Sketch the region, set up, but not evaluate an integral by using the method of

Cylindrical Shells to find the volume of the solid obtained by rotating the region

bounded by the given curves

y = x3, y = √x

about the lines

(1) y = −1

To find the volume of the solid obtained by rotating the region bounded by the curves \( y = x^3 \) and \( y = \sqrt{x} \) about the line \( y = -1 \), we use the method of cylindrical shells.

### Step 1: Sketch the Region

1. **Curves to Sketch:**
   - \( y = x^3 \): This is a cubic function passing through the origin. For \( x \geq 0 \), it increases slowly initially and then more steeply.
   - \( y = \sqrt{x} \): This is a square root function, starting from the origin and increasing rapidly initially, but then leveling off as \( x \) increases.

2. **Intersection Points:**
   - Set \( x^3 = \sqrt{x} \) to find the points of intersection.
   - Squaring both sides: \( x^6 = x \), or \( x(x^5 - 1) = 0 \).
   - Solutions: \( x = 0 \) or \( x = 1 \).

   Therefore, the curves intersect at \( x = 0 \) and \( x = 1 \).

3. **Region Bounded by the Curves:**
   - The region of interest is between \( x = 0 \) and \( x = 1 \) where \( y = \sqrt{x} \) is above \( y = x^3 \).

### Step 2: Set Up the Integral Using Cylindrical Shells

For the cylindrical shells method, we consider a vertical strip at a point \( x \) from \( x = 0 \) to \( x = 1 \).

1. **Height of the Shell:**
   - The height of the shell is the vertical distance between the two curves: 
     \[
     \text{Height} = \sqrt{x} - x^3.
     \]

2. **Radius of the Shell:**
   - Since we are rotating around \( y = -1 \), the radius is the distance from a point on the curve to the line \( y = -1 \).
   - At a point \( (x, y) \) on the curve, the radius is:
     \[
     \text{Radius} = y + 1.
     \]
   - Substituting for \( y \), we consider both curves:
     - For \( y = x^3 \), the radius is \( x^3 + 1 \).
     - For \( y = \sqrt{x} \), the radius is \( \sqrt{x} + 1 \).

3. **Volume Integral:**
   - The volume using cylindrical shells is given by:
     \[
     V = \int_{a}^{b} 2 \pi (\text{radius}) \times (\text{height}) \, dx.
     \]

   - Here, the region is bounded between \( x = 0 \) and \( x = 1 \), so:
     \[
     V = \int_{0}^{1} 2 \pi (\sqrt{x} + 1)(\sqrt{x} - x^3) \, dx.
     \]

### Final Integral Setup:

\[
V = 2 \pi \int_{0}^{1} (\sqrt{x} + 1)(\sqrt{x} - x^3) \, dx.
\]

This integral represents the volume of the solid obtained by rotating the region bounded by \( y = x^3 \) and \( y = \sqrt{x} \) around the line \( y = -1 \).

Would you like to know how to evaluate this integral, or do you have any questions about the setup?

### Follow-up Questions:

1. What would be the setup if the rotation was around \( y = 1 \) instead of \( y = -1 \)?
2. How would you set up the integral if rotating around the \( x \)-axis instead?
3. What changes in the method if you use the disk method instead of cylindrical shells?
4. How would the integral change if the region were bounded by different curves, like \( y = x^2 \) and \( y = 1 - x \)?
5. What are some advantages of using the cylindrical shell method compared to the disk method?

#### Tip:

When setting up integrals for volumes of rotation, always carefully identify the radius and height of the shell or disk to match the axis of rotation.

Sketch the region, set up, but not evaluate an integral by using the method of Cylindrical Shells to find the volume of the solid obtained by rotating the region bounded by the given curves y = x^3, y = √x about the lines (1) y = −1.

Learn how to set up an integral using the cylindrical shells method to find the volume of the solid formed by rotating the region bounded by the curves y = x^3 and y = √x around the line y = -1. This problem involves key calculus concepts, including volume of solids of revolution, integration, and curve intersection points. Suitable for undergraduate calculus students.

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