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−1(x),   y = pi/2
,   x = 0;
about y = 2. in terms of dy

To set up the integral for the volume of the solid obtained by rotating the region bounded by the curves \( y = \sin^{-1}(x) \), \( y = \frac{\pi}{2} \), and \( x = 0 \) about the line \( y = 2 \) in terms of \( dy \), we can proceed as follows:

### 1. **Identify the boundaries and functions:**

- The curve \( y = \sin^{-1}(x) \) can be rewritten as \( x = \sin(y) \) (since \( y \) is the dependent variable and \( x \) is the independent variable).
- The other boundary is \( y = \frac{\pi}{2} \), and the vertical line \( x = 0 \) (which corresponds to \( y \) on the \( y \)-axis).

### 2. **Set up the integral using the washer method:**

The volume \( V \) of the solid formed by rotating the region around the line \( y = 2 \) is calculated using the washer method. The formula for the volume is:

\[
V = \pi \int_{a}^{b} \left[ R_{\text{outer}}^2(y) - R_{\text{inner}}^2(y) \right] dy
\]

Where:
- \( R_{\text{outer}}(y) \) is the distance from the axis of rotation (\( y = 2 \)) to the outer curve.
- \( R_{\text{inner}}(y) \) is the distance from the axis of rotation (\( y = 2 \)) to the inner curve.

### 3. **Determine \( R_{\text{outer}}(y) \) and \( R_{\text{inner}}(y) \):**

- The outer radius \( R_{\text{outer}}(y) \) is the distance from \( y = 2 \) to \( x = 0 \), which is \( 2 - y \).
- The inner radius \( R_{\text{inner}}(y) \) is the distance from \( y = 2 \) to the curve \( x = \sin(y) \), which is \( 2 - y \).

Since \( x = \sin(y) \) represents the boundary in \( x \)-coordinates, the integral must reflect the changes in \( y \).

### 4. **Determine the bounds of integration:**

The bounds for \( y \) are from \( y = 0 \) to \( y = \frac{\pi}{2} \).

### 5. **Set up the integral:**

The volume \( V \) is given by:

\[
V = \pi \int_{0}^{\frac{\pi}{2}} \left[ (2 - 0)^2 - (2 - \sin(y))^2 \right] dy
\]

Simplify the expression:

\[
V = \pi \int_{0}^{\frac{\pi}{2}} \left[ 4 - (2 - \sin(y))^2 \right] dy
\]

This is the integral that represents the volume of the solid.

Would you like more details on how to evaluate this integral or have any questions?

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### Related Questions:
1. How do you evaluate the integral for the volume once it's set up?
2. Can you explain how the washer method works for solids of revolution?
3. What are the alternative methods for finding the volume of a solid of revolution?
4. How would the integral change if the axis of rotation were different, say \( y = 3 \)?
5. What if the region were rotated about the x-axis instead of \( y = 2 \)?

### Tip:
When setting up an integral for a volume of revolution, always ensure you carefully identify the outer and inner radii relative to the axis of rotation. This helps prevent errors in the integral setup.

Learn how to set up, but not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the curves y = sin^{-1}(x), y = pi/2, x = 0 about y = 2 in terms of dy. This problem involves the concepts of solid of revolution, volume integral using the washer method, and is suitable for college-level students.

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