Let's solve Part 1 (Area) and Part 2 (Centroid) step by step, with explanations after each step. --- ## **Step 1: Find the Shaded Area Using Integration** The curve is given as: \[ x = \frac{4y^2}{9} \] The area is bounded by \( x = \), \( x = 4 \), \( y = \), and the curve. First, solve for \( y \) when \( x = 4 \): \[ 4 = \frac{4y^2}{9} \implies y^2 = 9 \implies y = 3 \] Set up the integral for the area, integrating **with respect to \( y \)** from \( y = \) to \( y = 3 \): \[ A = \int_{y=}^{y=3} x \, dy = \int_^3 \frac{4y^2}{9} \, dy \] \[ A = \frac{4}{9} \int_^3 y^2 \, dy = \frac{4}{9} \left[ \frac{y^3}{3} \right]_^3 \] \[ = \frac{4}{9} \cdot \left( \frac{27}{3} \right) = \frac{4}{9} \cdot 9 = 4 \] **Explanation:** We integrated the function describing the curve from the bottom (\( y= \)) to the top (\( y=3 \)), yielding the area under the curve as **4 m²**. --- ## **Step 2: Find the Centroid (\( \bar{x}, \bar{y} \))** Recall: \[ \bar{x} = \frac{1}{A} \int x \, dA \] \[ \bar{y} = \frac{1}{A} \int y \, dA \] Here, \( dA = x \, dy \) as a vertical strip. ### **Find \( \bar{x} \):** \[ \bar{x} = \frac{1}{A} \int_^3 x^2 \, dy = \frac{1}{4} \int_^3 \left( \frac{4y^2}{9} \right)^2 dy = \frac{1}{4} \int_^3 \frac{16y^4}{81} dy \] \[ = \frac{1}{4} \cdot \frac{16}{81} \int_^3 y^4 dy = \frac{4}{81} \left[ \frac{y^5}{5} \right]_^3 = \frac{4}{81} \cdot \frac{243}{5} = \frac{4 \cdot 243}{81 \cdot 5} = \frac{972}{405} = \frac{108}{45} = 2.4 \] ### **Find \( \bar{y} \):** \[ \bar{y} = \frac{1}{A} \int_^3 y x \, dy = \frac{1}{4} \int_^3 y \cdot \frac{4y^2}{9} dy = \frac{1}{4} \int_^3 \frac{4y^3}{9} dy = \frac{1}{4} \cdot \frac{4}{9} \int_^3 y^3 dy \] \[ = \frac{1}{9} \left[ \frac{y^4}{4} \right]_^3 = \frac{1}{9} \cdot \frac{81}{4} = \frac{81}{36} = 2.25 \] **Explanation:** We used the definitions of centroid coordinates, integrating \( x \) and \( y \) times the area element, and divided by the total area. --- ## **Final Answers** - **Area:** \( 4 \) m² - **Centroid:** \( (\bar{x}, \bar{y}) = (2.4\ \text{m},\ 2.25\ \text{m}) \)

# Solution to the Problem ## **Step 1: Find the Shaded Area Using Integration** The curve is given by the equation: \[ x = \frac{4y^2}{9} \] To find the area under the curve from \( y = 0 \) to \( y = 3 \), we determine where the curve intersects with \( x = 4 \): \[ 4 = \frac{4y^2}{9} \implies y^2 = 9 \implies y = 3 \] Next, we set up the integral for the area \( A \): \[ A = \int_{0}^{3} x \, dy = \int_{0}^{3} \frac{4y^2}{9} \, dy \] Calculating the integral: \[ A = \frac{4}{9} \int_{0}^{3} y^2 \, dy = \frac{4}{9} \left[ \frac{y^3}{3} \right]_{0}^{3} \] \[ = \frac{4}{9} \cdot \left( \frac{27}{3} \right) = \frac{4}{9} \cdot 9 = 4 \] **Explanation:** We integrated the function from \( y = 0 \) to \( y = 3 \) to find the area beneath the curve, resulting in an area of **4 m²**. --- ## **Step 2: Find the Centroid (\( \bar{x}, \bar{y} \))** ### **Finding \( \bar{x} \)** The formula for \( \bar{x} \) is: \[ \bar{x} = \frac{1}{A} \int x \, dA \] Here, \( dA = x \, dy \). Thus: \[ \bar{x} = \frac{1}{4} \int_{0}^{3} x^2 \, dy = \frac{1}{4} \int_{0}^{3} \left( \frac{4y^2}{9} \right)^2 dy \] \[ = \frac{1}{4} \int_{0}^{3} \frac{16y^4}{81} dy = \frac{4}{81} \int_{0}^{3} y^4 \, dy \] Calculating the integral: \[ = \frac{4}{81} \left[ \frac{y^5}{5} \right]_{0}^{3} = \frac{4}{81} \cdot \frac{243}{5} = \frac{972}{405} = \frac{108}{45} = 2.4 \] ### **Finding \( \bar{y} \)** The formula for \( \bar{y} \) is: \[ \bar{y} = \frac{1}{A} \int y \, dA \] Using \( dA = x \, dy \): \[ \bar{y} = \frac{1}{4} \int_{0}^{3} y x \, dy = \frac{1}{4} \int_{0}^{3} y \cdot \frac{4y^2}{9} dy = \frac{1}{4} \int_{0}^{3} \frac{4y^3}{9} dy \] \[ = \frac{1}{9} \left[ \frac{y^4}{4} \right]_{0}^{3} = \frac{1}{9} \cdot \frac{81}{4} = \frac{81}{36} = 2.25 \] **Explanation:** We calculated the centroid coordinates by integrating \( x \) and \( y \) multiplied by the area element, then dividing by the total area. --- ## **Final Answers** - **Area:** \( 4 \) m² - **Centroid:** \( (\bar{x}, \bar{y}) = (2.4\ \text{m},\ 2.25\ \text{m}) \)