Question

The area of the region \[\left\{ \left( x, y \right) : x^2 + y^2 \leq 1 \leq x + y \right\}\] is __________ .

Options

• \[\frac{\pi}{5}\]

• \[\frac{\pi}{4}\]

• \[\frac{\pi}{2} - \frac{1}{2}\]

• \[\frac{\pi^2}{2}\]

• None of these

Answer 1

To find the points of intersection of the line and the circle substitute y = 1 − x in x² + y² = 1,we get A(0, 1) and B(1, 0).
Therefore, the required area of the shaded region,
\[A = \int_0^1 \left( y_1 - y_2 \right) d x ............\left(\text{Where, }y_1 = \sqrt{1 - x^2}\text{ and }y_2 = 1 - x \right)\]
\[ = \int_0^1 \left[ \left( \sqrt{1 - x^2} \right) - \left( 1 - x \right) \right] d x\]
\[ = = \int_0^1 \left( \sqrt{1 - x^2} - 1 + x \right) d x\]
\[ = \left[ \frac{x}{2}\sqrt{1 - x^2} + \frac{1}{2} \sin^{- 1} \left( x \right) - x + \frac{x^2}{2} \right]_0^1 \]
\[ = \left[ \frac{1}{2}\sqrt{1 - 1^2} + \frac{1}{2} \sin^{- 1} \left( 1 \right) - \left( 1 \right) + \frac{\left( 1 \right)^2}{2} \right] - \left[ \frac{\left( 0 \right)}{2}\sqrt{1 - \left( 0 \right)^2} + \frac{1}{2} \sin^{- 1} \left( 0 \right) - \left( 0 \right) + \frac{\left( 0 \right)^2}{2} \right]\]
\[ = \left( \frac{\pi}{4} - \frac{1}{2} \right)\text{ square units }\]