Using integration, find the area of the enclosed region: $\left\{(x, y): \frac{x^{2}}{9}+\frac{y^{2}}{4} \leq 1 \leq \frac{x}{3}+\frac{y}{2}\right\}$

$\frac{3}{2}(\pi-2)$ square units

Area Under Curves — CBSE Class 12 · 2010 Board Question with Solution

Using integration, find the area of the enclosed region:

{(x, y) : \frac{x ^{2}}{9} + \frac{y ^{2}}{4} \leq 1 \leq \frac{x}{3} + \frac{y}{2}}

Answer: $\frac{3}{2} (π - 2)$ square units

The region lies inside the ellipse

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

but on the far side of the chord

\frac{x}{3} + \frac{y}{2} = 1

(the line through

(3, 0)

and

(0, 2)

).
In the first quadrant the ellipse gives

y_{1} = \frac{2}{3} 9 - x^{2}

and the line gives

y_{2} = \frac{2}{3} (3 - x)

, so

A = \int_{0}^{3} (y_{1} - y_{2}) d x = \frac{2}{3} \int_{0}^{3} [9 - x^{2} - (3 - x)] d x .

\int_{0}^{3} 9 - x^{2} d x = \frac{9 π}{4}

(quarter circle) and

\int_{0}^{3} (3 - x) d x = \frac{9}{2}

, so

A = \frac{2}{3} (\frac{9 π}{4} - \frac{9}{2}) = \frac{3 π}{2} - 3 = \frac{3}{2} (π - 2) square units.

CBSE Class 12 Board · 2010 · Subjective