From the problem and the image, we are tasked to:

1. Set up an integral for the volume $V$ of the solid formed by rotating the region bounded by $y = \sqrt{x - 1}$, $y = 0$, and $x = 4$ about the $x$-axis.
2. Find the volume of the solid.

Step 1: Volume Formula for Solids of Revolution

Using the disk method (rotation about the $x$-axis), the volume of the solid is given by: $V = \pi \int_a^b \left[ f(x) \right]^2 dx$ Here:

• $f(x) = \sqrt{x - 1}$,
• Bounds are $x = 1$ (where $y = \sqrt{x - 1} = 0$) to $x = 4$.

Step 2: Write the Integral

The volume becomes: $V = \pi \int_1^4 \left( \sqrt{x - 1} \right)^2 dx$

Simplify $(\sqrt{x - 1})^2$: $V = \pi \int_1^4 (x - 1) dx$

Step 3: Solve the Integral

The integral of $(x - 1)$ is: $\int (x - 1) dx = \frac{x^2}{2} - x + C$

Evaluate the definite integral: $\int_1^4 (x - 1) dx = \left[ \frac{x^2}{2} - x \right]_1^4$

At $x = 4$: $\frac{4^2}{2} - 4 = \frac{16}{2} - 4 = 8 - 4 = 4$

At $x = 1$: $\frac{1^2}{2} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}$

Subtract: $\left[ \frac{x^2}{2} - x \right]_1^4 = 4 - \left(-\frac{1}{2}\right) = 4 + \frac{1}{2} = \frac{9}{2}$

Step 4: Multiply by $\pi$

The volume is: $V = \pi \cdot \frac{9}{2} = \frac{9\pi}{2} \approx 14.14 \text{ (units}^3\text{)}.$

Final Answer:

• The correct volume is approximately 14.14 units³.

Would you like me to provide further details, such as a graph or step breakdown?

Related Questions:

1. How does the disk method differ from the washer method?
2. What changes if the region is rotated about a line above the $x$-axis, like $y = 1$?
3. How would this setup look if the bounds changed (e.g., $x = 3$)?
4. What if we rotated the same region around the $y$-axis instead?
5. How do we handle functions with discontinuities in their bounds?

Tip:

Always verify the bounds and the function's square before applying the disk or washer method to avoid errors in volume calculations!