Question

Using integration, find the area of the region {(x, y) : x² + y² ≤ 1 ≤ x + y}.

Answer 1

x² + y² = 1 represents a circle with centre at (0, 0) and radius as 1 unit.
Since, x² + y² ≤ 1, so the region represents the interior of the circle x² + y² = 1.
x + y = 1 is the equation of a straight line cutting X and Y axes at (1, 0) and (0, 1) respectively.

\[x + y \geq 1\]b  represents the region lying above the line x + y = 1.
So, the shaded region is as follows:

\[\therefore \text { Area of the shaded region } = \int_0^1 \sqrt{1 - x^2}dx - \int_0^1 \left( 1 - x \right)dx\]

\[ = \left[ \frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2} \sin^{- 1} \left( \frac{x}{1} \right) - x + \frac{x^2}{2} \right]_0^1 \]

\[ = \left[ \frac{1}{2}\sqrt{1 - 1} + \frac{1}{2} \sin^{- 1} \left( \frac{1}{1} \right) - 1 + \frac{1}{2} \right] - \left[ \frac{0}{2}\sqrt{1 - 0} + \frac{1}{2} \sin^{- 1} \left( \frac{0}{1} \right) - 0 + \frac{0}{2} \right]\]

\[ = \frac{\pi}{4} - 1 + \frac{1}{2}\]

\[ = \frac{\pi - 4 + 2}{4}\]

\[ = \frac{\pi - 2}{4}\text { units }\]