Find the area lying above x-axis and included between the circle $x^{2} + y^{2} = 8 x$ and inside of the parabola $y^{2} = 4 x .$ is $m π + \frac{32}{3}$ .Find the value of $m$

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Solution

$= 2 \int_{0}^{1} \sqrt{2} d x + \int_{4}^{8} \sqrt{8 x - x^{2}} d x$
$= \frac{32}{2} + 4 π$
Hence $m = 4$

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