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.

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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.