Use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving the plane region about the x-axis.
y = 
1
x
The x y-coordinate plane is given. A curve, a region, and two representative rectangles are graphed.
The curve starts at (1, 1), goes down and right becoming less steep, and ends at (4, 1⁄4).
The region is below the curve, above the x-axis, to the right of x = 1, and to the left of x = 4.
The region is divided into two parts by a dashed horizontal line at y = 1⁄4.
The first horizontal representative rectangle goes from (1, 0.67) to (1.5, 0.67).
The second horizontal representative rectangle goes from (1, 0.13) to (4, 0.13).

To find the volume of the solid generated by revolving the region under the curve \( y = \frac{1}{x} \) about the **x-axis** using the **shell method**, we follow these steps:

### 1. Set up the Shell Method Formula:
The general formula for the volume of a solid of revolution using the shell method, when revolving around the **x-axis**, is:

\[
V = 2\pi \int_{y_1}^{y_2} (\text{radius}) (\text{height}) \, dy
\]

### 2. Identify Variables:
In this case, we revolve the region about the x-axis, and the height of the shells will be given in terms of \(y\), while the radius will simply be \( y \) because we're revolving around the x-axis. From the curve \( y = \frac{1}{x} \), we solve for \(x\) in terms of \(y\):

\[
y = \frac{1}{x} \implies x = \frac{1}{y}
\]

Thus, the radius is \(y\), and the height of the shell is the distance between \(x = 1\) and \(x = \frac{1}{y}\) at each \(y\), which is given by \( \frac{1}{y} - 1 \).

### 3. Determine the Limits of Integration:
From the given information, the region lies between \(x = 1\) and \(x = 4\), which corresponds to the values of \(y = \frac{1}{x}\), or \(y = 1\) to \(y = \frac{1}{4}\). Therefore, the limits of integration for \(y\) are:

\[
y_1 = \frac{1}{4}, \quad y_2 = 1
\]

### 4. Write the Integral:
Now, we substitute everything into the formula for the volume:

\[
V = 2\pi \int_{\frac{1}{4}}^{1} y \left( \frac{1}{y} - 1 \right) \, dy
\]

### 5. Simplify the Integral:
Distribute the \(y\) term inside the parentheses:

\[
V = 2\pi \int_{\frac{1}{4}}^{1} (1 - y) \, dy
\]

### 6. Evaluate the Integral:
First, integrate \(1 - y\):

\[
\int (1 - y) \, dy = y - \frac{y^2}{2}
\]

Now, evaluate this from \(y = \frac{1}{4}\) to \(y = 1\):

\[
V = 2\pi \left[ \left( 1 - \frac{1^2}{2} \right) - \left( \frac{1}{4} - \frac{\left(\frac{1}{4}\right)^2}{2} \right) \right]
\]

Simplifying the expressions:

\[
V = 2\pi \left[ \left( 1 - \frac{1}{2} \right) - \left( \frac{1}{4} - \frac{1}{32} \right) \right]
\]
\[
V = 2\pi \left[ \frac{1}{2} - \left( \frac{8}{32} - \frac{1}{32} \right) \right]
\]
\[
V = 2\pi \left[ \frac{1}{2} - \frac{7}{32} \right]
\]
\[
V = 2\pi \left[ \frac{16}{32} - \frac{7}{32} \right]
\]
\[
V = 2\pi \cdot \frac{9}{32}
\]

### 7. Final Answer:
\[
V = \frac{9\pi}{16}
\]

Thus, the volume of the solid generated by revolving the region under the curve \( y = \frac{1}{x} \) from \(x = 1\) to \(x = 4\) about the x-axis is \( \frac{9\pi}{16} \).

---

Would you like further details or have any questions?

Here are 5 related questions to consider:
1. How would the solution change if we revolved around the y-axis instead?
2. Can you derive the volume using the disk/washer method instead of the shell method?
3. What if the region was bounded between \(x = 2\) and \(x = 5\)?
4. How do you set up the shell method for different types of curves?
5. What is the geometric interpretation of the shell height and radius in this example?

**Tip:** The shell method is often easier for regions where the function is best described in terms of \(y\).

Use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving the plane region about the x-axis. y = 1/x. The x y-coordinate plane is given. A curve, a region, and two representative rectangles are graphed. The curve starts at (1, 1), goes down and right becoming less steep, and ends at (4, 1⁄4). The region is below the curve, above the x-axis, to the right of x = 1, and to the left of x = 4. The region is divided into two parts by a dashed horizontal line at y = 1⁄4. The first horizontal representative rectangle goes from (1, 0.67) to (1.5, 0.67). The second horizontal representative rectangle goes from (1, 0.13) to (4, 0.13).

Explore the step-by-step solution for finding the volume of a solid generated by revolving the region under y = 1/x around the x-axis using the shell method. This problem covers key calculus concepts, including integration and the shell method, suitable for 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