The area enclosed by y² = 8x and y = $$\sqrt2$$ x that lies outside the triangle formed by y = $$\sqrt2$$ x, x = 1, y = 2$$\sqrt2$$, is equal to:

The area of the bounded region enclosed by the curve

$$y = 3 - \left| {x - {1 \over 2}} \right| - |x + 1|$$ and the x-axis is :

The area of the region S = {(x, y) : y² $$\le$$ 8x, y $$\ge$$ $$\sqrt2$$x, x $$\ge$$ 1} is

The area of the region bounded by y² = 8x and y² = 16(3 $$-$$ x) is equal to: