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The problem asks to find the volume of the solid generated when the area in the first quadrant of the ellipse 

\[
\frac{x^2}{9} + \frac{y^2}{4} = 1
\]

is rotated through \(360^\circ\) about the axis.

### Steps to Solve:

The equation of the ellipse given is 

\[
\frac{x^2}{9} + \frac{y^2}{4} = 1
\]

which can be rewritten as:

\[
y = 2\sqrt{1 - \frac{x^2}{9}}
\]

This is the equation of the ellipse solved for \(y\), and it represents the upper half of the ellipse. We are only concerned with the first quadrant, meaning we'll rotate this upper part about the x-axis to calculate the volume.

#### Volume by the Disk Method

The volume \(V\) of a solid generated by rotating a curve around the x-axis can be calculated using the disk method, given by:

\[
V = \pi \int_{a}^{b} [f(x)]^2 \, dx
\]

Here, \(f(x) = 2\sqrt{1 - \frac{x^2}{9}}\) and we rotate this from \(x = 0\) to \(x = 3\) (the x-intercepts of the ellipse in the first quadrant). Thus, the volume will be:

\[
V = \pi \int_{0}^{3} \left(2\sqrt{1 - \frac{x^2}{9}}\right)^2 dx
\]

Simplifying:

\[
V = \pi \int_{0}^{3} 4\left(1 - \frac{x^2}{9}\right) dx
\]

Now, evaluate this integral:

\[
V = 4\pi \int_{0}^{3} \left(1 - \frac{x^2}{9}\right) dx
\]

Break this into two simpler integrals:

\[
V = 4\pi \left[\int_{0}^{3} 1 \, dx - \frac{1}{9}\int_{0}^{3} x^2 \, dx \right]
\]

1. The first integral:

\[
\int_{0}^{3} 1 \, dx = 3
\]

2. The second integral:

\[
\int_{0}^{3} x^2 \, dx = \left[\frac{x^3}{3}\right]_{0}^{3} = \frac{27}{3} = 9
\]

Substituting back:

\[
V = 4\pi \left[3 - \frac{1}{9}(9)\right] = 4\pi (3 - 1) = 4\pi \times 2 = 8\pi
\]

Thus, the volume of the solid generated is:

\[
V = 8\pi \, \text{cubic units}.
\]

Would you like further details on any of these steps?

---

Here are 5 related questions to expand your understanding:

1. How would the volume change if the ellipse were rotated around the y-axis instead?
2. What happens to the volume if only half of the first quadrant is rotated?
3. Can we use the shell method to calculate the same volume? How would the steps differ?
4. How would the problem change if the ellipse had different axis lengths?
5. What is the volume of a solid generated by rotating a circle around the x-axis?

**Tip:** When using the disk or shell method for volume, always sketch the region and determine the axis of rotation clearly to avoid confusion about the limits and function used for integration.

Find the volume of the solid generated when the area in the first quadrant of the ellipse x^2/9 + y^2/4 = 1 is rotated through 360 degrees about the axis.

Learn how to find the volume of a solid generated by rotating the area in the first quadrant of an ellipse, x^2/9 + y^2/4 = 1, through 360 degrees about the axis. This problem involves concepts of calculus, integration, and the disk method for finding volumes of solids of revolution. Suitable for students in Grades 11-12.

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