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

The Correct Option is A

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.)