Question

Find the area of the region {(x, y): x² + y² ≤ 4, x + y ≥ 2}.

Answer 1

\[\text{ Let R }= \left\{ \left( x, y \right): x^2 + y^2 \leq 4 , x + y \geq 2 \right\}\]
\[ R_1 = \left\{ \left( x, y \right): x^2 + y^2 \leq 4 \right\}\]
\[ R_2 = \left\{ \left( x, y \right): x + y \geq 2 \right\}\]
\[ \therefore R = R_1 \cap R_2\]
The region R₁ represents interior of the circle x² + y² = 4 with centre (0, 0) and has a radius 2.
The region R2 lies above the line x + y =2
The line x + y =2 and circle x² + y² = 4 intersect each other at (2, 0) and (0, 2).
Here, the length of the shaded region is given by

\[\left| y_2 - y_1 \right|\] where y₂ is y for the circle x² + y² = 4 and y₁ is y for the line x + y = 2 ; y₂ > y₁ and  the width of the shaded portion is dx.
Therefore the area,
\[A = \int_0^2 \left( y_2 - y_1 \right) d x\]
\[ = \int_0^2 \left[ \sqrt{4 - x^2} - \left( 2 - x \right) \right] d x\]
\[ = \left[ \frac{1}{2}x\sqrt{4 - x^2} + \frac{4}{2} \sin^{- 1} \left( \frac{x}{2} \right) \right]_0^2 - \left[ 2x - \frac{x^2}{2} \right]_0^2 \]
\[ = \left[ \frac{2}{2}\sqrt{4 - 2^2} + \frac{4}{2} \sin^{- 1} \left( \frac{2}{2} \right) - \frac{0}{2}\sqrt{4 - 0^2} - \frac{4}{2} \sin^{- 1} \left( \frac{0}{2} \right) \right] - \left[ 2\left( 2 \right) - \frac{2^2}{2} - 2\left( 0 \right) + \frac{0^2}{2} \right]\]
\[ = \left[ 0 + 2 \sin^{- 1} \left( 1 \right) - 0 - 0 \right] - \left[ 4 - 2 - 0 + 0 \right]\]
\[ = 2 \sin^{- 1} \left( 1 \right) - 2\]
\[ = 2 \times \frac{\pi}{2} - 2\]
\[ = \pi - 2\]