Using integration, find the smaller area enclosed by the circle x^{2} + y^{2} = 4  and the line x + y = 2.

 

Answers (1)

Circle: x^{2} + y^{2} = 4 ......(i)

Line: x + y = 2.....(ii)

Put value of y from equation (ii) in equation (i)

\\ x^{2} + (2-x)^{2} = 4 \\

\\ 2x^{2} - 4x = 0 \\

\\ x = 0 \ \ or \ \ x =2 \\

Intersection point A(0,2) and B(2,0)

\\$Required Area$ = $Area of arc AMBA$ \\\\ = \int _0^2 \sqrt{4-x^2}dx - \int _0^2 (2-x)dx

\\ = \left [\frac{x^2}{2} \sqrt{4-x^2}+ \frac{4}{2} \sin ^{-1} \frac{x}{2} \right ]_0^2 +\left[ \frac{(2-x)^2}{2}\right]_0^2 \\\\ = \left [0+ 2 \sin ^{-1}1 - 0-0 \right ] +\left[ 0 - 2\right] \\\\ = \pi -2 \ square \ units

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