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

Solving the curves, we get `x^(2)+4x=8xrArrx=0,4`

`"Required area ="int_(0)^(4) ""^(y)"parabola "dx+int_(4)^(8)""^(y)"circle "dx`
`"Circle is "(x-4)^(4)+y^(2)=4^(2),`
`"Area of quarter of circle "=(1)/(4)pi4^(2)=4pi`
`=int_(0)^(4)2sqrt(x) dx+4pi`
`=(4)/(3)[x^(3//2)]_(0)^(4)+4pi`
`=(4)/(3)xx4sqrt(4)+4pi" sq. units"`
`=(32)/(3)+4pi" sq. units"`