1. **State the problem:** Find the coordinates of the center of mass (centroid) $[\bar{x}, \bar{y}]$ of the lamina bounded by the curve $y^2 = 5x$ for $0 \leq x \leq 6$. 2. **Recall formulas:** For a lamina with uniform density, the centroid coordinates are given by: $$\bar{x} = \frac{1}{A} \int x \, dA, \quad \bar{y} = \frac{1}{A} \int y \, dA$$ where $A$ is the area of the lamina. 3. **Express area element:** Since the curve is $y^2 = 5x$, we have $y = \pm \sqrt{5x}$. The lamina is symmetric about the $x$-axis, so $\bar{y} = 0$ by symmetry. 4. **Calculate area $A$:** The area between $x=0$ and $x=6$ is $$A = \int_0^6 (\text{top } y - \text{bottom } y) \, dx = \int_0^6 (\sqrt{5x} - (-\sqrt{5x})) \, dx = 2 \int_0^6 \sqrt{5x} \, dx$$ 5. **Evaluate area integral:** $$2 \int_0^6 \sqrt{5x} \, dx = 2 \int_0^6 \sqrt{5} \sqrt{x} \, dx = 2 \sqrt{5} \int_0^6 x^{1/2} \, dx = 2 \sqrt{5} \left[ \frac{2}{3} x^{3/2} \right]_0^6 = \frac{4 \sqrt{5}}{3} (6^{3/2})$$ Calculate $6^{3/2} = (\sqrt{6})^3 = 6 \sqrt{6}$, so $$A = \frac{4 \sqrt{5}}{3} \times 6 \sqrt{6} = \frac{24 \sqrt{30}}{3} = 8 \sqrt{30}$$ 6. **Calculate $\bar{x}$:** $$\bar{x} = \frac{1}{A} \int x \, dA = \frac{1}{A} \int_0^6 x (\text{top } y - \text{bottom } y) \, dx = \frac{1}{A} \int_0^6 x (2 \sqrt{5x}) \, dx = \frac{2 \sqrt{5}}{A} \int_0^6 x^{3/2} \, dx$$ 7. **Evaluate $\bar{x}$ integral:** $$\int_0^6 x^{3/2} \, dx = \left[ \frac{2}{5} x^{5/2} \right]_0^6 = \frac{2}{5} 6^{5/2}$$ Calculate $6^{5/2} = (\sqrt{6})^5 = 36 \sqrt{6}$, so $$\bar{x} = \frac{2 \sqrt{5}}{8 \sqrt{30}} \times \frac{2}{5} \times 36 \sqrt{6} = \frac{4 \sqrt{5} \times 36 \sqrt{6}}{40 \sqrt{30}}$$ Simplify numerator and denominator: $$4 \times 36 = 144, \quad \sqrt{5} \times \sqrt{6} = \sqrt{30}$$ So $$\bar{x} = \frac{144 \sqrt{30}}{40 \sqrt{30}} = \frac{144}{40} = 3.6$$ 8. **Conclusion:** $$[\bar{x}, \bar{y}] = [3.6, 0]$$ The center of mass lies at $x=3.6$ on the $x$-axis due to symmetry.