find the volume of the solid obtained by rotating the region bounded by the given curves about the line $$ {\displaylines x=8\\ \text{ }x=y^6\text{ }x=1} $$

We are tasked with finding the volume of the solid formed by rotating the region bounded by the curves $ x = y^6 $, $ x = 1 $, and $ x = 8 $ about the line $ x = 8 $.

### Step 1: Set up the problem
The curves in question are:
- $ x = y^6 $, which describes a curve symmetric about the $ y $-axis.
- The vertical lines $ x = 1 $ and $ x = 8 $, which bound the region horizontally.

### Step 2: Define the method of rotation
Since we are rotating around a vertical line ($ x = 8 $), we will use the **shell method** to find the volume. In the shell method, we use cylindrical shells with height and radius as functions of $ y $, integrating with respect to $ y $.

### Step 3: Set up the shell method formula
The volume of a solid generated by rotating around a vertical line is given by the shell method formula:
\[
V = 2\pi \int_{a}^{b} \left( \text{radius} \right) \cdot \left( \text{height} \right) \, dy
\]

- **Radius**: The distance from the axis of rotation $ x = 8 $ to a point on the curve $ x = y^6 $ is $ 8 - y^6 $.
- **Height**: The height is the difference in $ x $-values between $ x = 1 $ and $ x = y^6 $, which gives $ 1 - y^6 $.

### Step 4: Determine the limits of integration
The curve $ x = y^6 $ intersects $ x = 1 $ when $ y^6 = 1 $, or $ y = 1 $. Thus, the region extends from $ y = 0 $ to $ y = 1 $.

### Step 5: Set up the integral
Now, we can set up the volume integral using the shell method:

\[
V = 2\pi \int_0^1 (8 - y^6)(1 - y^6) \, dy
\]

### Step 6: Expand and simplify the integrand
We first expand $ (8 - y^6)(1 - y^6) $:

\[
(8 - y^6)(1 - y^6) = 8(1 - y^6) - y^6(1 - y^6) = 8 - 8y^6 - y^6 + y^{12}
\]
\[
= 8 - 9y^6 + y^{12}
\]

Thus, the integral becomes:
\[
V = 2\pi \int_0^1 \left( 8 - 9y^6 + y^{12} \right) \, dy
\]

### Step 7: Compute the integral
Now, we compute the integral term by term:
\[
\int_0^1 8 \, dy = 8y \Big|_0^1 = 8
\]
\[
\int_0^1 9y^6 \, dy = 9 \cdot \frac{y^7}{7} \Big|_0^1 = \frac{9}{7}
\]
\[
\int_0^1 y^{12} \, dy = \frac{y^{13}}{13} \Big|_0^1 = \frac{1}{13}
\]

Thus, the volume is:
\[
V = 2\pi \left( 8 - \frac{9}{7} + \frac{1}{13} \right)
\]

### Step 8: Simplify the result
To simplify $ 8 - \frac{9}{7} + \frac{1}{13} $, we get a common denominator:
\[
8 = \frac{728}{91}, \quad \frac{9}{7} = \frac{117}{91}, \quad \frac{1}{13} = \frac{7}{91}
\]
\[
8 - \frac{9}{7} + \frac{1}{13} = \frac{728}{91} - \frac{117}{91} + \frac{7}{91} = \frac{728 - 117 + 7}{91} = \frac{618}{91}
\]

Thus, the volume is:
\[
V = 2\pi \cdot \frac{618}{91} = \frac{1236\pi}{91}
\]

### Final Answer:
The volume of the solid is $ \boxed{\frac{1236\pi}{91}} $.

---

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

Here are some related questions to consider:
1. How would the method change if the axis of rotation was the $ y $-axis?
2. How would you compute the volume if the region was rotated around the line $ x = 1 $ instead of $ x = 8 $?
3. Can you apply the disk method for this problem?
4. What happens if the curve $ x = y^6 $ was replaced by a different function, say $ x = y^4 $?
5. How would the limits of integration change if the region extended beyond $ y = 1 $?

**Tip:** When using the shell method, the radius is always the distance between the curve and the axis of rotation. Double-check your setup to avoid common mistakes!

Find the volume of the solid obtained by rotating the region bounded by the given curves about the line x=8, x=y^6, x=1.

Learn how to compute the volume of a solid formed by rotating the region bounded by the curves x = y^6, x = 1, and x = 8 around the line x = 8. This step-by-step solution uses the shell method, integration, and binomial expansion to solve the problem. Suitable for students in Grades 11-12 and college-level calculus.

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