1. **State the problem:** Find the coordinates of the centroid $(\bar{x}, \bar{y})$ of the shaded region bounded by the curves $f_1(x) = (x-4)^2$ and $f_2(x) = 4$ between $x=2$ and $x=6$. 2. **Recall formulas:** - The area (mass) $M$ of the region is given by $$M = \int_2^6 \bigl(f_2(x) - f_1(x)\bigr) \, dx = \int_2^6 \bigl(4 - (x-4)^2\bigr) \, dx$$ - The moments about the $x$-axis and $y$-axis are $$M_x = \int_2^6 \frac{1}{2} \bigl(f_2(x)^2 - f_1(x)^2\bigr) \, dx$$ $$M_y = \int_2^6 x \bigl(f_2(x) - f_1(x)\bigr) \, dx$$ - The centroid coordinates are $$\bar{x} = \frac{M_y}{M}, \quad \bar{y} = \frac{M_x}{M}$$ 3. **Given values:** - $M = \int_2^6 (4 - (x-4)^2) dx = \frac{32}{3}$ - $M_x = \int_2^6 \frac{1}{2} \bigl(16 - ((x-4)^2)^2\bigr) dx = 25.6$ - $M_y = \int_2^6 x (4 - (x-4)^2) dx = \frac{128}{3}$ 4. **Calculate centroid coordinates:** $$\bar{x} = \frac{M_y}{M} = \frac{\frac{128}{3}}{\frac{32}{3}} = \frac{128}{3} \times \frac{3}{32} = 4$$ $$\bar{y} = \frac{M_x}{M} = \frac{25.6}{\frac{32}{3}} = 25.6 \times \frac{3}{32} = 2.4$$ 5. **Interpretation:** The centroid is at $(4, 2.4)$ meters. **Final answer:** $$\boxed{(\bar{x}, \bar{y}) = (4.0, 2.4)}$$