotating about $y = 4$, for a point at $y = \frac{4}{5}x$, the radius is $|4 - \frac{4}{5}x| = 4 - \frac{4}{5}x$ (since the region is in the first quadrant and $\frac{4}{5}x \leq 4$ for $x$ between 0 and 5).
• The height of a cylindrical shell is simply the $y$-coordinate of the curve, which is $\frac{4}{5}x$.

Thus, the integral becomes:

$V = 2 \pi \int_{0}^{5} \left( 4 - \frac{4}{5}x \right) \cdot \frac{4}{5}x \, dx$

Step 3: Solve the integral

Let's simplify the integrand:

$V = 2 \pi \int_{0}^{5} \left( 4 \cdot \frac{4}{5}x - \frac{4}{5}x \cdot \frac{4}{5}x \right) \, dx$

$V = 2 \pi \int_{0}^{5} \left( \frac{16}{5}x - \frac{16}{25}x^2 \right) \, dx$

Now, we compute the integral:

$V = 2 \pi \left[ \int_{0}^{5} \frac{16}{5}x \, dx - \int_{0}^{5} \frac{16}{25}x^2 \, dx \right]$

1. $\int_{0}^{5} \frac{16}{5}x \, dx = \frac{16}{5} \cdot \frac{x^2}{2} \Bigg|_{0}^{5} = \frac{16}{5} \cdot \frac{25}{2} = 40$
2. $\int_{0}^{5} \frac{16}{25}x^2 \, dx = \frac{16}{25} \cdot \frac{x^3}{3} \Bigg|_{0}^{5} = \frac{16}{25} \cdot \frac{125}{3} = \frac{200}{3}$

Now substitute these into the volume formula:

$V = 2 \pi \left( 40 - \frac{200}{3} \right)$

$V = 2 \pi \left( \frac{120}{3} - \frac{200}{3} \right)$

$V = 2 \pi \cdot \frac{-80}{3}$

$V = \frac{-160 \pi}{3}$

Thus, the volume of the solid is:

$V = \frac{160 \pi}{3} \, \text{cubic units}$

Conclusion

The volume of the solid formed by revolving the region about the line $y = 4$ is $\frac{160\pi}{3}$ cubic units.

Would you like further details or help with other questions?

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Here are 5 related questions that could expand on this topic:

1. What is the method of disk/washer for finding volumes of solids of revolution, and how does it compare to cylindrical shells?
2. How do you calculate the volume of a solid formed by revolving a region about a horizontal line (e.g., $x = 5$)?
3. What are the steps to set up and solve a volume integral when revolving about a line other than the axis of the region?
4. Can you derive the formula for the volume of a cone using cylindrical shells?
5. What happens to the volume of a solid when the axis of rotation is moved farther away from the region?

Tip: When working with solids of revolution, always carefully examine the shape and bounds of the region before choosing the appropriate method (shells vs. disks/washer). This can significantly simplify the integral.