To find the volume $V$ of the solid obtained by rotating the region bounded by the curves $y = x - 1$, $y = 0$, and $x = 6$ about the x-axis, we can use the Disk Method.

Steps:

1. Set up the function to rotate:

   The function that represents the curve is: $y = x - 1$ This will be rotated around the x-axis. The region is bounded by $y = 0$ (the x-axis), $x = 6$, and starts at $x = 1$, because $y = 0$ when $x = 1$.

2. Volume formula using the Disk Method:

   The volume of the solid formed by rotating about the x-axis is given by: $V = \pi \int_{a}^{b} \left[ f(x) \right]^2 \, dx$ where $f(x)$ is the function representing the curve, and $[a, b]$ are the limits of integration.

3. Determine the limits of integration:

   The region starts at $x = 1$ (where $y = 0$) and ends at $x = 6$.

4. Set up the integral:

   The function is $f(x) = x - 1$, so we square the function and integrate: $V = \pi \int_{1}^{6} (x - 1)^2 \, dx$

5. Solve the integral:

   First, expand $(x - 1)^2$: $(x - 1)^2 = x^2 - 2x + 1$

   Now, integrate each term: $V = \pi \int_{1}^{6} (x^2 - 2x + 1) \, dx$

   The integrals of the terms are: $\int x^2 \, dx = \frac{x^3}{3}, \quad \int -2x \, dx = -x^2, \quad \int 1 \, dx = x$

   So, we have: $V = \pi \left[ \frac{x^3}{3} - x^2 + x \right]_{1}^{6}$

6. Evaluate the definite integral:

   Substitute $x = 6$ into the expression: $\frac{6^3}{3} - 6^2 + 6 = \frac{216}{3} - 36 + 6 = 72 - 36 + 6 = 42$

   Substitute $x = 1$ into the expression: $\frac{1^3}{3} - 1^2 + 1 = \frac{1}{3} - 1 + 1 = \frac{1}{3}$

   Now, subtract the two values: $42 - \frac{1}{3} = \frac{126}{3} - \frac{1}{3} = \frac{125}{3}$

7. Final volume:

   Multiply by $\pi$ to get the volume: $V = \pi \cdot \frac{125}{3} = \frac{125\pi}{3}$

Thus, the volume of the solid is: $V = \frac{125\pi}{3} \, \text{cubic units}.$

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Would you like any further details or explanations?

Here are some related questions:

1. How does the Disk Method differ from the Shell Method?
2. Can this problem be solved using the Shell Method?
3. What is the volume if the region is rotated about a different axis?
4. What happens if the curve changes to $y = x^2 - 1$?
5. How do you visualize the solid formed by rotation?

Tip: When setting up integrals for volume, always carefully identify the bounds of integration and the function being squared or rotated.