To find the volume of the solid generated by revolving the region bounded by the curves about the $x$-axis using the shell method, we follow these steps:

Step 1: Identify the Functions and Region

We are given the curves:

1. $y = \sqrt{x}$
2. $y = 0$ (the $x$-axis)
3. $y = \frac{x - 11}{10}$

We need to determine the limits of integration by solving for the points of intersection of the given curves.

Find Intersection Points

Solve for $x$ where $\sqrt{x} = \frac{x - 11}{10}$.

$\sqrt{x} = \frac{x - 11}{10}$

Square both sides:

$x = \left( \frac{x - 11}{10} \right)^2$

Multiply both sides by 100 to clear the fraction:

$100x = (x - 11)^2$

Expanding:

$100x = x^2 - 22x + 121$

Rearrange:

$x^2 - 122x + 121 = 0$

Solve this quadratic equation using the quadratic formula:

$x = \frac{122 \pm \sqrt{(-122)^2 - 4(1)(121)}}{2(1)}$

$x = \frac{122 \pm \sqrt{14884 - 484}}{2}$

$x = \frac{122 \pm \sqrt{14400}}{2}$

$x = \frac{122 \pm 120}{2}$

$x = \frac{122 + 120}{2} = \frac{242}{2} = 121$

$x = \frac{122 - 120}{2} = \frac{2}{2} = 1$

Thus, the region is bounded between $x = 1$ and $x = 121$.

Step 2: Shell Method Formula

The volume using the shell method is:

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

• Radius: The distance from the shell to the $x$-axis is just $y$.

• Height: The horizontal distance between the curves is:

  $x_{\text{right}} - x_{\text{left}} = 10y + 11 - y^2$

Rewriting in terms of $y$, we integrate from $y = 0$ to $y = \sqrt{121} = 11$:

$V = \int_0^{11} 2\pi y (10y + 11 - y^2) \, dy$

Expanding:

$V = 2\pi \int_0^{11} (10y^2 + 11y - y^3) \, dy$

Step 3: Compute the Integral

Compute each term separately:

$\int 10y^2 \, dy = \frac{10y^3}{3}$

$\int 11y \, dy = \frac{11y^2}{2}$

$\int y^3 \, dy = \frac{y^4}{4}$

Evaluate from 0 to 11:

$V = 2\pi \left[ \frac{10(11)^3}{3} + \frac{11(11)^2}{2} - \frac{(11)^4}{4} \right]$

$V = 2\pi \left[ \frac{10(1331)}{3} + \frac{11(121)}{2} - \frac{14641}{4} \right]$

$V = 2\pi \left[ \frac{13310}{3} + \frac{1331}{2} - \frac{14641}{4} \right]$

Now compute the numerical result:

$V \approx 2\pi \times 1036.33$

$V \approx 6510.27\pi$

Thus, the final volume is approximately:

$V \approx 20464.8$

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How does the shell method compare to the disk method for this problem?
2. What if the region were revolved around the y-axis instead?
3. Can we confirm this result using cylindrical shells by integrating with respect to $x$?
4. How do we determine whether the shell method or washer method is more efficient?
5. What happens if we change the limits of integration to $x = 0$ and $x = 121$?

Tip:

The shell method is particularly useful when integrating in terms of $y$ simplifies the expressions compared to integrating in terms of $x$. Always analyze the region before choosing the best method.