Find the area between the region of parabola $y^{2} = 2 x$ and circle $x^{2} + y^{2} = 8$ .

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Solution

$y^{2} = 2 x, x^{2} + y^{2} = 8$

Area $= 2 [\int_{0}^{1} \sqrt{2 x} d x + \int_{1}^{2 \sqrt{2}} \sqrt{8 - x^{2}} d x]$

$= 2 ⎧ ⎪ ⎨ ⎪ ⎩ \frac{2 \sqrt{2}}{3} {[x^{\frac{3}{2}}]}_{0}^{\frac{3}{2}} + {[\frac{x}{2} \sqrt{8 - x^{2}} + \frac{8}{2} {sin}^{- 1} (\frac{x}{2 \sqrt{2}})]}_{1}^{- 1} ⎫ ⎪ ⎬ ⎪ ⎭$

$= 2 {[\frac{2 \sqrt{2}}{3} - 0] + [\frac{2 \sqrt{2}}{2} \sqrt{8 - 8} + 4 {sin}^{- 1} (\frac{2 \sqrt{2}}{2 \sqrt{2}}) - \frac{1}{2} \sqrt{8 - 1} + \frac{8}{2} {sin}^{- 1} (\frac{1}{2 \sqrt{2}})]}$

$= 2 [\frac{2 \sqrt{2}}{3} + 4 {sin}^{- 1} 1 - \frac{\sqrt{7}}{2} + 4 {sin}^{- 1} (\frac{1}{2 \sqrt{2}})]$

$= 2 [\frac{2 \sqrt{2}}{3} + 2 π - \frac{\sqrt{7}}{2} + 4 {sin}^{- 1} (\frac{1}{2 \sqrt{2}})]$ sq. units

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