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JEE Main 2017 · Mathematics · Area Under Curves · hard

The area of the region lying inside the circle x^2+y^2=2 and above the parabola y=x^2 is

Single correct option

The area of the region lying inside the circle x2+y2=2x^2+y^2=2 and above the parabola y=x2y=x^2 is
  1. A. π2+23\dfrac\pi2+\dfrac23
  2. B. π+13\pi+\dfrac13
  3. C. π2−13\dfrac\pi2-\dfrac13
  4. D. π2+13\dfrac\pi2+\dfrac13
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Answer: (D) π2+13\dfrac\pi2+\dfrac13

Intersection: x2+x4=2x^2+x^4=2 gives x2=1x^2=1, so the curves meet at (±1,1)(\pm1,1). Between x=−1x=-1 and x=1x=1 the upper boundary is the circle y=2−x2y=\sqrt{2-x^2} and the lower boundary is the parabola. Area =∫−11(2−x2−x2)dx=\int_{-1}^{1}\left(\sqrt{2-x^2}-x^2\right)dx. Using ∫a2−x2dx=x2a2−x2+a22sin⁡−1xa\int\sqrt{a^2-x^2}dx=\tfrac x2\sqrt{a^2-x^2}+\tfrac{a^2}2\sin^{-1}\tfrac xa with a2=2a^2=2, the first part is 1+2sin⁡−112=1+π21+2\sin^{-1}\tfrac1{\sqrt2}=1+\tfrac\pi2. The second is 23\tfrac23. Area =π2+13=\tfrac\pi2+\tfrac13.
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