use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by x=0,  y=x^2 and y=18-3x about the line x=4.  round to 4 decimal places


To find the volume of the solid using the **method of cylindrical shells**, we follow these steps:

1. **Visualize the Problem**: 
   We are rotating the region bounded by:
   - \( y = x^2 \) (a parabola)
   - \( y = 18 - 3x \) (a line)
   - \( x = 0 \) (the y-axis)

   about the vertical line \( x = 4 \).

2. **Determine the Limits of Integration**:
   To find the limits of integration, we need to find the points where \( y = x^2 \) and \( y = 18 - 3x \) intersect. These are the boundaries for the region being rotated.

   Set \( x^2 = 18 - 3x \) and solve for \( x \):
   \[
   x^2 + 3x - 18 = 0
   \]
   Solve this quadratic equation using the quadratic formula:
   \[
   x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-18)}}{2(1)}
   \]
   \[
   x = \frac{-3 \pm \sqrt{9 + 72}}{2} = \frac{-3 \pm \sqrt{81}}{2} = \frac{-3 \pm 9}{2}
   \]
   So the solutions are \( x = 3 \) and \( x = -6 \).

   Thus, the region of interest is between \( x = 0 \) and \( x = 3 \).

3. **Set up the Volume Integral**:
   Using the method of cylindrical shells, the volume is given by the formula:
   \[
   V = 2\pi \int_{a}^{b} ( \text{radius} ) ( \text{height} ) \, dx
   \]
   - The radius of each cylindrical shell is the distance from \( x \) to the line of rotation \( x = 4 \), which is \( 4 - x \).
   - The height is the difference between the two curves, \( (18 - 3x) - x^2 \).

   Thus, the volume is:
   \[
   V = 2\pi \int_{0}^{3} (4 - x) \left( (18 - 3x) - x^2 \right) \, dx
   \]

4. **Simplify the Integral**:
   Expand the integrand:
   \[
   V = 2\pi \int_{0}^{3} (4 - x) \left( 18 - 3x - x^2 \right) \, dx
   \]
   Multiply out:
   \[
   V = 2\pi \int_{0}^{3} \left[ (4)(18 - 3x - x^2) - x(18 - 3x - x^2) \right] \, dx
   \]
   \[
   V = 2\pi \int_{0}^{3} \left[ 72 - 12x - 4x^2 - 18x + 3x^2 + x^3 \right] \, dx
   \]
   Combine like terms:
   \[
   V = 2\pi \int_{0}^{3} \left( 72 - 30x - x^2 + x^3 \right) \, dx
   \]

5. **Compute the Integral**:
   Now, integrate each term:
   \[
   V = 2\pi \left[ 72x - 15x^2 - \frac{x^3}{3} + \frac{x^4}{4} \right]_{0}^{3}
   \]
   Evaluate at \( x = 3 \) and \( x = 0 \):
   \[
   V = 2\pi \left[ 72(3) - 15(3^2) - \frac{3^3}{3} + \frac{3^4}{4} \right]
   \]
   \[
   V = 2\pi \left[ 216 - 135 - 9 + \frac{81}{4} \right]
   \]
   \[
   V = 2\pi \left[ 72 - 9 + 20.25 \right]
   \]
   \[
   V = 2\pi (83.25) = 166.5\pi
   \]

6. **Final Answer**:
   Now, compute the numerical value of the volume:
   \[
   V \approx 166.5 \times 3.1416 = 523.5988
   \]

Thus, the volume of the solid is approximately \( \boxed{523.5988} \) cubic units.

---

Would you like further clarification on any step, or additional details?

Here are five related questions for further exploration:
1. How would the result change if the region was rotated about the y-axis?
2. What is the geometric interpretation of cylindrical shells in this context?
3. How would you compute the volume if the region was rotated around \( x = -2 \)?
4. Can you apply the disk method for this problem? How does it compare?
5. How do you handle situations where the boundaries are defined by parametric equations?

**Tip**: Always sketch the region to understand its boundaries and how it will be rotated for a better grasp of the problem.

Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by x=0, y=x^2 and y=18-3x about the line x=4. Round to 4 decimal places.

Discover how to calculate the volume of a solid obtained by rotating the region bounded by y=x^2, y=18-3x, and x=0 about the line x=4 using the method of cylindrical shells. This advanced calculus problem involves integration and the quadratic formula, ideal for Grades 11-12 students.

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