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 = squareroot of x, y = 0, x = 4; about x = 9. the problem is asking for the top bound and the integral set up in terms of y

We are tasked with setting up an integral to calculate the volume of a solid obtained by rotating a region about a vertical line, \( x = 9 \). The region is bounded by the curves:

- \( y = \sqrt{x} \)
- \( y = 0 \)
- \( x = 4 \)

Since the problem asks for the integral setup in terms of \( y \), we will:

### Step 1: Rewrite the boundary curve in terms of \( y \)
The curve \( y = \sqrt{x} \) can be rewritten as \( x = y^2 \), as we are solving for \( x \) in terms of \( y \).

### Step 2: Identify the bounds of \( y \)
Given the curve \( y = \sqrt{x} \) and the upper bound \( x = 4 \), the corresponding \( y \)-value when \( x = 4 \) is:

\[
y = \sqrt{4} = 2
\]

Thus, the region of interest for \( y \) is from \( y = 0 \) to \( y = 2 \).

### Step 3: Set up the integral using the washer method
We are rotating the region about the vertical line \( x = 9 \), so the distance from the line of rotation to the curve is \( 9 - x \). 

In terms of \( y \), this becomes:
\[
9 - x = 9 - y^2
\]

The outer radius is the distance from the line \( x = 9 \) to the \( x = 0 \) boundary, which is \( 9 \), and the inner radius is the distance from \( x = 9 \) to the curve \( x = y^2 \), which is \( 9 - y^2 \).

The volume of a solid of revolution using the washer method is given by the integral:

\[
V = \pi \int_{y=0}^{y=2} \left[ R_{\text{outer}}^2 - R_{\text{inner}}^2 \right] \, dy
\]

Substituting the radii:
\[
R_{\text{outer}} = 9 \quad \text{and} \quad R_{\text{inner}} = 9 - y^2
\]

The integral becomes:
\[
V = \pi \int_{0}^{2} \left[ 9^2 - (9 - y^2)^2 \right] \, dy
\]

This is the integral setup for the volume of the solid.

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Would you like to go over the steps in detail, or do you have any specific questions?

### Related Questions:
1. How do you calculate the volume when rotating around horizontal lines?
2. What happens if the curve is rotated about the y-axis instead of a vertical line?
3. Can you apply the shell method for this problem, and how would the integral change?
4. How does the washer method work when the rotation line is within the region?
5. What if the boundary curve were \( y = x^2 \) instead of \( y = \sqrt{x} \)?

### Tip:
When setting up integrals for solids of revolution, always carefully visualize the axis of rotation and the distances to the boundaries.

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 = squareroot of x, y = 0, x = 4; about x = 9. The problem is asking for the top bound and the integral set up in terms of y.

Learn how to set up the integral for calculating the volume of a solid obtained by rotating the region bounded by y = sqrt(x), y = 0, and x = 4 around the line x = 9. This detailed solution uses the washer method and is suitable for undergraduate calculus students.

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