IIT-JEE 1995Question 1
Consider a square with vertices at $(1,1),(-1,1),(-1,-1)$ and $(1,-1)$. Let $S$ be the region consisting of all points inside the square which are nearer to the origin than to any edge. Sketch the region $S$ and find its area.
Application of Integrals
Answer$\dfrac{16\sqrt2-20}{3}$
Solution
Nearer to the origin than the edge $x=1$: $x^2+y^2<(1-x)^2\Rightarrow x<\frac{1-y^2}2$ (four such parabolic arcs). In the octant $0\le y\le x$ the arc meets $y=x$ at $x=\sqrt2-1$, so area $=8\int_0^{\sqrt2-1}\left(\frac{1-y^2}2-y\right)dy=\frac{16\sqrt2-20}3$.