Find the volume of the solid obtained by rotating the region bounded by the graphs y = 1/x, y = 0, x = 1, ﻿and x = 7 about y=9

To find the volume of the solid obtained by rotating the region bounded by the curves \( y = \frac{1}{x} \), \( y = 0 \), \( x = 1 \), and \( x = 7 \) about the line \( y = 9 \), we can use the method of cylindrical shells.

### Step 1: Set up the cylindrical shells formula
The volume \( V \) using the method of cylindrical shells is given by:

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

### Step 2: Determine the radius and height
- **Radius:** The distance from the curve \( y = \frac{1}{x} \) to the axis of rotation \( y = 9 \). Since the axis of rotation is above the curve, the radius is:

\[
\text{Radius} = 9 - \frac{1}{x}
\]

- **Height:** The height of each cylindrical shell is the value of the function \( y = \frac{1}{x} \).

\[
\text{Height} = \frac{1}{x}
\]

### Step 3: Set up the integral
The limits of integration are from \( x = 1 \) to \( x = 7 \). Therefore, the volume \( V \) is:

\[
V = 2\pi \int_{1}^{7} \left(9 - \frac{1}{x}\right) \cdot \frac{1}{x} \, dx
\]

### Step 4: Simplify the integrand
Distribute \( \frac{1}{x} \) across the terms inside the integral:

\[
V = 2\pi \int_{1}^{7} \left(\frac{9}{x} - \frac{1}{x^2}\right) \, dx
\]

### Step 5: Integrate
Now, integrate each term separately:

\[
\int \frac{9}{x} \, dx = 9\ln|x|
\]

\[
\int \frac{1}{x^2} \, dx = -\frac{1}{x}
\]

So the integral becomes:

\[
V = 2\pi \left[9\ln|x| + \frac{1}{x} \right]_{1}^{7}
\]

### Step 6: Evaluate the definite integral
Substitute the limits \( x = 7 \) and \( x = 1 \):

At \( x = 7 \):

\[
9\ln(7) + \frac{1}{7}
\]

At \( x = 1 \):

\[
9\ln(1) + \frac{1}{1} = 0 + 1 = 1
\]

So the volume \( V \) is:

\[
V = 2\pi \left[\left(9\ln(7) + \frac{1}{7}\right) - 1\right]
\]

Simplify:

\[
V = 2\pi \left(9\ln(7) + \frac{1}{7} - 1\right)
\]

\[
V = 2\pi \left(9\ln(7) - \frac{6}{7}\right)
\]

Thus, the volume of the solid is:

\[
\boxed{V = 2\pi \left(9\ln(7) - \frac{6}{7}\right)}
\]

Would you like more details or have any questions?

### Relative Questions:
1. How would the volume change if the region were rotated about the line \( y = -9 \)?
2. What would be the effect on the volume if the upper bound for \( x \) were increased to 10?
3. How does the method of cylindrical shells compare to the method of washers for this problem?
4. What would be the volume if the region were rotated around the y-axis instead of the line \( y = 9 \)?
5. Can the integral approach used here be applied if the function was \( y = \frac{1}{x^2} \) instead? How?

### Tip:
When rotating a region about a horizontal line, always carefully assess whether the line is above or below the region, as this affects the radius calculation for cylindrical shells.

Learn how to find the volume of a solid obtained by rotating the region bounded by y = 1/x, y = 0, x = 1, and x = 7 about the line y = 9 using the method of cylindrical shells. Step-by-step solution with integral calculation and evaluation. Suitable for college-level students studying calculus and volume of revolution.

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