what is the area enclosed by the equation `x^2+y^2=2`

Options

A. `4 pi` square units
B. `2 pi` square units
C. `4pi^(2)` square units
D. `4` square units

Answer

B

Explanation

Given equation of circle is `x^(2) + y^(2) = 2`
`rArr y = sqrt(2-x^(2))`

Required aera `= 4 xx " Area of shaded poration "`
`= 4 int_(0)^(sqrt(2)) sqrt(2+x^(2)) dx`
`sqrt(2-x^(2)) dx`
Let `x sqrt(2) sin t rArr t = sin^(-1) (x/(sqrt(2)))`
`dx = sqrt(2) cos t . dt`
`:.int sqrt(2-x^(2)).dx = intsqrt(2-2sin^(2)t).sqrt(2) cost.dt`
`= intsqrt(2cos^(2))t.sqrt(2) cott.dt`
`= 2 int cos^(2)t.dt `
We know,
`int cos^(n)x. dx = (n-1)/(n) int cos^(n-2) dx +(cos^(n-1) x . sin x)/(n)`
`=2[(cost.sint)/(2).(1)/(2)int1.dt] = 2[(cost.sint)/(2)+t/2]`
`=costsint+t`
`=cos(sin^(-1) 'x/(sqrt(2))) xx sin (sin^(-1)'x/(sqrt(2)))+sin^(-1) '(x)/(sqrt(2))`
`= sqrt(1-(x^(2))/(2)).(x)/(sqrt(2)) +sin^(-1)'(x)/(sqrt(2))`
`= x/2sqrt(2-x^(2)) + sin^(-1)'(x)/(sqrt(2))`
`:. 4 . int_(0)^(sqrt(2)) sqrt(2-x^(2)).dx = 4 [x/2 sqrt(2-x^(2)) +sin^(-1)'x/(sqrt(2))]_(0)^(sqrt(2))`
`= 4 [sin^(-1) 1-sin^(-1) 0] = 4(pi/2 - 0) = 2pi`sq units.