The area enclosed between the parabolas $$y^2 = 16(1 + x )$$ and $$y^2 = 16(1 - x)$$ is
$$y^2 = 16(1 + x )$$ and $$y^2 = 16(1 - x)$$
Equating the parabolas ,we will get the point of intersection of bothÂ
16(1+x) = 16(1-x)
x=0
Substitute the value of x in the parabola $$y^2 = 16(1 + x )$$
y = $$\pm$$ 4
Point of intersection = (0,4),(0,-4)
=$$2(\int_{-4}^{4} (\frac{y^{2}-16}{16})-(\frac{16-y^{2}}{16}))$$ dy
= $$2(\int_{0}^{4} (\frac{y^{2}-16}{16})-(\frac{16-y^{2}}{16}))$$ dy           [$$\because$$ $$\int_{0}^{2a} F(x)dx$$=2$$(\int_{0}^{a} F(x)dx)$$]
=$$2(\int_{0}^{4} (\frac{y^{2}}{16})-1-1+(\frac{y^{2}}{16}))$$
=$$2(\int_{0}^{4} (\frac{y^{2}}{8})-2))$$
=2$$(\frac{y^{3}}{3*8}-2y))$$ y varies from 0-4Â
=2(8/3 - 8)
=2(16/3)Â Â ($$\because$$ Area cant be negative)
Hence C is the correct answer
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