Step 1: Understanding the Concept:
We need to find the area between an ellipse and a straight line.
The line connects the x-intercept and y-intercept of the ellipse in the first quadrant.
The bounded area is the difference between the area of the quarter-ellipse and the area of the right-angled triangle formed by the line.
Step 2: Key Formula or Approach:
The ellipse equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a=3$ and $b=2$.
The line equation is $\frac{x}{a} + \frac{y}{b} = 1$.
Area of ellipse $= \pi a b$. Area of its quadrant in first quadrant $= \frac{1}{4} \pi a b$.
Area of triangle formed by intercepts $a, b$ on axes $= \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} a b$.
Required bounded Area $= (\text{Area of quarter ellipse}) - (\text{Area of triangle})$.
Step 3: Detailed Explanation:
The given curve is an ellipse $\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1$. Its semi-major axis is $a=3$ and semi-minor axis is $b=2$.
It intersects the axes at $(3,0), (-3,0), (0,2), (0,-2)$.
The given line is $\frac{x}{3} + \frac{y}{2} = 1$. Its x-intercept is 3 and y-intercept is 2.
Thus, the line passes through the points $(3,0)$ and $(0,2)$.
The region bounded by the ellipse and the line lies entirely in the first quadrant.
The area under the ellipse in the first quadrant is:
$A_{\text{ellipse\_quadrant}} = \frac{1}{4} \cdot \pi \cdot a \cdot b = \frac{1}{4} \cdot \pi \cdot 3 \cdot 2 = \frac{6\pi}{4} = \frac{3\pi}{2}$ sq. units.
The area of the right-angled triangle formed by the line and the coordinate axes is:
$A_{\text{triangle}} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 3 \cdot 2 = 3$ sq. units.
The required bounded area $A$ is the difference between these two areas:
$A = A_{\text{ellipse\_quadrant}} - A_{\text{triangle}}$
$A = \frac{3\pi}{2} - 3$
Taking $\frac{3}{2}$ common:
$A = \frac{3}{2}(\pi - 2)$ sq. units.
Step 4: Final Answer:
The area of the region is $\frac{3}{2}(\pi - 2)$ sq. units.