To find the volume of the solid obtained by rotating the region bounded by the curves $y = x^4$ and $y = 1$ about the line $y = 4$, we use the method of cylindrical shells.

Step 1: Visualize the problem

We are rotating the region between the curves around the line $y = 4$, which is horizontal. The shell method is best suited for this scenario since it involves integration with respect to the variable $x$.

The volume of a cylindrical shell is given by the formula: $V = 2\pi \int_{a}^{b} \text{(radius of the shell)} \times \text{(height of the shell)} \, dx$

Step 2: Define the variables

1. Height of the shell: This is the difference between the two curves, which is $y = 1$ and $y = x^4$. The height is: $\text{Height} = 1 - x^4$

2. Radius of the shell: The radius is the horizontal distance from the line of rotation $y = 4$ to a point on the curve. Since the line of rotation is $y = 4$, the radius of a shell at a point $y = x^4$ is $4 - x^4$.

Step 3: Set up the integral

To find the volume, we need to integrate with respect to $x$, where $x$ ranges from 0 to 1 because the curves intersect at $x = 0$ and $x = 1$.

The volume integral becomes: $V = 2\pi \int_{0}^{1} (4 - x^4) \times (1 - x^4) \, dx$

Step 4: Simplify the integrand

Now, we expand the integrand: $(4 - x^4)(1 - x^4) = 4(1 - x^4) - x^4(1 - x^4)$ $= 4 - 4x^4 - x^4 + x^8$ $= 4 - 5x^4 + x^8$

Step 5: Set up the final integral

The volume integral is now: $V = 2\pi \int_{0}^{1} (4 - 5x^4 + x^8) \, dx$

Step 6: Compute the integral

We can now integrate each term separately: $\int_0^1 4 \, dx = 4x \Big|_0^1 = 4$ $\int_0^1 5x^4 \, dx = 5 \times \frac{x^5}{5} \Big|_0^1 = 1$ $\int_0^1 x^8 \, dx = \frac{x^9}{9} \Big|_0^1 = \frac{1}{9}$

Thus, the total volume is: $V = 2\pi \left( 4 - 1 + \frac{1}{9} \right)$ $V = 2\pi \left( 3 + \frac{1}{9} \right)$ $V = 2\pi \times \frac{28}{9}$ $V = \frac{56\pi}{9}$

Final Answer:

The volume of the solid is $\frac{56\pi}{9} \, \text{cubic units}$.

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Would you like a detailed explanation of any of the steps, or do you have any further questions?

Here are some related questions for you:

1. How does the volume change if the region is rotated about a different axis, such as $y = 0$?
2. What is the difference between using the disk method and the shell method for finding volumes of revolution?
3. Can you find the volume of a solid obtained by rotating a region bounded by more than two curves?
4. How would you apply the shell method if the region were rotated about a vertical line, say $x = 2$?
5. Can you explain the method of washers in the context of finding volumes of solids of revolution?

Tip: When setting up the integral for volume problems, always carefully check the limits of integration and the expressions for the radius and height of the shells or disks.