Question

Find the area of the region \[\left\{ \left( x, y \right): \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \leq \frac{x}{a} + \frac{y}{b} \right\}\]

Answer 1

\[\text{ Let }R = \left\{ \left( x, y \right) : \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \leq \frac{x}{a} + \frac{y}{b} \right\}\]
\[ \Rightarrow R_1 = \left\{ \left( x, y \right) : \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \right\}\]
\[\text{ and }R_{2 =} \left\{ \left( x, y \right) : 1 \leq \frac{x}{a} + \frac{y}{b} \right\}\]
Then, 
\[R = R_1 \cap R_2 \]
\[\text{ Consider }\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 . \text{ This represents an ellipse, symmetrical about both axis and cutting }x - \text{ axis at A(a, 0) and A'( - a, 0) and }y - \text{ axis at B(0, b), B'(0, - b)}\]
\[ \Rightarrow R_1 = \left( \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \right)\text{ represents the area inside the ellipse }\]
\[\frac{x}{a} + \frac{y}{b} = 1 = \text{ represents a straight line cutting }x - \text{ axis at A(a, 0) and }y - \text{ axis at B(0, b)}\]
\[ \Rightarrow R_2 = \left( \frac{x}{a} + \frac{y}{b} \geq 1 \right)\text{ represents the area above the straight line }\]
\[ \Rightarrow R = R_1 \cap R_2\text{ represents the smaller shaded area bounded by the line and the ellipse }\]
\[\text{ In the shaded region, consider a vertical strip with length }= \left| y_2 - y_1 \right|\text{ and width = dx, such that }P(x, y_2 ) \text{ lies on ellipse and }Q(x, y_1 )\text{ lies on the straight line }\]
\[\text{ Area of approximating rectangle }= \left| y_2 - y_1 \right| dx \]
\[\text{ The approximating rectangle moves from }x = 0\text{ to }x = a\]
\[ \therefore \text{ Area of the shaded region }= \int_0^a \left| y_2 - y_1 \right| dx = \int_0^a \left( y_2 - y_1 \right) dx ..................\left[ As, y_2 > y_1 , \left| y_2 - y_1 \right| = y_2 - y_1 \right] \]
\[ \Rightarrow A = \int_0^a \left( \frac{b}{a}\sqrt{a^2 - x^2} - \frac{b}{a}\left( a - x \right) \right) dx\]
\[ \Rightarrow A = \int_0^a \left( \frac{b}{a}\sqrt{a^2 - x^2} \right)dx - \int_0^a \frac{b}{a}\left( a - x \right) dx\]
\[ \Rightarrow A = \frac{b}{a} \left[ \left\{ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \sin^{- 1} \left( \frac{x}{a} \right) \right\} \right]_0^a - \frac{b}{a}\left[ ax - \frac{x^2}{2} \right]\]
\[ \Rightarrow A = \frac{b}{a}\left[ 0 + \frac{1}{2} a^2 \sin^{- 1} 1 - \left( a^2 - \frac{a^2}{2} \right) \right]\]
\[ \Rightarrow A = \frac{b}{a}\left[ \frac{1}{2} a^2 \times \frac{\pi}{2} - \frac{a^2}{2} \right]\]
\[ \Rightarrow A = \frac{ab}{2}\left[ \frac{\pi}{2} - 1 \right]\]
\[ \Rightarrow A = \frac{ab}{4}\left[ \pi - 2 \right]\text{ sq . units } \]