We are tasked with finding the volume of a solid obtained by rotating the region bounded by $y = 6x + 54$, $y = 0$, and $x = 0$ about the y-axis. This can be solved using the method of cylindrical shells.

Step-by-Step Process:

1. Identify the boundaries:

   • The curve $y = 6x + 54$.
   • The line $y = 0$ is the x-axis.
   • The line $x = 0$ is the y-axis.

2. Find the limits of integration:

   • When $x = 0$, $y = 6(0) + 54 = 54$.
   • When $y = 0$, solving for $x$ from $y = 6x + 54$: $0 = 6x + 54 \quad \Rightarrow \quad x = -9.$
   • So the region is bounded between $y = 0$ and $y = 54$, and $x$ runs from $-9$ to $0$.

3. Set up the cylindrical shell formula:

   • The formula for the volume of a solid by cylindrical shells is: $V = 2\pi \int_{y_1}^{y_2} x(y) \cdot h(y) \, dy$ where $x(y)$ is the radius of the shell (which depends on $y$) and $h(y)$ is the height of the shell.

4. Express $x$ in terms of $y$:

   • From $y = 6x + 54$, solve for $x$: $x = \frac{y - 54}{6}.$

5. Set up the integral:

   • The radius is $x = \frac{y - 54}{6}$.
   • The height is the distance from $x = 0$, which is given by $x$.
   • So the volume integral becomes: $V = 2\pi \int_{0}^{54} \left( \frac{y - 54}{6} \right) \cdot \left( \frac{y - 54}{6} \right) \, dy.$

6. Simplify the integral: $V = 2\pi \int_{0}^{54} \frac{(y - 54)^2}{36} \, dy.$

7. Evaluate the integral: $V = \frac{\pi}{18} \int_{0}^{54} (y - 54)^2 \, dy.$ Let $u = y - 54$, so when $y = 0$, $u = -54$, and when $y = 54$, $u = 0$. The integral becomes: $V = \frac{\pi}{18} \int_{-54}^{0} u^2 \, du.$ Now, use the power rule to integrate: $\int u^2 \, du = \frac{u^3}{3}.$ So, $V = \frac{\pi}{18} \left[ \frac{u^3}{3} \right]_{-54}^{0} = \frac{\pi}{18} \left( 0 - \frac{(-54)^3}{3} \right).$ Calculate $(-54)^3$: $(-54)^3 = -157464,$ so the volume is: $V = \frac{\pi}{18} \left( \frac{157464}{3} \right) = \frac{\pi}{18} \times 52488 = 2916\pi.$

Final Answer:

The volume of the solid is $2916\pi$ cubic units.

Would you like more details on any step or have questions?

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Here are five related questions to further explore the concept:

1. How does the cylindrical shell method differ from the disk/washer method?
2. What is the geometric interpretation of the cylindrical shell method?
3. How would the problem change if the region were rotated about the x-axis instead?
4. Can the same result be obtained using the disk method? Why or why not?
5. What happens to the volume if the line $y = 0$ is replaced by $y = 10$?

Tip: When dealing with rotation problems, always carefully choose the method (disk, washer, or cylindrical shells) based on the axis of rotation and the shape of the region.