To find the volume of the solid formed by revolving the region bounded by $y = 14 - x$, $y = 14$, and $y = x$ about the $y$-axis, we can use the shell method or disk/washer method. Here, the shell method is the most convenient because the boundaries are expressed in terms of $x$ and $y$, and the axis of rotation is vertical.

Step 1: Understand the Boundaries

• $y = 14 - x$: Line with negative slope, intercepting the $y$-axis at $y = 14$.
• $y = x$: Line with positive slope passing through the origin.
• $y = 14$: A horizontal line.

The region of interest lies in the first quadrant, where $x$ and $y$ are non-negative.

Step 2: Set up the Shell Method Formula

For the shell method, the volume is calculated as: $V = 2\pi \int_{x_{\text{min}}}^{x_{\text{max}}} \text{(radius)} \cdot \text{(height)} \, dx$

• Radius: The distance from the axis of rotation ($y$-axis) is $x$.
• Height: The difference between the upper and lower bounds of $y$ for a given $x$, which is $(14 - x) - x = 14 - 2x$.

Step 3: Find the Limits of Integration

The region spans from $x = 0$ (where $y = 14 - x$ intersects $y = 14$) to $x = 7$ (where $y = 14 - x$ intersects $y = x$).

Step 4: Write the Integral

The volume is: $V = 2\pi \int_{0}^{7} x (14 - 2x) \, dx$

Step 5: Simplify and Solve

Expand the integrand: $x (14 - 2x) = 14x - 2x^2$

The integral becomes: $V = 2\pi \int_{0}^{7} (14x - 2x^2) \, dx$

Evaluate term by term: $\int 14x \, dx = 7x^2, \quad \int 2x^2 \, dx = \frac{2x^3}{3}$

Substitute the limits: $V = 2\pi \left[ \left( 7x^2 - \frac{2x^3}{3} \right) \Big|_{0}^{7} \right]$

At $x = 7$: $7x^2 = 7(7^2) = 343, \quad \frac{2x^3}{3} = \frac{2(7^3)}{3} = \frac{686}{3}$

At $x = 0$: $7x^2 = 0, \quad \frac{2x^3}{3} = 0$

Subtract: $V = 2\pi \left[ 343 - \frac{686}{3} \right]$

Simplify: $343 - \frac{686}{3} = \frac{1029}{3} - \frac{686}{3} = \frac{343}{3}$

$V = 2\pi \cdot \frac{343}{3} = \frac{686\pi}{3}$

Final Answer:

$\boxed{\frac{686\pi}{3}}$

Let me know if you'd like further clarification or details!

Related Questions:

1. How does the shell method differ from the washer method in calculating volumes?
2. What happens if the axis of rotation changes to $x = 0$?
3. How would the result change if $y = 14 - x$ is replaced with a quadratic function?
4. What is the significance of the limits of integration in this problem?
5. Can this volume be computed using numerical methods instead of integration?

Tip:

For regions involving rotation around the $y$-axis, the shell method is typically simpler when boundaries are given in terms of $x$.