Problem Statement
Find the area of the smaller region bounded by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the straight line $\frac{x}{3} + \frac{y}{2} = 1$ using calculus.
Verified Solution & Marking Scheme
Find points of intersection
The ellipse intersects the coordinate axes at $(\pm 3, 0)$ and $(0, \pm 2)$. The line $\frac{x}{3} + \frac{y}{2} = 1$ passes through $(3, 0)$ and $(0, 2)$.
Thus, the smaller region lies in the first quadrant between $x = 0$ and $x = 3$.
Express curves as y in terms of x
For the ellipse in quadrant 1: $y_1 = 2\sqrt{1 - \frac{x^2}{9}} = \frac{2}{3}\sqrt{9 - x^2}$.
For the line: $y_2 = 2\left(1 - \frac{x}{3}\right) = \frac{2}{3}(3 - x)$.
Set up the integral for the bounded area
$\text{Area } A = \int_{0}^{3} (y_1 - y_2) \, dx = \frac{2}{3} \int_{0}^{3} \left( \sqrt{9 - x^2} - (3 - x) \right) \, dx$
Evaluate standard integrals
$\int_{0}^{3} \sqrt{9 - x^2} \, dx = \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) \right]_{0}^{3} = 0 + \frac{9}{2}\sin^{-1}(1) - 0 = \frac{9}{2} \cdot \frac{\pi}{2} = \frac{9\pi}{4}$
$\int_{0}^{3} (3 - x) \, dx = \left[ 3x - \frac{x^2}{2} \right]_{0}^{3} = 9 - \frac{9}{2} = \frac{9}{2}$
Substituting back:
$A = \frac{2}{3} \left( \frac{9\pi}{4} - \frac{9}{2} \right) = \frac{3}{2}\pi - 3 = \frac{3}{2}(\pi - 2) \text{ sq. units}$