The area enclosed between the curves $$y =2x^2$$ and $$y = 6$$ is

$$y =2x^2$$ and $$y = 6$$

6 = $$2x^2$$

x = -$$\sqrt{3}$$ and $$\sqrt{3}$$

$$\int_{-\sqrt{3}}^{\sqrt{3}} 2 *x^2$$

Since x^2 is an even function

$$\int_{-\sqrt{3}}^{\sqrt{3}} 2 *x^2$$ 

= 4$$\int_{0}^{\sqrt{3}} x^2$$

= 4*$$\frac{x^3}{3}$$ where x varies from 0 to $$\sqrt{3}$$

= $$4\sqrt3$$.

Now this is the area between the parabola and the x-axis. The area between the line y=6 and x-axis is $$6\times\ 2\sqrt{\ 3}$$

Hence, the area between the curves is $$8\sqrt{\ 3}$$.