The area of the region, enclosed by the circle x2 + y2 = 2 which is not common to the region bounded by the parabola y2 = x and the straight line y = x, is :

Question

The area of the region, enclosed by the circle x2 + y2 = 2 which is not common to the region bounded by the parabola y2 = x and the straight line y = x, is : (A) (1/3)(12π - 1) (B) (1/6)(12π - 1) (C) (1/3)(6π - 1) (D) (1/6)(24π - 1)

Tardigrade Admin · Accepted Answer

A = ∫10(√x-x)dx = [(3/2)x3/2 -(x2/2)]10 = (1/6) Required Area: π r2 -(1/6) = (1/6)(12π-1)

Q. The area of the region, enclosed by the circle $x^{2} + y^{2} = 2$ which is not common to the region bounded by the parabola $y^{2} = x$ and the straight line $y = x$ , is :

$\frac{1}{3} (12 π - 1)$

$\frac{1}{6} (12 π - 1)$

$\frac{1}{3} (6 π - 1)$

$\frac{1}{6} (24 π - 1)$

$A = \int_{0}^{1} (x - x) d x$
$= [\frac{3}{2} x^{3/2} - \frac{x ^{2}}{2}]_{0}^{1} = \frac{1}{6}$
Required Area : $π r^{2} - \frac{1}{6} = \frac{1}{6} (12 π - 1)$

Q. The area of the region, enclosed by the circle x2+y2=2 which is not common to the region bounded by the parabola y2=x and the straight line y=x, is :

31​(12π−1)

61​(12π−1)

31​(6π−1)

61​(24π−1)