find the volume of the solid obtained by rotating the region bounded by y=6x+54, y=0, x=0 about the y axis

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.

Find the volume of the solid formed by rotating the region bounded by y = 6x + 54, y = 0, and x = 0 around the y-axis. This detailed solution uses the cylindrical shell method and covers key calculus concepts such as integration and the method of solids of revolution. Suitable for Grades 11-12 and college calculus students.

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