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: $V = 648\pi \text{ cubic units.}$

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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.