https://file.aiquickdraw.com/mathbot/file/172914922986808kmtmtm.jpg

We are tasked with finding the volume of a solid generated by revolving the region bounded by the curves \( x = 2\sqrt{y} \), \( x = -y \), and \( y = 2 \) about the x-axis using the shell method.

### Step-by-Step Approach:

1. **Understand the Region**:
   - \( x = 2\sqrt{y} \) represents a parabola opening to the right.
   - \( x = -y \) is a straight line with a slope of -1.
   - \( y = 2 \) represents the upper boundary of the region.

2. **Shell Method Formula**:
   The shell method formula for a volume of revolution about the x-axis is given by:
   \[
   V = 2\pi \int_{a}^{b} \text{radius}(y) \cdot \text{height}(y) \, dy
   \]
   Here, the radius is simply \( y \) (since we are rotating around the x-axis), and the height is the difference between the rightmost function and the leftmost function, \( 2\sqrt{y} - (-y) = 2\sqrt{y} + y \).

3. **Set Up the Integral**:
   We integrate from \( y = 0 \) to \( y = 2 \), the limits of the region along the y-axis:
   \[
   V = 2\pi \int_0^2 y \cdot (2\sqrt{y} + y) \, dy
   \]
   Simplify the expression inside the integral:
   \[
   V = 2\pi \int_0^2 y (2\sqrt{y} + y) \, dy = 2\pi \int_0^2 (2y^{3/2} + y^2) \, dy
   \]

4. **Compute the Integral**:
   We now integrate term by term:
   \[
   \int 2y^{3/2} \, dy = \frac{2 \cdot 2}{5} y^{5/2} = \frac{4}{5} y^{5/2}
   \]
   \[
   \int y^2 \, dy = \frac{y^3}{3}
   \]
   So, the volume integral becomes:
   \[
   V = 2\pi \left[ \frac{4}{5} y^{5/2} + \frac{y^3}{3} \right]_0^2
   \]

5. **Evaluate the Definite Integral**:
   Substituting \( y = 2 \):
   \[
   \frac{4}{5} (2^{5/2}) = \frac{4}{5} \cdot 4\sqrt{2} = \frac{16\sqrt{2}}{5}
   \]
   \[
   \frac{2^3}{3} = \frac{8}{3}
   \]
   Now, evaluate at \( y = 0 \) (both terms go to zero):
   \[
   V = 2\pi \left( \frac{16\sqrt{2}}{5} + \frac{8}{3} \right)
   \]

6. **Combine the Terms**:
   We need a common denominator to combine the two terms:
   \[
   V = 2\pi \left( \frac{48\sqrt{2}}{15} + \frac{40}{15} \right) = 2\pi \cdot \frac{48\sqrt{2} + 40}{15}
   \]

7. **Final Answer**:
   The volume of the solid is:
   \[
   V = \frac{2\pi (48\sqrt{2} + 40)}{15} \, \text{cubic units}.
   \]

Let me know if you'd like any more details or have further questions!

---

Here are five follow-up questions to expand on this:

1. How would the setup change if the region were rotated about the y-axis instead?
2. Could we use the disk or washer method instead of the shell method for this problem?
3. What is the geometrical interpretation of the height in the shell method?
4. How does the integral setup change if one of the boundary curves is a different shape, such as a sine curve?
5. Can we approximate the volume using numerical integration if we can't find an analytical solution?

**Tip**: Always sketch the region of revolution when using methods like shell or washer, as it helps visualize the setup more clearly!

Use the shell method to find the volume of the solid generated by revolving the region bounded by x = 2√y, x = -y, and y = 2 about the x-axis.

Learn how to find the volume of a solid generated by revolving the region bounded by x = 2√y, x = -y, and y = 2 about the x-axis using the shell method. This problem involves integrating functions of y and is suitable for advanced high school or 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