The area of the region $R=\{(x, y): |x| ≤ y\, and\, x^2 + y^2 ≤1\}$, in square units, is
Answer & explanation
Correct answer: option 3
Let A be the required area. Then,
A = 2 (Area of the shaded region in first quadrant)
$⇒A=2\int\limits_{0}^{1/\sqrt{2}}(y_1-y_2)dx$
$⇒A=2\int\limits_{0}^{1/\sqrt{2}}(\sqrt{1-x^2}-x)dx$
$⇒A=2\left[\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x-\frac{x^2}{2}\right]_{0}^{1/\sqrt{2}}$
$⇒A=2\left[\frac{1}{2\sqrt{2}}×\frac{1}{\sqrt{2}}+\frac{1}{2}×\frac{п}{4}-\frac{1}{4}\right]=\frac{п}{4}$ sq. units.