Use the shell method to find the volume of the solid generated by revolving the regions bounded by the curves and lines about the​ x-axis.
yequalsStartRoot x EndRoot​,   yequals​0,   yequalsStartFraction x minus 11 Over 10 EndFraction

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.

Use the shell method to find the volume of the solid generated by revolving the regions bounded by the curves and lines about the x-axis. y = √x, y = 0, y = (x - 11)/10

Learn how to find the volume of a solid using the shell method for a region bounded by the curves y = √x, y = 0, and y = (x - 11)/10. This step-by-step solution involves calculus, definite integrals, and solving a quadratic equation for the limits of integration. Suitable for high school students in grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation