To set up the integral for the volume of the solid obtained by rotating the region bounded by $y = \sin(x)$, $y = 0$, and $0 \leq x \leq \pi$ about the line $y = -3$, we will use the method of cylindrical shells or disk/washer method. Let's use the washer method here.

Step 1: Set up the washer method formula for rotation about a horizontal line.

When rotating about a horizontal line like $y = -3$, the volume is given by the integral: $V = \pi \int_{a}^{b} \left[ R_{\text{outer}}(x)^2 - R_{\text{inner}}(x)^2 \right] \, dx$ where:

• $R_{\text{outer}}(x)$ is the distance from the axis of rotation to the upper curve,
• $R_{\text{inner}}(x)$ is the distance from the axis of rotation to the lower curve.

Step 2: Determine $R_{\text{outer}}(x)$ and $R_{\text{inner}}(x)$.

• The upper boundary curve is $y = \sin(x)$, and the lower boundary curve is $y = 0$.
• The axis of rotation is $y = -3$, which is below both curves.

For any $x$ in $[0, \pi]$:

• The distance from the axis $y = -3$ to the upper boundary $y = \sin(x)$ is $R_{\text{outer}}(x) = \sin(x) - (-3) = \sin(x) + 3$.
• The distance from the axis $y = -3$ to the lower boundary $y = 0$ is $R_{\text{inner}}(x) = 0 - (-3) = 3$.

Step 3: Set up the volume integral.

The volume is thus given by: $V = \pi \int_0^\pi \left[ (\sin(x) + 3)^2 - 3^2 \right] \, dx$ Simplifying the terms inside the integral: $V = \pi \int_0^\pi \left[ (\sin(x) + 3)^2 - 9 \right] \, dx$

Final setup:

The integral for the volume is: $V = \pi \int_0^\pi \left[ \sin^2(x) + 6\sin(x) + 9 - 9 \right] \, dx$ which simplifies to: $V = \pi \int_0^\pi \left[ \sin^2(x) + 6\sin(x) \right] \, dx$ This is the setup for the volume of the solid obtained by rotating the given region about the line $y = -3$.

Would you like to evaluate the integral or need further clarification on any part?

---

Here are 5 related questions to expand this concept:

1. How would the setup change if the region was rotated about the x-axis instead of $y = -3$?
2. What happens if the region is rotated about $y = 3$ instead of $y = -3$?
3. Can we apply the shell method to this problem? How would that integral look?
4. How would you handle the integral if the curve was $y = \cos(x)$ instead of $y = \sin(x)$?
5. What if the region was bounded by $y = \sin(x)$ and $y = \cos(x)$?

Tip: When rotating around a line different from the axis, always adjust the radius to reflect the new distance from the curve to the line.