https://r2.erweima.ai/mathbot/image/1726525403667zz203db2.jpg

To solve this problem, we will use the **Disk Method** to calculate the volume of the solid generated by revolving the region bounded by \( f(x) = 2 - x^2 \) and \( g(x) = 1 \) around the line \( y = 1 \).

### Step-by-step procedure:

1. **Identify the curves and region:**
   - The upper boundary is \( f(x) = 2 - x^2 \), a parabola.
   - The lower boundary is the line \( g(x) = 1 \).
   - The solid is generated by rotating the shaded area around \( y = 1 \).

2. **Set up the disk method formula:**
   The formula for the volume generated by revolving around a horizontal axis \( y = 1 \) is:

   \[
   V = \pi \int_{a}^{b} \left( R(x)^2 - r(x)^2 \right) \, dx
   \]
   Where:
   - \( R(x) \) is the outer radius (the distance from the parabola \( f(x) \) to the line \( y = 1 \)).
   - \( r(x) \) is the inner radius (the distance from the line \( g(x) = 1 \) to the axis of rotation \( y = 1 \), which is just 0 since \( g(x) = y = 1 \)).
   - \( a \) and \( b \) are the points of intersection between the curves.

3. **Find the points of intersection:**
   To find the limits \( a \) and \( b \), we set \( f(x) = g(x) \):

   \[
   2 - x^2 = 1 \implies x^2 = 1 \implies x = \pm 1
   \]

   So the region is bounded between \( x = -1 \) and \( x = 1 \).

4. **Set up the radii:**
   - The outer radius \( R(x) \) is the distance from the curve \( f(x) = 2 - x^2 \) to the axis \( y = 1 \), which is \( R(x) = (2 - x^2) - 1 = 1 - x^2 \).
   - The inner radius \( r(x) = 0 \) because \( g(x) = y = 1 \) is on the axis of rotation.

5. **Write the integral:**
   \[
   V = \pi \int_{-1}^{1} \left( (1 - x^2)^2 \right) \, dx
   \]

6. **Solve the integral:**
   Expand \( (1 - x^2)^2 \):

   \[
   (1 - x^2)^2 = 1 - 2x^2 + x^4
   \]

   So the integral becomes:

   \[
   V = \pi \int_{-1}^{1} (1 - 2x^2 + x^4) \, dx
   \]

   Since the integrand is even, we can double the integral from 0 to 1:

   \[
   V = 2\pi \int_0^{1} (1 - 2x^2 + x^4) \, dx
   \]

   Now integrate term by term:

   \[
   \int_0^1 1 \, dx = 1, \quad \int_0^1 x^2 \, dx = \frac{1}{3}, \quad \int_0^1 x^4 \, dx = \frac{1}{5}
   \]

   Therefore:

   \[
   V = 2\pi \left[ 1 - 2\left(\frac{1}{3}\right) + \frac{1}{5} \right]
   \]
   \[
   V = 2\pi \left( 1 - \frac{2}{3} + \frac{1}{5} \right)
   \]
   Find a common denominator:

   \[
   V = 2\pi \left( \frac{15}{15} - \frac{10}{15} + \frac{3}{15} \right) = 2\pi \left( \frac{8}{15} \right)
   \]
   \[
   V = \frac{16\pi}{15}
   \]

Thus, the volume of the solid generated by the revolution is \( \frac{16\pi}{15} \).

Would you like further details or have any questions?

Here are some related questions:
1. What is the general formula for the disk method for solids of revolution?
2. How would the solution change if the region were revolved around another horizontal line?
3. What is the significance of even and odd functions when computing integrals over symmetric intervals?
4. Could the washer method be used here instead of the disk method? Why or why not?
5. How can we apply this method to find the volume of solids with more complex boundaries?

**Tip:** For functions symmetric about the y-axis, you can often simplify integrals by only calculating over half the region and then doubling the result!

Calcular el volumen del sólido generado al girar la región limitada por f(x) = 2 - x^2 y g(x) = 1 en torno a la recta y = 1.

Calculate the volume of a solid formed by rotating the region bounded by the parabola f(x) = 2 - x^2 and the line g(x) = 1 around the line y = 1 using the Disk Method. This detailed solution covers key calculus concepts and is suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation