To find the volume of the solid generated by rotating the region bounded by the curve $y = \sqrt{x}$, the line $y = 1$, and $x = 4$, about the line $y = 1$, we will use the method of cylindrical shells or washer method.

Step-by-step approach using the Washer Method:

We rotate the region about the line $y = 1$. First, let's describe the boundaries of the region:

1. Curve: $y = \sqrt{x}$, so $x = y^2$.
2. Line: $y = 1$.
3. Vertical boundary: $x = 4$, which gives $y = \sqrt{4} = 2$.

Thus, the region is bounded between $y = 0$ and $y = 2$.

Now, when we rotate the region about the line $y = 1$, we will calculate the volume using the washer method. The formula for the volume is:

$V = \pi \int_{a}^{b} \left( R_{\text{outer}}^2 - R_{\text{inner}}^2 \right) \, dy$

Where:

• $R_{\text{outer}}$ is the distance from the line of rotation to the curve $y = 1$,
• $R_{\text{inner}}$ is the distance from the line $y = 1$ to the curve $y = \sqrt{x}$.

For this problem:

• The outer radius is from $y = 1$ to $y = 0$, which is $1$.
• The inner radius is $1 - y$, because the distance between $y = 1$ and the curve $y = \sqrt{x}$.

The limits of integration are from $y = 0$ to $y = 2$. Therefore, the volume is:

$V = \pi \int_0^2 \left( 1^2 - (1 - y)^2 \right) \, dy$

Simplify the integrand:

$1^2 = 1$ $(1 - y)^2 = 1 - 2y + y^2$ Thus, the integrand becomes: $1 - (1 - 2y + y^2) = 2y - y^2$

Set up the integral:

$V = \pi \int_0^2 (2y - y^2) \, dy$

Solve the integral:

$V = \pi \left[ y^2 - \frac{y^3}{3} \right]_0^2$ Evaluating this at the limits: $V = \pi \left[ \left( 2^2 - \frac{2^3}{3} \right) - \left( 0^2 - \frac{0^3}{3} \right) \right]$ $V = \pi \left( 4 - \frac{8}{3} \right)$ $V = \pi \left( \frac{12}{3} - \frac{8}{3} \right)$ $V = \pi \times \frac{4}{3}$

Thus, the volume of the solid is:

$V = \frac{4\pi}{3}$

Do you need further clarification or a breakdown of any step?

Here are 5 related questions:

1. How does the Washer method differ from the Shell method for finding volume?
2. How would the volume change if the curve were rotated around the x-axis instead of $y = 1$?
3. What happens if the region is bounded by a different upper line, say $y = 2$?
4. How would we handle the integral if the curve were more complex, such as $y = \sin(x)$?
5. How does the method change if the axis of rotation is $y = -1$?

Tip: Always identify the radii (inner and outer) carefully when using the washer method; this ensures you subtract the correct areas.