Let's break down the problem step by step: --- ### **Given:** - The reservoir is a right prism with triangular bases: - Height: \( 44\,\text{m} \) - Width: \( 66\,\text{m} \) - Length: \( 33\,\text{m} \) - Water needs to be pumped to a height \( 11\,\text{m} \) **above the reservoir**. - Density of water, \( \rho = 1000\,\text{kg/m}^3 \) - \( g = 9.8\,\text{m/s}^2 \) --- ### **(i) Express \( P(x) \) (Weight of thin water layer at height \( x \))** Let \( x \) be the height from the base. - The cross-sectional area of the reservoir (base) is a right triangle: \[ \text{Area}_{\text{triangle}} = \frac{1}{2} \times 66 \times 44 = 1452\,\text{m}^2 \] - The length of the prism is \( 33\,\text{m} \), so the volume of a thin layer of thickness \( \Delta x \) is: \[ \text{Volume} = \text{Area}_{\text{base}} \times \Delta x = 1452 \times 33 \times \Delta x \] But since it's a prism, the length is already included in the base area calculation if you take the total volume. For a thin horizontal slice at height \( x \), the area of the slice is the same as the base area (since the prism has uniform cross-section): \[ \text{Volume of slice} = (\text{Area of base}) \cdot \Delta x = (0.5 \times 66 \times 33) \cdot \Delta x = 1089\,\Delta x \] Correction: The height of the triangle is 44 m, the width is 66 m, and the length (depth into the page) is 33 m. So, for a slice at height \( x \), the area is \( 0.5 \times 66 \times 44 \) and for the whole depth, you multiply by 33. But for a thin layer at height \( x \), the volume is: \[ \text{Volume} = (\text{Area of base}) \cdot \Delta x = (0.5 \times 66 \times 44) \cdot 33 \cdot \Delta x \] Actually, the correct way is: the cross-sectional area is \( 0.5 \times \text{width} \times \text{height} \), and the prism is 33 m long, so the total volume is \( 0.5 \times 66 \times 44 \times 33 \). However, for a layer at height \( x \), the area is \( \frac{66}{44} x \) (since the triangle base and height are proportional). The length is 33 m. So at height \( x \), the width is \( \frac{66}{44} x \), so the area at height \( x \) is: \[ \text{Area}(x) = \frac{1}{2} \cdot \frac{66}{44} x \cdot x = \frac{33}{44} x^2 \] The volume of a thin slab at height \( x \) is: \[ dV = \text{Area}(x) \cdot 33 \cdot dx = \frac{33}{44} x^2 \cdot 33 \cdot dx = \frac{1089}{44} x^2 dx \] The weight of this thin layer is: \[ P(x) = \rho g \, dV = 1000 \cdot 9.8 \cdot \frac{1089}{44} x^2 dx \] **Final formula:** \[ \boxed{P(x) = 1000 \cdot 9.8 \cdot \frac{1089}{44} x^2} \] where \( x \) is in meters, and this gives the weight in Newtons per \( dx \). --- ### **(ii) Work to pump a thin layer to 11 m above the reservoir** The work to move this thin layer to 11 m above the reservoir is: \[ \text{Distance lifted} = 44 + 11 - x = 55 - x \] So: \[ w(x) = P(x) \cdot (55 - x) \] **Final formula:** \[ \boxed{w(x) = 1000 \cdot 9.8 \cdot \frac{1089}{44} x^2 (55-x)} \] --- ### **(iii) Total work to pump all water to 11 m above the reservoir** Integrate \( w(x) \) from \( x = 0 \) to \( x = 44 \) (entire height of the reservoir): \[ W = \int_0^{44} 1000 \cdot 9.8 \cdot \frac{1089}{44} x^2 (55-x) dx \] Let's compute the constants: \[ 1000 \cdot 9.8 \cdot \frac{1089}{44} = 1000 \cdot 9.8 \cdot 24.75 \approx 242,550 \] So, \[ W = 242{,}550 \int_0^{44} x^2 (55-x) dx \] \[ = 242{,}550 \int_0^{44} (55x^2 - x^3) dx \] \[ = 242{,}550 \left[ \frac{55}{3} x^3 - \frac{1}{4} x^4 \right]_0^{44} \] \[ = 242{,}550 \left( \frac{55}{3} \cdot 44^3 - \frac{1}{4} \cdot 44^4 \right) \] Calculate \( 44^3 \) and \( 44^4 \): - \( 44^3 = 44 \times 44 \times 44 = 85,184 \) - \( 44^4 = 44 \times 85,184 = 3,748,096 \) Plug these values in: \[ = 242{,}550 \left( \frac{55}{3} \cdot 85,184 - \frac{1}{4} \cdot 3,748,096 \right) \] \[ = 242{,}550 \left( 1,563,374.67 - 937,024 \right) \] \[ = 242{,}550 \times 626,350.67 \] \[ = 151,943,931,868.5\, \text{J} \] \[ \approx \boxed{1.52 \times 10^{11}\, \text{J}} \] --- ### **Summary of Answers:** #### (i) \[ \boxed{P(x) = 1000 \cdot 9.8 \cdot \frac{1089}{44} x^2} \] #### (ii) \[ \boxed{w(x) = 1000 \cdot 9.8 \cdot \frac{1089}{44} x^2 (55-x)} \] #### (iii) \[ \boxed{1.52 \times 10^{11} \text{ J}} \] Let me know if you need further clarification or breakdown!

The total work to pump water from a reservoir involves integrating the weight of each water layer multiplied by the distance it must be lifted to reach the outlet.