Use the shell method to write and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the x-axis.
x + y2 = 36

To find the volume of the solid generated by revolving the plane region around the x-axis using the **shell method**, we follow these steps:

### 1. Set up the equation of the region:
We are given the equation:
\[
x + y^2 = 36
\]
We can rearrange this as:
\[
x = 36 - y^2
\]
This describes a sideways parabola. We will revolve this region around the x-axis using the shell method.

### 2. Shell Method Setup:
The shell method formula for revolving a region around the x-axis is:
\[
V = 2\pi \int_{a}^{b} y \cdot \text{(radius)} \cdot \text{(height)} \, dy
\]
Here:
- **y** represents the distance from the x-axis (the shell's radius).
- **(36 - y^2)** represents the shell's height (the x-coordinate of the curve).
- The bounds for \(y\) are determined by the region. Solving for \(y\) when \(x = 0\):
  \[
  36 - y^2 = 0 \quad \Rightarrow \quad y = \pm 6
  \]

So, the limits of integration are from \( y = -6 \) to \( y = 6 \).

### 3. Write the integral:
The volume integral becomes:
\[
V = 2\pi \int_{-6}^{6} y \cdot (36 - y^2) \, dy
\]

### 4. Simplify the integrand:
Expanding the expression inside the integral:
\[
y \cdot (36 - y^2) = 36y - y^3
\]
Thus, the integral becomes:
\[
V = 2\pi \int_{-6}^{6} (36y - y^3) \, dy
\]

### 5. Evaluate the integral:
We can now evaluate the integral term by term:
\[
V = 2\pi \left[ \int_{-6}^{6} 36y \, dy - \int_{-6}^{6} y^3 \, dy \right]
\]
- The integral of \(36y\) is:
  \[
  \int 36y \, dy = 18y^2
  \]
- The integral of \(y^3\) is:
  \[
  \int y^3 \, dy = \frac{y^4}{4}
  \]
Evaluating both integrals from \(-6\) to \(6\), note that the integral of \(36y\) over a symmetric interval cancels out because it's an odd function, leaving only the \(y^3\) term:

\[
V = 2\pi \left[ 0 - \left( \frac{6^4}{4} - \frac{(-6)^4}{4} \right) \right]
\]
\[
V = 2\pi \left[ -\frac{1296}{4} \right]
\]
\[
V = 2\pi \left[ -324 \right]
\]
Thus, the volume is:
\[
V = 648\pi \text{ cubic units.}
\]

---

### Conclusion:
The volume of the solid generated by revolving the region around the x-axis is \( 648\pi \) cubic units.

Would you like more details or have any questions?

### Related Questions:
1. How would the result change if the region were revolved around the y-axis?
2. Can you explain how to use the disk method for this problem?
3. What if the region was bounded by \( x + y^2 = 25 \)? How would the volume change?
4. How would you modify the shell method for revolving around \( y = 3 \)?
5. What are the general differences between the shell and disk methods?

**Tip:** When using the shell method, remember that it often simplifies problems where the radius and height are functions of the same variable.

Use the shell method to write and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the x-axis. x + y^2 = 36

Learn how to find the volume of a solid generated by revolving the region defined by x + y^2 = 36 around the x-axis using the shell method. This guide includes step-by-step calculations and explains key calculus concepts related to volumes of solids of revolution. Ideal for college students studying Calculus I or II.

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