https://file.aiquickdraw.com/mathbot/file/1726029137336z4l3cztv.jpg

We are asked to find the volume of the solid formed by rotating the region bounded by the curves \( y = -x^2 + 11x - 30 \) and \( y = 0 \) about the y-axis.

This problem can be solved using the **shell method** since the axis of rotation is vertical (the y-axis). The shell method formula for rotation around the y-axis is:

\[
V = 2\pi \int_{x_1}^{x_2} x \cdot f(x) \, dx
\]

### Step-by-Step Solution:

1. **Identify the curve**: 
   The equation of the curve is \( y = -x^2 + 11x - 30 \). We need to find the points where this curve intersects the x-axis, which is when \( y = 0 \).

   \[
   -x^2 + 11x - 30 = 0
   \]

   Solve this quadratic equation using the quadratic formula:

   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \( a = -1 \), \( b = 11 \), and \( c = -30 \).

   Substituting into the formula:

   \[
   x = \frac{-11 \pm \sqrt{11^2 - 4(-1)(-30)}}{2(-1)}
   \]
   \[
   x = \frac{-11 \pm \sqrt{121 - 120}}{-2}
   \]
   \[
   x = \frac{-11 \pm \sqrt{1}}{-2}
   \]
   \[
   x = \frac{-11 \pm 1}{-2}
   \]
   \[
   x_1 = \frac{-11 + 1}{-2} = 5, \quad x_2 = \frac{-11 - 1}{-2} = 6
   \]

   So, the region of interest is between \( x = 5 \) and \( x = 6 \).

2. **Set up the integral**: 
   Now we can set up the volume integral using the shell method:

   \[
   V = 2\pi \int_{5}^{6} x \cdot (-x^2 + 11x - 30) \, dx
   \]

3. **Expand the integrand**: 
   Multiply \( x \) through the polynomial:

   \[
   V = 2\pi \int_{5}^{6} (-x^3 + 11x^2 - 30x) \, dx
   \]

4. **Integrate**: 
   Now, we find the antiderivative of each term:

   \[
   \int -x^3 \, dx = -\frac{x^4}{4}, \quad \int 11x^2 \, dx = \frac{11x^3}{3}, \quad \int -30x \, dx = -15x^2
   \]

   Substituting these into the integral:

   \[
   V = 2\pi \left[ -\frac{x^4}{4} + \frac{11x^3}{3} - 15x^2 \right]_{5}^{6}
   \]

5. **Evaluate the definite integral**:
   We need to calculate the expression at \( x = 6 \) and \( x = 5 \), then subtract.

   First, evaluate at \( x = 6 \):

   \[
   -\frac{6^4}{4} + \frac{11 \cdot 6^3}{3} - 15 \cdot 6^2 = -\frac{1296}{4} + \frac{11 \cdot 216}{3} - 15 \cdot 36
   \]
   \[
   = -324 + 792 - 540 = -72
   \]

   Now, evaluate at \( x = 5 \):

   \[
   -\frac{5^4}{4} + \frac{11 \cdot 5^3}{3} - 15 \cdot 5^2 = -\frac{625}{4} + \frac{11 \cdot 125}{3} - 15 \cdot 25
   \]
   \[
   = -156.25 + 458.33 - 375 = -72.92
   \]

6. **Subtract the results**:

   \[
   V = 2\pi (-72 - (-72.92)) = 2\pi (0.92)
   \]

   Therefore, the volume is:

   \[
   V = 1.84\pi \ \text{cubic units}
   \]

The volume of the solid is approximately \( 1.84\pi \).

---

Would you like more details or have any questions about this process?

### Related Questions:
1. How do we apply the disk method for this kind of problem?
2. What are the main differences between the shell method and disk method?
3. Can you explain how to use the washer method?
4. How would the result change if we rotated around the x-axis instead?
5. How do you solve quadratic equations for intersections efficiently?

### Tip:
When working with rotation problems, always visualize the region and axis of rotation to determine the best method (disk, washer, or shell).

The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.
 y = -x^2 + 11x - 30, y = 0; about the y-axis

Find the volume of the solid formed by rotating the region bounded by y = -x^2 + 11x - 30 and y = 0 about the y-axis using the shell method. This problem involves concepts of rotational volumes, integration, and solving quadratic equations. Suitable for college-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation