(1) To find the points of itersection of the circle ` x^2 + y ^2 = 16 ` and the line ` y = x `
Substituting ` y = x ` in ` x^2 + y ^2 = 16`, we get,
` 2x^2 = 16 " " therefore x ^2 = 8 `
` therefore x = 2 sqrt 2 `, as point is in the first quadrant
` therefore y = x = 2 sqrt 2 `
` therefore ` the point of intersection of the circle and the line in the first quadrant is ` B ( 2 sqrt2, 2 sqrt2) `.
Requied area = area of the region OADBO
= area of ` Delta ` OCB + area of the region CADBC
Now, area of ` Delta OCB ` = area under the line ` y = x `
` = int_ (0 ) ^ ( 2sqrt2 ) y dx, ` where ` y= x `
` = int _ 0 ^ ( 2sqrt2 ) x dx = [ ( x^2)/( 2 ) ] _0 ^ 2 sqrt2 `
` = ( 1)/ ( 2 ) ( 8 - 0 ) = 4 `
Area of the region CADBC = area under the circle ` x ^2 + y ^2 = 16 `
` = int _ ( 2 sqrt 2 ) ^ ( 4 ) y dx ` where ` x ^ 2 + y ^2 = 16 ` , i.e., ` y = sqrt ( 16 - x ^2)`
` = [ ( x ) /(2) sqrt ( 16 - x ^(2)) + ( 16) /(2) sin ^ ( - 1) ((x ) /(4)) ] _ ( 2 sqrt 2 ) ^ 4 `
` = 0 + 8 xx ( pi ) /(2) - 4 - 8 xx ( pi )/ ( 4 ) `
` = 2 pi - 4 `
` therefore `required area = ` 4 + 2pi - 4 = 2pi ` sq units.
(2)
To find the points of intersection of the curves ` 4 x ^2 + 4y ^2 = 9 and y ^2 = 4 x `. From the equations of the curves.
` ( 9 - 4 x ^2)/ ( 4 ) = 4 x `
` therefore 9 - 4 x^2 = 16 x `
` therefore 4 x^2 + 16 x - 9 = 0 `
` therefore 4 x ^2 + 18 x - 2 x - 9 = 0 `
` therefore 2x ( 2x + 9 ) - 1 ( 2x + 9 ) = 0 `
` therefore ( 2x - 1 ) ( 2x + 9 ) = 0 `
` therefore x = ( 1) / ( 2 ) , x = ( - ( 9 ) / ( 2 ) ) `
If ` x = - ( 9 ) /(2) ` , then from the equation ` y ^2 = 4 x `.
` y^2 = 4 ( - ( 9 ) /(2)) = - 18 lt 0 `, which is not possible
` therefore x ne - ( 9 ) / ( 2 ) `
` therefore x = ( 1 ) / ( 2 ) `
Where ` x = ( 1 ) / (2), y^2 = 4 xx ( 1 ) / ( 2 ) = 2 `
` therefore y = pm sqrt 2 `
` therefore ` the points of intersection are ` B ((1 ) / ( 2 ) , sqrt 2 ) and C ( ( 1 )/ (2), -sqrt 2 ) `.
Required area = area of the region OCABO
= 2 ( area of the region OMABO )
= 2 ( area of region OMBO + area of the region BMAB)
Now, area of the region OMBO
= area under the parabola ` y ^2 = 4 x`, i.e., ` y = 2 sqrt x `
` = int _ 0^ ( 1//2) 2 sqrt x dx = 2 int _ 0 ^ (1//2) x ^((1)/(2)) dx `
` = 2 * [ (x ^ ( ( 3 ) /(2)) ) /( 3 //2) ] _ 0 ^ ( (1) /(2)) = ( 4 ) / ( 3 ) [ ( ( 1 ) /(2)) ^ ((3) /(2)) - 0 ] `
` = ( 4 ) / ( 3) xx ( 1 ) / ( 2sqrt 2) = ( 2 ) / ( 3 sqrt 2) ` sq units.
Area of the region BMAB = area under the circle ` 4 x ^2 + 4y ^2 = 9`, i.e., ` y = sqrt (( 9 - 4 x^2)/(4)) `i.e., ` y = ( 1 ) /(2) sqrt ( 9 - 4 x ^2)`
` = int _ ( 1//2)^(3//2) (1) /(2) sqrt ( 9 - 4 x ^2)dx = (2) /(2) int _ ( 1//2) ^( 3//2) sqrt ( ( 9 ) /(4) - x ^(2) ) dx `
` = [ ( x ) /(2) sqrt ( (9)/(4) - x ^(2) ) + ( ( 9 //4) ) /( 2) sin ^ ( - 1 ) ((x) /( 3//2))] _ ( 1//2) ^( 3//2) `
` = [ ( 3 ) /(4) sqrt ( (9)/(4) - (9) /( 4 ) ) + ( 9) / ( 8 ) sin ^ ( - 1 ) ( 1) ]- [ ( 1) / (4) sqrt ( ( 9) / (4) - (1 ) / (4)) + (9) / (8) sin ^(-1) (( 1 )/ (3)) ] `
` = 0 + (9) / (8) xx ( pi ) / ( 2) - ( 1)/ (4) * (2sqrt2) / (2 ) - (9) / ( 8) sin ^ ( - 1 ) ( (1) / (3)) `
` = [ ( 9pi ) / ( 16) - ( 1 ) /( 2sqrt2 ) - ( 9 ) / ( 8 ) sin ^(-1 ) ((1)/ (3))] ` sq units.
` therefore ` required area ` = 2[ ( 2 ) / ( 3 sqrt 2 ) + ( 9pi ) / ( 16) - ( 1 ) / ( 2 sqrt 2 ) - ( 9) / (8 ) sin ^(-1) ((1) / (3))] `
` = ( 2 sqrt 2 ) / ( 3 ) + ( 9pi ) / ( 8 ) - ( 1 )/ ( sqrt 2 ) - ( 9 ) / ( 4 ) sin ^( - 1 ) (( 1 ) / ( 3))`
` = (( 2sqrt 2 ) /(3) - ( 1 ) / ( sqrt 2) ) + ( 9 pi )/ ( 8 ) - ( 9) / ( 4 ) sin ^ ( - 1 ) (( 1) / ( 3)) `
` = [ ( 1 ) / ( 3 sqrt 2 ) + ( 9pi ) / ( 8 ) - ( 9) / ( 4 ) sin ^( - 1) ((1) / ( 3 )) ] ` sq units.
