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 both 

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 0-4 

=2(8/3 - 8)

=2(16/3)   ($$\because$$ Area cant be negative)

=32/3

Hence C is the correct answer