Find the volume of the solid obtained by rotating the region bounded by
y = x2, x = 1, x = 8 and y = 0 about the line x = 1.


To find the volume of the solid obtained by rotating the region bounded by the curves around the line \( x = 1 \), we will apply the **cylindrical shell method**. This is the best approach because we are rotating around a vertical line not on the boundary of the region.

### Step-by-Step Solution

We are given the following bounds:
- The curve: \( y = x^2 \)
- Vertical lines: \( x = 1 \) and \( x = 8 \)
- The x-axis: \( y = 0 \)
- The line of rotation: \( x = 1 \)

The formula for the volume using the **cylindrical shell method** is:

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

### 1. **Determine the radius and height:**
- The **radius** of a typical cylindrical shell is the distance from the axis of rotation \( x = 1 \) to a point on the curve, which is \( r(x) = x - 1 \).
- The **height** of the shell is given by the function \( y = x^2 \), so \( h(x) = x^2 \).

### 2. **Set up the integral:**
We are rotating the region around \( x = 1 \), so the limits of integration are from \( x = 1 \) to \( x = 8 \) (the bounds of the region along the x-axis).

\[
V = 2\pi \int_{1}^{8} (x - 1) \cdot x^2 \, dx
\]

### 3. **Solve the integral:**
First, expand the integrand:

\[
(x - 1) \cdot x^2 = x^3 - x^2
\]

Now, integrate term by term:

\[
V = 2\pi \int_{1}^{8} (x^3 - x^2) \, dx
\]

\[
V = 2\pi \left[ \frac{x^4}{4} - \frac{x^3}{3} \right]_{1}^{8}
\]

### 4. **Evaluate the integral:**
Substitute the upper and lower limits:

For \( x = 8 \):
\[
\frac{8^4}{4} - \frac{8^3}{3} = \frac{4096}{4} - \frac{512}{3} = 1024 - \frac{512}{3} = 1024 - 170.67 = 853.33
\]

For \( x = 1 \):
\[
\frac{1^4}{4} - \frac{1^3}{3} = \frac{1}{4} - \frac{1}{3} = \frac{3}{12} - \frac{4}{12} = -\frac{1}{12}
\]

Now subtract the two values:

\[
853.33 - \left( -\frac{1}{12} \right) = 853.33 + \frac{1}{12} = 853.33 + 0.0833 = 853.4167
\]

### 5. **Multiply by \( 2\pi \):**
\[
V = 2\pi \times 853.4167 = 1706.83\pi
\]

Thus, the volume of the solid is approximately:

\[
V \approx 1706.83\pi \, \text{cubic units}
\]

Would you like more details on any part of the solution? 

### Related Questions:
1. How do you apply the washer method for this type of problem?
2. What would the volume be if we rotated around the y-axis instead?
3. Can we use the disk method for this problem? Why or why not?
4. How would the volume change if the region were rotated around \( x = 8 \)?
5. What if the region was defined by \( y = \sqrt{x} \) instead of \( y = x^2 \)?

**Tip**: The cylindrical shell method is particularly useful when rotating around a vertical line, especially when the region's height is easier to express in terms of \( x \).

Find the volume of the solid obtained by rotating the region bounded by y = x^2, x = 1, x = 8 and y = 0 about the line x = 1.

This problem demonstrates how to find the volume of a solid obtained by rotating the region bounded by y = x^2, x = 1, x = 8, and y = 0 about the line x = 1 using the cylindrical shell method. The step-by-step solution includes integral setup, expansion, and evaluation. Suitable for advanced high school or college-level calculus students.

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