Question

Find the smaller area enclosed by the circle $x^2+y^2=4$ and the line $x+y=2$.

Answer 1

As per point of intersection equation of circle is taken as
$x^2+y^2=4$
Equations of the circle and the line are
$x^2+y^2=4$ and $x+y=2$.
The curves intersect at $(2,0)$ and $(0,2)$.
Shaded area is the smaller area enclosed by the two curves and is
${\int }_0^2\sqrt{4-x^2}dx-{\int }_0^2(2-x)dx$ ...$\left(i\right)$

$={\left[\begin{array}{c}\frac{x.\sqrt{4-x^2}}{2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}\end{array}\right]}_0^2-{\left[\begin{array}{c}2x-\frac{x^2}{2}\end{array}\right]}_0^2$

$={\left[\begin{array}{c}\frac{x.\sqrt{4-x^2}}{2}+2{\sin }^{-1}\frac{x}{2}-2x+\frac{x^2}{2}\end{array}\right]}_0^2$

$=\frac{2.\sqrt{4-4}}{2}+2{\sin }^{-1}\left(1\right)-4+2$

$=2.\frac{\pi }{2}-2$

$=(\pi -2)$ sq. unit.