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