To find the volume of the solid obtained by rotating the region bounded by the curves $x = y^4$ and $x = 1$ about the line $x = 2$, we can use the method of cylindrical shells.

Step 1: Setup the integral

The method of cylindrical shells involves integrating the lateral surface area of cylindrical shells created by revolving vertical slices of the region around the line $x = 2$.

• Radius of each shell: The distance from the line $x = 2$ to the curve $x = y^4$ is $2 - y^4$.
• Height of each shell: The height is the difference in the $x$-values of the two curves, which is $x_{\text{right}} - x_{\text{left}} = 1 - y^4$.
• Thickness of each shell: The thickness is $dy$, as we are integrating with respect to $y$.

Step 2: Determine the bounds

We need to find the values of $y$ that correspond to the intersections of the curves. The curves intersect when:

$y^4 = 1 \quad \Rightarrow \quad y = \pm 1.$

So, the region is bounded by $y = -1$ and $y = 1$.

Step 3: Volume integral

The volume $V$ is given by:

$V = 2\pi \int_{y=-1}^{y=1} \text{(radius)} \times \text{(height)} \, dy.$

Substituting the expressions for the radius and height:

$V = 2\pi \int_{-1}^{1} (2 - y^4)(1 - y^4) \, dy.$

Step 4: Simplify the integrand

Now, expand the integrand:

= 2 - 2y^4 - y^4 + y^8 = 2 - 3y^4 + y^8.$$ Thus, the volume integral becomes: $$V = 2\pi \int_{-1}^{1} (2 - 3y^4 + y^8) \, dy.$$ ### Step 5: Compute the integral We can now compute the integral term by term: $$\int_{-1}^{1} 2 \, dy = 2[y]_{-1}^{1} = 2(1 - (-1)) = 4,$$ $$\int_{-1}^{1} -3y^4 \, dy = -3 \left[\frac{y^5}{5}\right]_{-1}^{1} = -3\left(\frac{1^5}{5} - \frac{(-1)^5}{5}\right) = 0 \quad (\text{since the powers of } y \text{ are odd}),$$ $$\int_{-1}^{1} y^8 \, dy = \left[\frac{y^9}{9}\right]_{-1}^{1} = \frac{1^9}{9} - \frac{(-1)^9}{9} = \frac{1}{9} - \left(-\frac{1}{9}\right) = \frac{2}{9}.$$ ### Step 6: Combine results Now substitute the results back into the volume expression: $$V = 2\pi \left(4 + 0 + \frac{2}{9}\right) = 2\pi \left(4 + \frac{2}{9}\right) = 2\pi \left(\frac{36}{9} + \frac{2}{9}\right) = 2\pi \times \frac{38}{9} = \frac{76\pi}{9}.$$ ### Final Answer: The volume of the solid is $$\boxed{\frac{76\pi}{9}}$$. --- Would you like further clarification on any part of this solution? Here are some related questions: 1. How does the method of cylindrical shells differ from the disk method for volume calculations? 2. How can you compute the volume of a solid of revolution using the disk method? 3. What happens if the axis of rotation is not vertical or horizontal? 4. How do you set up integrals for more complex solids with multiple bounds? 5. Can you apply the cylindrical shells method to solids revolving around non-vertical lines? **Tip:** When performing integrals with symmetry (like in this case), always check if the function or the limits allow simplification, such as using even or odd properties of the integrand!