Exclude the Enclosed

Calculus Level 3

Find the area that is enclosed by the circle x 2 + y 2 = 2 x^2 + y^2 = 2 and excluding the common area enclosed by y = x y = x and y 2 = x y^2 = x .

12 π − 1 6 \frac{12\pi - 1}6 12 π − 1 12 \frac{12\pi - 1}{12} 6 π − 1 6 \frac{6\pi - 1}6 6 π − 1 12 \frac{6\pi - 1}{12}

Ram Mohith by Ram Mohith , 18, India

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1 solution

The word 'circles' is confusing, since there is only one circle in the problem. The straight line and the parabola intersect at points ( 0 , 0 ) (0,0) and ( 1 , 1 ) (1,1) . The area enclosed by them is ∫ 0 1 ( x − x ) d x = 1 6 \int_0^1 {(\sqrt x-x) dx}=\dfrac{1}{6} . Area of the circle is 2 π 2π . Hence the required area is 2 π − 1 6 = 12 π − 1 6 2π-\dfrac{1}{6}=\boxed {\dfrac{12π-1}{6}}

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