1. **State the problem:** We have a circular gate 125 cm in diameter immersed in water. The vertical distance from the top of the gate to the water surface is 60 cm, and from the bottom to the surface is 150 cm. We want to find the total force due to water on one side of the gate. The density of water is 1000 kg/m³. 2. **Convert all lengths to meters:** Diameter $d = 125$ cm = $1.25$ m Top depth $h_{top} = 60$ cm = $0.6$ m Bottom depth $h_{bottom} = 150$ cm = $1.5$ m 3. **Find the radius and the coordinates for integration:** Radius $r = \frac{d}{2} = 0.625$ m We set vertical coordinate $y$ with $y=0$ at top of the gate going downward. So the gate extends from $y=0.6$ m to $y=1.5$ m below water surface. Water pressure at depth $y$ is $p = \rho g y$ where: $\rho = 1000$ kg/m³ (density) $g = 9.8$ m/s² (gravity) 4. **Express width of the gate at depth $y$:** Since the gate is circular, circle centered at $y=r + 0.6=1.225$ m vertical from surface (middle of gate) Width at depth $y$ is twice the horizontal half-width: $w(y) = 2\sqrt{r^{2} - (y - 1.225)^2}$ 5. **Total hydrostatic force:** The differential force $dF = p(y) \times w(y) dy = \rho g y \times w(y) dy$ Total force: $$F = \int_{0.6}^{1.5} \rho g y \cdot 2 \sqrt{r^{2} - (y-1.225)^2} dy$$ 6. **Numerical evaluation:** Given complexity, standard formula for force on vertical curved surface: $$F = \rho g A \bar{h}$$ where $A$ is area of gate and $\bar{h}$ is depth to center of pressure. Area of circle: $$A = \pi r^2 = \pi \times 0.625^{2} = 1.227$ m$^{2}$$ Depth to centroid $\bar{h} = 1.05$ m (average of 0.6 and 1.5) 7. **Calculate total force:** $$F = 1000 \times 9.8 \times 1.227 \times 1.05 = 12628.65 \text{ N}$$ **Answer:** The total force due to the water on one side of the gate is approximately **12629 N**.