1. **State the problem:** We want to find the area enclosed between the curves defined by $$x = 2 \sqrt{y}$$ and $$y = 2 \sqrt{x}$$ which intersect each other. 2. **Rewrite the equations:** To find the points of intersection and set up the integral, express both curves in terms of $$y$$ or $$x$$. From $$x = 2 \sqrt{y}$$ we get $$\sqrt{y} = \frac{x}{2}$$ so $$y = \left(\frac{x}{2}\right)^2 = \frac{x^2}{4}$$. From $$y = 2 \sqrt{x}$$ we get $$\sqrt{x} = \frac{y}{2}$$ so $$x = \left(\frac{y}{2}\right)^2 = \frac{y^2}{4}$$. But it is simpler to use the forms: - $$y = \frac{x^2}{4}$$ (from the first) - $$y = 2 \sqrt{x}$$ 3. **Find the intersection points:** Solve $$\frac{x^2}{4} = 2 \sqrt{x}$$. Multiply both sides by 4: $$x^2 = 8 \sqrt{x}$$ Let $$t = \sqrt{x}$$, so $$x = t^2$$. Substitute: $$t^4 = 8 t$$ Divide both sides by $$t$$ (note $$t \geq 0$$ since square roots): $$t^3 = 8$$ So $$t = 2$$. Since $$t = \sqrt{x}$$, $$\sqrt{x} = 2 \Rightarrow x = 4$$. Corresponding $$y$$ value: $$y = \frac{x^2}{4} = \frac{16}{4} = 4$$. Also, the curves intersect at $$x=0, y=0$$ (origin) since both curves equal zero there. 4. **Set up the integral for the area:** Between $$x=0$$ and $$x=4$$, the upper curve is $$y=2 \sqrt{x}$$ and the lower curve is $$y=\frac{x^2}{4}$$. So area $$A$$ is: $$A = \int_0^4 \left(2 \sqrt{x} - \frac{x^2}{4}\right) dx$$. 5. **Compute the integral:** First express integrand: $$2 \sqrt{x} = 2 x^{1/2}$$, $$\frac{x^2}{4} = \frac{1}{4} x^2$$. Integral becomes: $$\int_0^4 \left(2 x^{1/2} - \frac{1}{4} x^2\right) dx = 2 \int_0^4 x^{1/2} dx - \frac{1}{4} \int_0^4 x^2 dx$$. Evaluate the integrals: $$\int_0^4 x^{1/2} dx = \left[ \frac{2}{3} x^{3/2} \right]_0^4 = \frac{2}{3} \times (4)^{3/2} = \frac{2}{3} \times 8 = \frac{16}{3}$$. $$\int_0^4 x^2 dx = \left[ \frac{x^3}{3} \right]_0^4 = \frac{64}{3}$$. So area: $$A = 2 \times \frac{16}{3} - \frac{1}{4} \times \frac{64}{3} = \frac{32}{3} - \frac{64}{12} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}$$. 6. **Interpretation:** The area enclosed between the curves is $$\frac{16}{3}$$ square units. **Final answer:** $$\boxed{\frac{16}{3}}$$