Question

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

Answer 1

\[\text{ Let R }= \left\{ \left( x, y \right) : \frac{x^2}{9} + \frac{y^2}{4} \leq 1 \leq \frac{x}{3} + \frac{y}{2} \right\}\]
\[ R_1 = \left\{ \left( x, y \right) : \frac{x^2}{9} + \frac{y^2}{4} \leq 1 \right\}\]
\[\text{ and }R_2 = \left\{ \left( x, y \right) : 1 \leq \frac{x}{3} + \frac{y}{2} \right\}\]
\[ \therefore R = R_{1 \cap} R_2 \]
\[\frac{x^2}{9} + \frac{y^2}{4} = 1\text{ represents an ellipse, with centre at O(0, 0) , cutting the coordinate axis at A }\left( 3, 0 \right), A'\left( - 3, 0 \right), B\left( 0, 2 \right)text{ and B' }\left( 0, - 2 \right)\]
\[\text{ Hence, } R_1\text{ is area interior to the ellipse }\]
\[\frac{x}{3} + \frac{y}{2} = 1\]
\[ \Rightarrow 2x + 3y = 6 \text{ represents a straight line cutting the coordinate axis at A }\left( 3, 0 \right)\text{ and }B\left( 0, 2 \right)\]
\[\text{ Hence, }R_2 \text{will be area above the line }\]
\[ \Rightarrow A\left( 3, 0 \right)\text{ and B }\left( 0, 2 \right) \text{ are points of intersection of ellipse and straight line .} \]
Area of shaded region, 
\[A = \int_0^3 \left[ \sqrt{4\left( 1 - \frac{x^2}{9} \right)} - 2\left( 1 - \frac{x}{3} \right) \right] dx\]
\[ = \int_0^3 \left[ \sqrt{\frac{36 - 4 x^2}{9}} - \left( \frac{6 - 2x}{3} \right) \right]dx\]
\[ = \frac{1}{3} \int_0^3 \left[ 2\sqrt{9 - x^2} - \left( 6 - 2x \right) \right] dx\]
\[ = \frac{1}{3} \left[ 2 \times \left\{ \frac{1}{2}x\sqrt{9 - x^2} + \frac{1}{2} \times 9 \sin^{- 1} \left( \frac{x}{3} \right) \right\} - \left( 6x - 2\frac{x^2}{2} \right) \right]_0^3 \]
\[ = \frac{1}{3} \left[ x\sqrt{9 - x^2} + 9 \sin^{- 1} \left( \frac{x}{3} \right) - 6x + x^2 \right]_0^3 \]
\[ = \frac{1}{3}\left[ 0 + 9 \sin^{- 1} \left( 1 \right) - 18 + 9 - 0 \right] \]
\[ = \frac{1}{3}\left[ 9\frac{\pi}{2} - 9 \right]\]
\[ = \left( \frac{3\pi}{2} - 3 \right)\text{ sq units }\]