find the volume generated by revolving the ellipse x²/a² y²/b² = 1 about the y axis maths problem 

To find the volume generated by revolving the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) about the \(y\)-axis, we'll use the disk method.

### Step-by-Step Solution

1. **Equation of the Ellipse:**
   \[
   \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
   \]

2. **Solving for \(x\):**
   \[
   x = a \sqrt{1 - \frac{y^2}{b^2}}
   \]

3. **Volume of a Thin Disk:**
   When the ellipse is revolved around the \(y\)-axis, each thin horizontal strip becomes a disk with radius \(x\) and thickness \(dy\).
   
   The volume \(dV\) of a thin disk at height \(y\) is:
   \[
   dV = \pi x^2 dy
   \]

4. **Expressing \(x^2\) in Terms of \(y\):**
   \[
   x^2 = a^2 \left(1 - \frac{y^2}{b^2}\right)
   \]

5. **Volume Element \(dV\):**
   \[
   dV = \pi \left( a^2 \left(1 - \frac{y^2}{b^2}\right) \right) dy = \pi a^2 \left(1 - \frac{y^2}{b^2}\right) dy
   \]

6. **Integrate to Find Total Volume:**
   Integrate \(dV\) from \(y = -b\) to \(y = b\):
   \[
   V = \int_{-b}^{b} \pi a^2 \left(1 - \frac{y^2}{b^2}\right) dy
   \]

7. **Simplify the Integral:**
   \[
   V = \pi a^2 \int_{-b}^{b} \left(1 - \frac{y^2}{b^2}\right) dy
   \]

8. **Separate the Integral:**
   \[
   V = \pi a^2 \left( \int_{-b}^{b} 1 \, dy - \int_{-b}^{b} \frac{y^2}{b^2} \, dy \right)
   \]

   Evaluate the first term:
   \[
   \int_{-b}^{b} 1 \, dy = y \Big|_{-b}^{b} = b - (-b) = 2b
   \]

   Evaluate the second term:
   \[
   \int_{-b}^{b} \frac{y^2}{b^2} \, dy = \frac{1}{b^2} \int_{-b}^{b} y^2 \, dy
   \]

   Using the integral of \(y^2\):
   \[
   \int_{-b}^{b} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{-b}^{b} = \frac{b^3}{3} - \left(-\frac{b^3}{3}\right) = \frac{2b^3}{3}
   \]

   So, the second term becomes:
   \[
   \frac{1}{b^2} \cdot \frac{2b^3}{3} = \frac{2b}{3}
   \]

9. **Combine Both Terms:**
   \[
   V = \pi a^2 \left( 2b - \frac{2b}{3} \right) = \pi a^2 \cdot \frac{4b}{3} = \frac{4 \pi a^2 b}{3}
   \]

### Final Volume:
\[
V = \frac{4 \pi a^2 b}{3}
\]

### Diagram

To visualize this, here is a conceptual diagram of the ellipse being revolved around the \(y\)-axis, forming a 3D solid of revolution:

```
           ^
         / | \
       /   |   \
     /     |     \
   /       |       \
 /         |         \
 ------------->
       y-axis
```

In this diagram, imagine rotating the ellipse around the \(y\)-axis to generate the solid of revolution.

---

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### Relative Questions:

1. What is the disk method in calculating volumes of revolution?
2. How do you solve for \(x\) in the ellipse equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)?
3. What is the integral of \(y^2\) over \([-b, b]\)?
4. How does the washer method differ from the disk method?
5. What are the applications of finding volumes of revolution in real life?
6. How do you derive the volume of an ellipse revolved around the x-axis?
7. How would the volume change if the ellipse was revolved around the x-axis instead?
8. Can you use cylindrical shells to find the volume of the same solid?

### Tip:

When integrating to find volumes, always ensure you have the correct bounds and integrand to simplify your work and avoid errors.

Learn how to calculate the volume generated by revolving the ellipse x^2/a^2 + y^2/b^2 = 1 around the y-axis using the disk method in integral calculus. Detailed step-by-step solution included.

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