The given problem asks to set up an integral for the volume of a solid obtained by rotating the region bounded by the curves $y = \sin(x)$, $y = 0$, over the interval $0 \leq x \leq \pi$, around the line $y = -5$.

To solve this problem, we need to use the method of cylindrical shells (or alternatively, the washer method if specified). The volume of the solid can be found using the washer method formula for rotation about a horizontal line $y = c$:

$V = \pi \int_{a}^{b} \left[ R(x)^2 - r(x)^2 \right] dx$

Where:

• $R(x)$ is the distance from the axis of rotation to the outer curve,
• $r(x)$ is the distance from the axis of rotation to the inner curve,
• $a$ and $b$ are the limits of integration.

Outer and Inner Radii

In this case, since we are rotating around $y = -5$:

• The outer radius $R(x)$ is the distance from the line $y = -5$ to $y = \sin(x)$, which is $\sin(x) - (-5) = \sin(x) + 5$.
• The inner radius $r(x)$ is the distance from the line $y = -5$ to $y = 0$, which is $0 - (-5) = 5$.

Setting Up the Integral

Thus, the integral for the volume is:

$V = \pi \int_{0}^{\pi} \left[ (\sin(x) + 5)^2 - (5)^2 \right] dx$

This expression represents the integral setup. You are not required to evaluate it for this problem.

Would you like further clarification or details on the reasoning? Here are some related questions to expand on this topic:

1. What is the washer method and when is it typically used?
2. How does rotating around a different line, such as $y = -5$, affect the setup of the volume integral?
3. Can the same problem be solved using the method of cylindrical shells?
4. How would the integral change if the region were rotated about a vertical line?
5. What are the geometric implications of the integral in terms of volume?

Tip:

Always identify whether the problem involves rotation around a horizontal or vertical axis, as it dictates which method (washer or cylindrical shell) is more convenient.