Find the area of the region bounded by the ellipse `(x^(2))/(9)+(y^(2))/(4)=1` in fourth quadrant.

Options

A. `(3pi)/(4)` sq unit
B. `(3pi)/(2)` sq unit
C. `(9pi)/(4)` sq unit
D. `(4pi)/(7)` sq unit

Explanation

To find the area of the region bounded by the ellipse \(\frac{x^2}{9} + \frac{y^2}{4} = 1\) in the fourth quadrant, we can follow these steps:

### Step 1: Understand the Ellipse
The given equation of the ellipse is \(\frac{x^2}{9} + \frac{y^2}{4} = 1\). This ellipse is centered at the origin (0,0) with semi-major axis \(a = 3\) along the x-axis and semi-minor axis \(b = 2\) along the y-axis.

### Step 2: Find the Area in the First Quadrant
To find the area in the fourth quadrant, we can first find the area in the first quadrant and then use symmetry. The equation of the ellipse can be rewritten to express \(y\) in terms of \(x\):
\[ y = 2\sqrt{1 - \frac{x^2}{9}} \]
We will calculate the area under this curve from \(x = 0\) to \(x = 3\).

### Step 3: Set Up the Integral
The area \(A\) in the first quadrant is given by the integral:
\[ A = \int_{0}^{3} 2\sqrt{1 - \frac{x^2}{9}} \, dx \]
Since the area in the fourth quadrant will be the same, the total area in the fourth quadrant will be:
\[ \text{Area in fourth quadrant} = A \]

### Step 4: Solve the Integral
We can simplify the integral:
\[ A = 2 \int_{0}^{3} \sqrt{1 - \frac{x^2}{9}} \, dx \]
Let \(a = 3\), then the integral becomes:
\[ A = 2 \int_{0}^{3} \sqrt{1 - \left(\frac{x}{3}\right)^2} \, dx \]
Using the substitution \(x = 3\sin(\theta)\), we have \(dx = 3\cos(\theta) d\theta\). The limits change from \(x = 0\) to \(\theta = 0\) and from \(x = 3\) to \(\theta = \frac{\pi}{2}\).

Now, substituting:
\[ A = 2 \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \sin^2(\theta)} \cdot 3\cos(\theta) \, d\theta = 2 \int_{0}^{\frac{\pi}{2}} 3\cos^2(\theta) \, d\theta \]
Using the identity \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\):
\[ A = 2 \cdot 3 \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta = 3 \left[ \theta + \frac{1}{2}\sin(2\theta) \right]_{0}^{\frac{\pi}{2}} \]
Evaluating this gives:
\[ A = 3 \left[ \frac{\pi}{2} + 0 - (0 + 0) \right] = \frac{3\pi}{2} \]

### Step 5: Final Area Calculation
Since the area in the fourth quadrant is the same as in the first quadrant, the area in the fourth quadrant is:
\[ \text{Area in fourth quadrant} = \frac{3\pi}{2} \]

### Conclusion
Thus, the area of the region bounded by the ellipse in the fourth quadrant is:
\[ \boxed{\frac{3\pi}{2}} \]