The problem asks to set up (but not evaluate) an integral for the volume of the solid obtained by rotating the region bounded by the curves $y = x^3$, $y = 0$, $x = 1$, and rotating about the line $x = 2$.

Step-by-step solution:

1. Identify the method to use:

   • Since we are rotating about a vertical line $x = 2$, we use the shell method.

   The shell method formula is: $V = 2\pi \int_{a}^{b} ( \text{radius} )(\text{height}) \, dx$

2. Radius of the shell:

   • The radius is the distance from the curve $x = 2$ to the current $x$-coordinate.
   • This distance is $2 - x$.

3. Height of the shell:

   • The height is the value of the function $y = x^3$, which is just $x^3$.

4. Bounds of integration:

   • The region is bounded by $x = 0$ and $x = 1$, so these will be the limits of integration.

5. Setting up the integral: $V = 2\pi \int_{0}^{1} (2 - x)(x^3) \, dx$

This is the required integral to compute the volume of the solid obtained by rotating the region around the line $x = 2$.

Would you like further details or have any questions?

Related Questions:

1. What is the shell method, and when is it used instead of the disk/washer method?
2. How would the setup change if we were rotating about the $y$-axis instead of $x = 2$?
3. How would we compute the volume if we used the disk/washer method (if applicable)?
4. What would be the integral setup if the function were $y = x^2$ instead of $y = x^3$?
5. How does the choice of the axis of rotation affect the integral setup?

Tip:

Always double-check the axis of rotation and whether the shell or disk method is more appropriate based on the problem's geometry.