Find the area lying above x-axis and included between the circle `x^(2) + y^(2) = 8x` and inside in the parabola `y^(2) = 4x.`

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"`