1. **Problem statement:** Find the centroid of the area bounded by the curves $x=2\sqrt{y}$ and $y=2\sqrt{x}$.\n\n2. **Rewrite curves in terms of \(y\):**\nFrom $x = 2\sqrt{y}$, square both sides: $$x^2 = 4y \implies y = \frac{x^2}{4}.$$\nFrom $y = 2\sqrt{x}$, square both sides: $$y^2 = 4x \implies x = \frac{y^2}{4}.$$ Alternatively, $y=2\sqrt{x} \implies y^2=4x$. In terms of $y$, $x = \frac{y^2}{4}$.\n\n3. **Find intersection points:**\nSet $y = \frac{x^2}{4}$ and $y = 2\sqrt{x}$. Equate: $$\frac{x^2}{4} = 2\sqrt{x}.$$\nMultiply both sides by 4: $$x^2 = 8\sqrt{x}.$$\nLet $t = \sqrt{x} \Rightarrow x = t^2$, rewrite as: $$t^4 = 8t \implies t^4 - 8t = 0 \implies t(t^3 - 8) = 0.$$\nSo, $t=0$ or $t^3=8 \Rightarrow t=2$.\nThus, $x=0$ or $x=4$.\nCompute corresponding $y$: at $x=0$, $y = \frac{0^2}{4} = 0$; at $x=4$, $y = \frac{4^2}{4} = 4$.\nIntersections are $(0,0)$ and $(4,4)$.\n\n4. **Set up the area and centroid integrals:**\nArea $A$ between curves from $x=0$ to $x=4$ is:\n$$A = \int_0^4 \left(y_{upper} - y_{lower}\right) dx.$$\nHere, $y_{upper} = 2\sqrt{x}$ and $y_{lower} = \frac{x^2}{4}$.\nSo, $$A = \int_0^4 \left(2\sqrt{x} - \frac{x^2}{4}\right) dx.$$\n\n5. **Calculate area $A$:$$\n\int_0^4 2x^{1/2} dx = 2 \cdot \frac{2}{3} x^{3/2} \Big|_0^4 = \frac{4}{3} (4)^{3/2} = \frac{4}{3} \cdot 8 = \frac{32}{3}.$$ \n$$\int_0^4 \frac{x^2}{4} dx = \frac{1}{4} \cdot \frac{x^3}{3}\Big|_0^4 = \frac{1}{4} \cdot \frac{64}{3} = \frac{16}{3}.$$\n\nTherefore, $$A = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}.$$\n\n6. **Calculate centroid coordinates $(\bar{x},\bar{y})$ formulas:**\n$$\bar{x} = \frac{1}{A} \int_0^4 x \left(y_{upper} - y_{lower}\right) dx, \quad \bar{y} = \frac{1}{2A} \int_0^4 \left(y_{upper}^2 - y_{lower}^2\right) dx.$$\n\n7. **Compute $\bar{x}$:**\n$$\int_0^4 x \left(2\sqrt{x} - \frac{x^2}{4}\right) dx = \int_0^4 \left(2x x^{1/2} - \frac{x^3}{4}\right) dx = \int_0^4 \left(2x^{3/2} - \frac{x^3}{4}\right) dx.$$\n\nCalculate integrals:\n$$\int_0^4 2x^{3/2} dx = 2 \cdot \frac{2}{5} x^{5/2} \Big|_0^4 = \frac{4}{5} (4)^{5/2} = \frac{4}{5} \cdot 32 = \frac{128}{5}.$$\n$$\int_0^4 \frac{x^3}{4} dx = \frac{1}{4} \cdot \frac{x^4}{4} \Big|_0^4 = \frac{1}{4} \cdot 256/4 = 16.$$\n\nTherefore, $$\int_0^4 x(y_{upper} - y_{lower}) dx = \frac{128}{5} - 16 = \frac{128}{5} - \frac{80}{5} = \frac{48}{5}.$$\n\nFinally, $$\bar{x} = \frac{1}{A} \cdot \frac{48}{5} = \frac{3}{16} \cdot \frac{48}{5} = \frac{144}{80} = \frac{9}{5} = 1.8.$$\n\n8. **Compute $\bar{y}$:**\nCalculate:\n$$y_{upper}^2 = (2\sqrt{x})^2 = 4x,$$\n$$y_{lower}^2 = \left(\frac{x^2}{4}\right)^2 = \frac{x^4}{16}.$$\n\nIntegral for $\bar{y}$ numerator:\n$$\int_0^4 (4x - \frac{x^4}{16}) dx = \int_0^4 4x dx - \int_0^4 \frac{x^4}{16} dx = 2x^2 \Big|_0^4 - \frac{1}{16} \cdot \frac{x^5}{5} \Big|_0^4.$$\n\nCalculate:\n$$2 \cdot 16 - \frac{1}{16} \cdot \frac{1024}{5} = 32 - \frac{1024}{80} = 32 - 12.8 = 19.2.$$\n\nDivide by $2A$:\n$$\bar{y} = \frac{19.2}{2 \cdot \frac{16}{3}} = \frac{19.2}{\frac{32}{3}} = 19.2 \cdot \frac{3}{32} = \frac{57.6}{32} = 1.8.$$\n\n**Answer:** The centroid is at $\boxed{\left(\frac{9}{5}, \frac{9}{5}\right)}$ or $(1.8, 1.8)$.