Question

The area of the region in the first quadrant which is above the parabola $y = x^2$ and enclosed by the circle $x^2 +y^2 = 2$ and the y-axis is

• A. $\dfrac{1}{6}+\dfrac{π}{4}$
• B. $\dfrac{1}{12}+\dfrac{π}{4}$
• C. $\dfrac{-1}{6}+\dfrac{\pi}{4}$
• D. $\dfrac{-1}{6}-\dfrac{π}{4}$
• E. $\dfrac{-π^2}{2} +4$

Answer 1

Answer: (A)

$\dfrac{1}{6}+\dfrac{π}{4}$

Answer 2

Given that

$y = x^2$ and $x^2 + y^2 = 2$

$⇒ x^2 + y^2 = 2$

$⇒y + y^2 = 2$

$⇒(y + 2)(y − 1) = 0$

$\therefore (y + 2)(y − 1) = 0$

$\therefore y = −2, 1$

Then the area,

$A = ∫_0^1 √ ydy + ∫_1^{√2}( 2 - y^2)dy$

$= \dfrac{2}{3} [y^{3/2} ]_0^1 + [\dfrac{ty}{2} √( 2 - y^2) + sin−1(\dfrac{ y}{ √ 2}]_1^{√ 2 }$

$= \dfrac{2}{3} + 0 + \dfrac{\pi}{2}-(\dfrac{1}{2}+\dfrac{\pi }{4} )$

$= \dfrac{\pi}{4}+\dfrac{1}{6}$(_Ans.)