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

We are tasked with finding the volume of a solid generated by revolving the plane region defined by the equation \( x + y^2 = 36 \) around the \(x\)-axis using the **shell method**.

### Step 1: Express the equation in terms of \(x\)

The given equation is:
\[
x + y^2 = 36
\]
Solving for \(x\), we get:
\[
x = 36 - y^2
\]

### Step 2: Set up the shell method

The shell method involves integrating along the axis perpendicular to the axis of revolution. Since we are revolving around the \(x\)-axis, we'll be integrating with respect to \(y\).

The volume element for the shell method is:
\[
dV = 2\pi \cdot \text{(radius)} \cdot \text{(height)} \cdot \text{(thickness)}
\]
- **Radius**: The distance from the axis of revolution (the \(x\)-axis) to the shell, which is \(y\).
- **Height**: The height of the shell, which is the value of \(x = 36 - y^2\).
- **Thickness**: This is \(dy\).

Thus, the volume element becomes:
\[
dV = 2\pi y \cdot (36 - y^2) \, dy
\]

### Step 3: Set up the limits of integration

To determine the limits of integration, we find the values of \(y\) where the region intersects the \(y\)-axis. From the equation \(x + y^2 = 36\), when \(x = 0\), we have:
\[
0 + y^2 = 36 \quad \Rightarrow \quad y = \pm 6
\]
Thus, the region goes from \(y = -6\) to \(y = 6\).

### Step 4: Write the integral

The integral to find the volume is:
\[
V = 2\pi \int_{-6}^{6} y(36 - y^2) \, dy
\]

### Step 5: Evaluate the integral

We can now expand and evaluate the integral:
\[
V = 2\pi \int_{-6}^{6} (36y - y^3) \, dy
\]
This integral can be split into two parts:
\[
V = 2\pi \left( \int_{-6}^{6} 36y \, dy - \int_{-6}^{6} y^3 \, dy \right)
\]

Now, let's calculate the integral values. We'll evaluate each term.

#### First term:
\[
\int_{-6}^{6} 36y \, dy
\]
Since this is an odd function integrated over a symmetric interval, its value is zero:
\[
\int_{-6}^{6} 36y \, dy = 0
\]

#### Second term:
\[
\int_{-6}^{6} y^3 \, dy
\]
Similarly, \(y^3\) is also an odd function, and its integral over a symmetric interval is also zero:
\[
\int_{-6}^{6} y^3 \, dy = 0
\]

### Step 6: Conclusion

Both integrals evaluate to zero, which means the volume is:
\[
V = 0
\]

### Why is the result zero?

The result of zero implies that the shape is symmetric and cancels out when revolving around the \(x\)-axis. This suggests that the given region, when rotated around the \(x\)-axis, does not enclose any volume.

---

Would you like further clarification on any step? Here are some questions that expand on this topic:

1. How does the shell method differ from the disk/washer method?
2. What happens if we revolve around the \(y\)-axis instead?
3. Can you explain how symmetry affects the volume calculation in more detail?
4. What are some practical applications of the shell method in real-life problems?
5. Could we use a different approach to solve this integral?

**Tip:** When dealing with functions that are symmetric about an axis, integrating odd functions over symmetric intervals results in zero due to cancellation.

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 plane region defined by the equation x + y^2 = 36 around the x-axis using the shell method. This detailed solution covers key concepts from integral calculus and explores the impact of symmetry on the volume of revolution. Suitable for Grades 11-12 and early college students.

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