Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = sin(x),   y = 0,   0 ≤ x ≤ 𝜋;
about y = −3

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?

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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.

Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. y = sin(x), y = 0, 0 ≤ x ≤ 𝜋; about y = −3

Learn how to set up the integral to find the volume of a solid obtained by rotating the region bounded by y = sin(x) and y = 0, from 0 to π, about the line y = -3. This solution uses the washer method and covers key concepts in integral calculus, including volume of solids of revolution. Suitable for advanced high school or undergraduate students.

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