Find the area, lying above the x=axis and included between the circle `x^2+y^2=8x` and the parabola `y^2=4xdot`

Explanation

The centre of the circle ` x^2 + y ^2 = 8x `.
i.e., ` ( x - 4 ) ^2 + y ^ 2 = 1 6 ` is at ` A ( 4, 0 ) ` and radius is 4.
The vertex of the parabola ` y^ 2 = 4 x ` is at the origin ` O ( 0, 0 ) `
We have to find the area in the first quadrant.
` therefore x ge 0 and y ge 0 ` .
To find the points of intersection of the circle and the parabola.
Put ` y^2 = 4x ` in ` x ^2 + y ^2= 8 x `, we get,
` x ^2 + 4x = 8 x " " therefore x ^2 - 4 x = 0 `
` therefore x (x - 4 ) = 0 `
` therefore x = 0 , x = 4 `

When ` x = 0, y = 0 `
When ` x = 4, y ^2 = 16 " " therefore y = 4 `
as point is in the first quadrant.
` therefore ` points of intersection are `O(0, 0) and B(4, 4)`.
Required area = area of the region ODBCO
= ( area of the region OABCO) - ( area of the region OABDO)
Now, area of the region OABCO = area under of the circle
` = int _ 0 ^ 4 y dx `, where ` x ^2 + y ^ 2 = 8x ` i.e., ` y = sqrt ( 8 x - x ^2) `
` = int _ 0 ^4 sqrt ( 8 x - x ^2 ) dx `
` = int_ 0 ^4 sqrt ( 16 - ( x ^2 - 8 x + 16 ) ) dx `
` = int _ 0 ^4 sqrt ( 4 ^ 2 - ( x - 4 ) ^2) dx`
` [ ((x - 4 ) ) /( 2 ) sqrt ( 16 - ( x - 4 ) ^2) + ( 16) /(2) sin ^ ( - 1 ) ((x - 4) /( 4)) ] _ 0 ^4 `
` = [ 0 + 8 sin ^ ( -1 ) 0 ] - [ ( - 4) /(2) sqrt ( 16 - 16 ) + 8 sin ^ ( - 1 ) ( - 1 ) ] `
` = 0 + 0 + 0 - 8 (- ( pi ) /(2)) = 4 pi `
Area of the region OABDO
= area under the parabola
` = int _ 0 ^ 4 y dx,` where ` y ^ 2 = 4x , `i.e., ` y = 2 sqrt x `
` = int_ 0 ^ 4 2 sqrt x dx = 2 int_ 0 ^ 4 x ^ ((1) /(2)) dx `
` = 2 [ (x ^((3)/(2))) / ( 3//2 ) ] _ 0^4 = ( 4 ) /( 3 ) [ x ^((3)/(2))] _ 0 ^ 4 `
` = ( 4 ) / ( 3 ) [8 -0] = ( 32 )/ ( 3 ) `
` therefore ` required area ` = ( 4pi - ( 32 ) / ( 3 ) ) ` sq units.