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Using integration, find the area lying above x-axis and included between the circle x^{2} + y^{2} = 8x and inside the parabola y^{2} = 4x.

 

 

Answers (1)
S safeer

Given:

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

y^{2} = 4x

For intersection point

x^{2} + 4x = 8x

x^{2} = 4x

x(x -4)=0

x =0 , x = 4 

y= \pm 4

 

Required Area = Shaded zone

\\ \text { Required Area }=\int_{0}^{4} 2 \sqrt{x} \mathrm{d} x+\int_{4}^{8} \sqrt{16-(x-4)^{2}} \mathrm{d} \mathrm{x} \\

\\ =\left|\frac{4}{3} x^{\frac{3}{2}}\right|_{0} ^{4}+\left |\frac{x-4}{2} \sqrt{16-(x-4)^{2}}+8 \sin ^{-1} \frac{x-4}{4}\right|_{4} ^{8} \\

\\=\frac{32}{3}-0+4 \pi-0-0\\\\ =\frac{32}{3}+4 \pi\\

 

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