Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about the​ x-axis.
yequalsStartRoot x EndRoot​, yequals​0, and xequals36
36
x
y
y equals StartRoot x EndRoot
R



A coordinate system has a horizontal x-axis labeled from 0 to 36 plus and an unlabeled positive vertical y-axis. From left to right, a curve labeled y = StartRoot x EndRoot starts on the origin and rises at a decreasing rate in quadrant 1. A dashed vertical line segment extends from the x-axis at 36 to the curve. The region below the curve, above the x-axis, and to the left of the vertical line segment is shaded and labeled R.
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Part 1
Set up the integral that gives the volume of the solid. Use increasing limits of integration. Select the correct choice below and fill in the answer boxes to complete your choice.

The shell method is a technique used to find the volume of a solid of revolution, especially when the solid is generated by revolving a region about an axis. In this problem, we are revolving the region \( R \) about the **x-axis**.

### Shell Method Formula
When revolving around the **x-axis**, the formula for the shell method is:

\[
V = \int_{a}^{b} 2\pi \cdot y \cdot f(y) \, dy
\]

Where:
- \( y \) is the radius of the shell (distance from the axis of rotation to the shell).
- \( f(y) \) represents the length of the shell at height \( y \), which is the function in terms of \( y \).
- The limits \( a \) and \( b \) represent the bounds for \( y \).

### Given Curves:
- \( y = \sqrt{x} \) (or equivalently, \( x = y^2 \)).
- \( y = 0 \) (the x-axis).
- The region is bounded by \( x = 36 \), which corresponds to \( y = \sqrt{36} = 6 \).

### Steps to Set Up the Integral

1. **Radius of the shell:** The distance from the x-axis (where the rotation occurs) to the point on the curve is \( y \).
   
2. **Height of the shell:** Since \( y = \sqrt{x} \) or equivalently \( x = y^2 \), the height of the shell at a given value of \( y \) is the value of \( x = y^2 \).

3. **Limits of integration:** The region \( R \) extends from \( y = 0 \) (where the curve meets the x-axis) to \( y = 6 \) (since \( \sqrt{36} = 6 \)).

### Integral Setup
Using the shell method for revolving around the **x-axis**, we integrate with respect to \( y \). The height of each cylindrical shell is \( y^2 \), and the radius of the shell is \( y \), so the integral becomes:

\[
V = \int_0^6 2\pi \cdot y \cdot (y^2) \, dy = \int_0^6 2\pi \cdot y^3 \, dy
\]

This is the correct setup for the integral.

### Would you like to compute the volume or need further clarification?

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Here are 5 related questions to expand on this concept:

1. How would the setup change if the region were revolved around the y-axis instead of the x-axis?
2. Can the disc/washer method be used for this problem? If so, how would you set it up?
3. What is the geometric interpretation of the shell method versus the disc method?
4. How do you determine the bounds of integration when using the shell method?
5. How would the integral change if the curve was \( y = x^2 \) instead of \( y = \sqrt{x} \)?

**Tip:** The shell method is especially useful when it's easier to express the height of the shell as a function of the axis of rotation!

Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about the x-axis. y = √x, y = 0, and x = 36.

This problem uses the shell method to calculate the volume of a solid generated when the region bounded by y = √x, y = 0, and x = 36 is revolved about the x-axis. Learn step-by-step how to set up the integral and apply the shell method formula.

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