1. Problem statement: Find the centroid of the planar region bounded by $x=2\sqrt{y}$ and $y=2\sqrt{x}$, which meet at $(0,0)$ and $(4,4)$.\n2. Convert to functions of $x$: $x=2\sqrt{y}$ gives $y=\frac{x^{2}}{4}$, and the other curve is $y=2\sqrt{x}$. We use $y_{l}=\frac{x^{2}}{4}$ as the lower curve and $y_{u}=2\sqrt{x}$ as the upper curve for $0\le x\le4$.\n3. Compute the area $A$ of the region: \n$$A=\int_{0}^{4}\left(2\sqrt{x}-\frac{x^{2}}{4}\right)\,dx$$\nIntegrate and evaluate: \n$$A=\left[\frac{4}{3}x^{3/2}-\frac{x^{3}}{12}\right]_{0}^{4}=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}$$\n4. Compute the centroid $\bar{x}$: \n$$\bar{x}=\frac{1}{A}\int_{0}^{4}x\left(2\sqrt{x}-\frac{x^{2}}{4}\right)\,dx$$\nEvaluate the integral: \n$$\int_{0}^{4}\left(2x^{3/2}-\frac{x^{3}}{4}\right)dx=\left[\frac{4}{5}x^{5/2}-\frac{x^{4}}{16}\right]_{0}^{4}=\frac{128}{5}-16=\frac{48}{5}$$\nThus \n$$\bar{x}=\frac{1}{A}\cdot\frac{48}{5}=\frac{3}{16}\cdot\frac{48}{5}=\frac{9}{5}$$\n5. Compute the centroid $\bar{y}$: \n$$\bar{y}=\frac{1}{2A}\int_{0}^{4}\left(y_{u}^{2}-y_{l}^{2}\right)dx$$\nSubstitute $y_{u}^{2}=4x$ and $y_{l}^{2}=\frac{x^{4}}{16}$ and integrate: \n$$\int_{0}^{4}\left(4x-\frac{x^{4}}{16}\right)dx=\left[2x^{2}-\frac{x^{5}}{80}\right]_{0}^{4}=32-\frac{1024}{80}=\frac{96}{5}$$\nTherefore \n$$\bar{y}=\frac{1}{2A}\cdot\frac{96}{5}=\frac{3}{32}\cdot\frac{96}{5}=\frac{9}{5}$$\n6. Final answer: The centroid is at $(\frac{9}{5},\frac{9}{5})$ which is $(1.8,1.8)$.\n