To find the volume of the solid obtained by rotating the region bounded by the curves $y = x^3$, $y = 1$, and $x = 2$ about the line $y = -5$, we can use the method of cylindrical shells. Here's how we'll proceed:

Step 1: Identify the boundaries

• The curve is $y = x^3$, which is the lower boundary.
• The horizontal line $y = 1$ is the upper boundary.
• The vertical line $x = 2$ serves as the right boundary.

Step 2: Set up the formula for cylindrical shells

The formula for the volume using cylindrical shells is:

$V = 2\pi \int_{a}^{b} ( \text{radius of shell} ) ( \text{height of shell} ) \, dx$

For this problem:

• The height of each shell is the distance between the curve $y = x^3$ and $y = 1$, which is $1 - x^3$.
• The radius of each shell is the distance from the curve to the axis of rotation $y = -5$. This is $(1 - (-5)) = (1 + 5) = 6$ for the upper boundary, and $(x^3 - (-5)) = x^3 + 5$ for the lower boundary.

Step 3: Set up the integral

We now integrate from $x = 0$ to $x = 2$ (since the region is bounded by $x = 2$):

$V = 2\pi \int_0^2 (6)(1 - x^3) \, dx$

Step 4: Solve the integral

First, compute the integral of $6(1 - x^3)$:

$V = 2\pi \int_0^2 (6 - 6x^3) \, dx$ $= 2\pi \left[ 6x - \frac{6x^4}{4} \right]_0^2$ $= 2\pi \left[ 6(2) - \frac{6(2)^4}{4} - (0) \right]$ $= 2\pi \left[ 12 - \frac{6 \times 16}{4} \right]$ $= 2\pi \left[ 12 - 24 \right]$ $= 2\pi (-12)$ $= -24\pi$

However, volume cannot be negative, so we take the absolute value:

$V = 24\pi$

Thus, the volume $V$ of the solid is $24\pi$.

Would you like to go over any of the steps in more detail or have any specific questions?

Here are 5 related questions to further your understanding:

1. How would the volume change if the region was rotated around $y = 0$ instead of $y = -5$?
2. Can the method of disks be used for this problem, and if so, how?
3. How does the choice of axis of rotation affect the complexity of the integral?
4. What is the difference between using shells and disks in volume of revolution problems?
5. How would the problem change if the boundary $x = 2$ were extended to a different value?

Tip: For solids of revolution, always choose the method (disks or shells) that simplifies the radius and height expressions in the integral.