The area of the region bounded by the curves `x^2 + y^2 = 8 and y^2=2x` (in sq. unit) is

Options

A. ` 2pi + (1 ) /(3 ) `
B. ` pi + ( 1 ) /( 3 ) `
C. ` 2pi + ( 4 ) /(3 ) `
D. ` pi + ( 4 ) /( 3 ) `

Answer

C

Explanation

Given curves ,
` x ^ 2 + y ^ 2 = 8 " " `… (i)
and ` y^ 2 = 2x " " `… (ii)

On solving Eqs. (i) and (ii), we get
` x^ 2 + 2x - 8 = 0 `
` x ^ 2 + 4x - 2x - 8 = 0 `
` x ( x + 4 ) - 2 ( x + 4 ) = 0 `
` (x- 2) ( x + 4 ) = 0 `
` therefore x = 2 and y =pm 2 `
` therefore ` Required area
` = 2 ` [Area of OAP + Area of PAB]
= `2 [ int _ 0 ^ 2 sqrt ( 2x ) dx + int _ ( 2) ^ ( 2sqrt2 ) sqrt ( 8 - x^ 2 )) dx] `
` = 2 [ sqrt 2 ( x ^ (3//2) * ( 2 ) /( 3 ) ) _ 0 ^ 2 + ( ( x ) /( 2 ) sqrt ( 8 -x ^ 2 ) + ( 8 ) /( 2 ) sin ^( - 1 ) "" ( x )/ ( 2 sqrt 2 ) ) _ 2 ^( 2 sqrt 2 ) ] `
` = 2[ ( 2sqrt 2 )/( 3 ) ( 2 ^( 3//2 ) ) + 4 xx ( pi ) /( 2 ) - 2 - 4xx ( pi ) /( 4 ) ] `
` = 2 [ ( 2sqrt2 )/( 3 ) * 2sqrt 2 + 2pi - 2 - pi] `
` = 2 [ (8)/( 3) - 2 + pi ] = 2 ( ( 2 ) /( 3 ) + pi ) = 2pi + (4 )/( 3 ) `