Use the shell method to find the volume of the solid generated by revolving the shaded region about the​ x-axis. Set up the integral that gives the volume of the solid. upper = y=sqrt3, lower = x=y^2

To find the volume of the solid generated by revolving the shaded region about the x-axis using the shell method, we need to set up the integral for the volume. The shell method involves integrating with respect to the variable perpendicular to the axis of revolution, which in this case is the \(y\)-axis, since we are revolving around the \(x\)-axis.

### Step-by-Step Solution:

1. **Identify the bounds and functions:**
   - Upper boundary: \( y = \sqrt{3} \)
   - Lower boundary: \( x = y^2 \)
   - We are rotating about the \(x\)-axis, so the radius of each shell will be \(y\).

2. **Determine the radius and height of the shell:**
   - **Radius** of the shell: \( y \) (distance from the \(y\)-axis to the \(x\)-axis).
   - **Height** of the shell: \( y^2 \) (given by \(x = y^2\)).

3. **Set up the integral:**
   - The formula for the volume of a solid using the shell method is:
     \[
     V = 2\pi \int_a^b (\text{radius}) \times (\text{height}) \, dy
     \]
   - Substituting in the radius and height:
     \[
     V = 2\pi \int_0^{\sqrt{3}} y \cdot (y^2) \, dy
     \]

4. **Simplify the integral:**
   - Combine the terms inside the integral:
     \[
     V = 2\pi \int_0^{\sqrt{3}} y^3 \, dy
     \]

5. **Evaluate the integral:**
   - Integrate \(y^3\):
     \[
     V = 2\pi \left[\frac{y^4}{4}\right]_0^{\sqrt{3}}
     \]
   - Substitute the limits:
     \[
     V = 2\pi \left(\frac{(\sqrt{3})^4}{4} - \frac{0^4}{4}\right)
     \]
   - Simplify:
     \[
     V = 2\pi \left(\frac{9}{4}\right) = \frac{9\pi}{2}
     \]

### Final Integral Setup:
\[
V = 2\pi \int_0^{\sqrt{3}} y^3 \, dy
\]

The evaluated volume is:
\[
V = \frac{9\pi}{2}
\]

Would you like further details on any of these steps or have any questions?

Here are some related questions for further practice:

1. How would you set up the shell method integral if the region was rotated around the y-axis?
2. What changes in the setup if the upper function was \(x = y + 1\) instead?
3. How would the volume change if the boundaries were between \( y = 1 \) and \( y = \sqrt{3} \)?
4. Can you set up the volume integral using the disk/washer method instead?
5. How do you determine the limits of integration for the shell method?

**Tip:** When using the shell method, always ensure your radius and height expressions are in terms of the integration variable for accuracy.

Use the shell method to find the volume of the solid generated by revolving the shaded region about the​ x-axis. Set up the integral that gives the volume of the solid. Upper = y=sqrt3, Lower = x=y^2.

Learn how to find the volume of a solid generated by revolving a region around the x-axis using the shell method. The problem involves setting up an integral from 0 to sqrt(3) and applying the shell method formula. This detailed explanation is suitable for high school calculus or college-level students.

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