Let's break it down step by step: ### **Given Data** - **Diameter of pool** (\(d\)) = 10 m - **Radius of pool** (\(r\)) = \( \frac{10}{2} = 5 \) m - **Depth of water** (\(h\)) = 2 m - **Gravity** (\(g\)) = 9.8 m/s² - **Density of water** (\(\rho\)) = 1000 kg/m³ - **Work required to pump all water over the side (3 m high)** --- ### **Step 1: Volume of Water** \[ \text{Volume} = \text{area of base} \times \text{depth} = \pi r^2 h \] \[ = \pi \times (5^2) \times 2 = \pi \times 25 \times 2 = 50\pi \text{ m}^3 \] --- ### **Step 2: Find the Work Required** Since the water must be pumped to the top of the sides, which is 3 m high, but the water level is only 2 m deep, the water at the bottom must be lifted 3 m, and the water at the top must be lifted 1 m (since it's already 2 m high). So, we need to find the *average* height each "layer" of water is lifted. Let \(y\) be the depth below the top (0 to 2 m). The "slice" at depth \(y\) must be lifted \(3 - y\) meters. - **Volume of a thin slice** at depth \(y\): \(dV = \pi r^2 dy = 25\pi dy\) - **Mass of thin slice**: \(dm = \rho dV = 1000 \times 25\pi dy = 25000\pi dy\) - **Work to lift this slice**: \(dW = \text{force} \times \text{distance} = dm \cdot g \cdot (3 - y)\) \[ dW = (25000\pi dy) \cdot 9.8 \cdot (3 - y) \] \[ dW = 245000\pi (3 - y) dy \] --- ### **Step 3: Integrate from \(y = 0\) (top) to \(y = 2\) (bottom)** \[ W = \int_0^2 245000\pi (3 - y) dy \] Expand and integrate: \[ W = 245000\pi \int_0^2 (3 - y) dy \] \[ = 245000\pi \left[3y - \frac{y^2}{2}\right]_0^2 \] \[ = 245000\pi \left[3(2) - \frac{(2)^2}{2} - (0 - 0)\right] \] \[ = 245000\pi [6 - 2] \] \[ = 245000\pi \times 4 \] \[ = 980000\pi \] --- ### **Step 4: Calculate the Final Answer** \[ 980000\pi \approx 980000 \times 3.1416 \approx 3,079,200 \text{ Joules} \] --- ### **Final Answer** \[ \boxed{3,079,200 \text{ Joules}} \] **(rounded to the nearest whole number, as requested)** --- #### **Summary of Steps** 1. Calculate the volume of water. 2. Set up the integral for work, considering the varying distance each layer must be lifted. 3. Integrate and evaluate. 4. Plug in values and calculate the numerical answer.

Let's go through the problem step-by-step with detailed calculations and explanations. --- ### **Given Data:** - Diameter of the pool, \(d = 10\, \text{m}\) - Radius of the pool, \(r = \frac{d}{2} = 5\, \text{m}\) - Depth of water, \(h = 2\, \text{m}\) - Height to which water must be pumped over the side, \(H = 3\, \text{m}\) - Acceleration due to gravity, \(g = 9.8\, \text{m/s}^2\) - Density of water, \(\rho = 1000\, \text{kg/m}^3\) --- ### **Step 1: Calculate the volume of water in the tank** The volume of water, \(V\), is the volume of a cylinder: \[ V = \pi r^2 h \] Substitute the values: \[ V = \pi \times (5)^2 \times 2 = \pi \times 25 \times 2 = 50 \pi\, \text{m}^3 \] --- ### **Step 2: Conceptualize the work required** - The water is at a height of 2 meters. - To pump all water over the side, which is 3 meters above the bottom, the lowest layer (at the bottom of the tank) must be lifted 3 meters. - The top layer of water (at the top of the water surface, 2 m high) needs only to be lifted **0 meters** (since it is already at the top level). - The work is done by lifting thin slices of water at different depths, each requiring different amounts of work. --- ### **Step 3: Set up the integral for work** **Coordinate system:** - Let \(y\) be the depth measured from the top surface of the water downward, with \(y=0\) at the surface and \(y=2\, \text{m}\) at the bottom. However, it's more convenient to consider the vertical position \(x\) measured from the bottom, ranging from 0 to 2 m: - **At the bottom (x=0):** water needs to be lifted \(3\, \text{m}\) (since the side is 3 m above the bottom). - **At height \(x\) from the bottom:** water needs to be lifted \(3 - x\) meters. --- ### **Step 4: Differential volume of a thin slice** - Thickness of each slice: \(dx\) - Cross-sectional area: \(\pi r^2 = 25 \pi\, \text{m}^2\) Thus, \[ dV = \text{area} \times dx = 25\pi\, dx \] - Mass of each slice: \[ dm = \rho dV = 1000 \times 25\pi\, dx = 25,000 \pi\, dx \] --- ### **Step 5: Differential work to lift each slice** - The work to lift this slice: \[ dW = \text{force} \times \text{distance} = dm \times g \times (\text{height to lift}) \] \[ dW = 25,000 \pi\, g\, (3 - x) dx \] Substitute \(g = 9.8\, \text{m/s}^2\): \[ dW = 25,000 \pi \times 9.8 \times (3 - x) dx = 245,000 \pi (3 - x) dx \] --- ### **Step 6: Integrate over the entire water depth** Integrate from \(x=0\) (bottom) to \(x=2\) (top): \[ W = \int_{0}^{2} 245,000 \pi (3 - x) dx \] --- ### **Step 7: Perform the integral** \[ W = 245,000 \pi \int_{0}^{2} (3 - x) dx \] Compute the integral: \[ \int_{0}^{2} (3 - x) dx = \left[ 3x - \frac{x^2}{2} \right]_0^{2} \] Evaluate at the bounds: \[ = \left(3 \times 2 - \frac{(2)^2}{2}\right) - (0 - 0) = (6 - 2) = 4 \] Now, multiply by the constant: \[ W = 245,000 \pi \times 4 = 980,000 \pi \] --- ### **Step 8: Numerical calculation** Using \(\pi \approx 3.1416\): \[ W \approx 980,000 \times 3.1416 = 3,078,528\, \text{J} \] **Final rounded result:** \[ \boxed{3,078,528\, \text{J}} \] --- ### **### **Final answer:** \[ \boxed{\textbf{Approximately } 3,078,528\, \text{J}} \] **This is the work required to pump all the water over the side of the tank to a height of 3 meters.**