Let's solve the problem step by step: **Problem recap:** A tank is full of water. Dimensions: - Height = 4 m - Width = 18 m - Depth (front to back) = 4 m - Water is to be pumped out of a spout at the top. - Use \(g = 9.8 \, \text{m/s}^2\) - Water density (\(\rho\)) = \(1000 \, \text{kg/m}^3\) --- ## Step 1: **Visualize the Problem** The tank is a rectangular prism, standing 4 m tall, 18 m wide, and 4 m deep, filled to the top with water. We want to pump all the water to the top of the tank (the spout). --- ## Step 2: **Set up the Work Integral** The work required to lift a small slab of water (\(dV\)) at depth \(y\) to the top: - **Distance to lift:** \(4-y\) meters (since y=0 at the bottom, y=4 at the top) - **Volume of thin slab:** \(dV = 18 \times 4 \times dy = 72 \, dy\) - **Mass of slab:** \(dm = \rho \, dV = 1000 \times 72 \, dy = 72000 \, dy\) - **Weight of slab:** \(dW = dm \times g = 72000 \times 9.8 \, dy = 705600 \, dy\) - **Work to lift it:** \(d\text{Work} = \text{weight} \times \text{distance to lift} = 705600 (4-y) \, dy\) --- ## Step 3: **Set up and Evaluate the Integral** Integrate from \(y=0\) (bottom) to \(y=4\) (top): \[ W = \int_{y=0}^{4} 705600 (4-y) \, dy \] Expand: \[ W = 705600 \int_{0}^{4} (4-y) \, dy \] \[ = 705600 \left[ 4y - \frac{y^2}{2} \right]_{0}^{4} \] Calculate at bounds: - When \(y=4\): \(4y - \frac{y^2}{2} = 16 - 8 = 8\) - When \(y=0\): \(0\) So, \[ W = 705600 \times (8 - 0) = 5,644,800 \text{ Joules} \] --- ## **Final Answer** \[ \boxed{5,644,800 \text{ Joules}} \] ### **Explanation Recap** - Set up a variable \(y\) for depth. - Found the volume, mass, and weight of a thin slab. - Distance each slab must be lifted = \(4-y\). - Integrated over the height of the tank. - **Final work required:** \(\boxed{5,644,800 \text{ Joules}}\)