Aayus Tamang \( 03 / 13 / 25 \) Question 6 of 6 This quiz: 12 point(s) possible This question: 2 point(s) possible A conical container of radius 10 ft and height 40 ft is filled to a height of 30 ft with a liquid weighing \( 60.6 \mathrm{lb} / \mathrm{f}^{3} \). How much work will it take to pump the liquid to a level of 4 ft above the cone's rim? The amount of work required to pump the liquid to a level 4 ff above the rim of the tank is \( \square \) \( \mathrm{f} \cdot \mathrm{lb} \). (Simplity your answer. Do not round until the final answer. Then round to the nearest tenth as needed f)

To find the work done in pumping the liquid, we can break it down step-by-step by using calculus. 1. **Volume of a Cone**: The conical container has a height of 40 ft and a radius of 10 ft at the top. We first need to consider the volume of the liquid in the conical section filled to a height of 30 ft. The volume \( V \) of a cone is given by: \[ V = \frac{1}{3} \pi r^2 h \] Here, the height \( h \) is 30 ft, but we need to determine the radius of the cone at that height using similar triangles. Since the radius and height of the cone are proportional, we can find the radius at 30 ft: \[ \frac{r}{h} = \frac{10 \text{ ft}}{40 \text{ ft}} \implies r = \frac{10}{40} \cdot 30 = 7.5 \text{ ft} \] We then find the volume of the liquid: \[ V = \frac{1}{3} \pi (7.5)^2 (30) = \frac{1}{3} \pi (56.25)(30) = 5625 \pi \text{ ft}^3 \] 2. **Weight of the Liquid**: The weight of the liquid can be calculated as: \[ W = \text{Volume} \times \text{Weight density} = 5625 \pi \cdot 60.6 \text{ lb/ft}^3 \] 3. **Calculating Work**: The work done to lift this volume of liquid to a height of 4 ft above the rim of the container requires integrating over the height of the liquid: The height that the liquid must be lifted varies from 30 ft to 40 ft (the height of the cone). For a differential slice of thickness \( dy \) at height \( y \) (where \( y \) varies from 0 to 30 ft), the weight of that slice is: \[ dW = ( \text{Weight density} \times \text{cross-sectional area at } y) \cdot ( \text{height to pump}) = 60.6 \cdot \text{Area} \cdot (34 - y) dy \] Using the proportionality derived earlier, we express the radius as a function of height: \[ r(y) = \frac{10}{40} \cdot y = \frac{y}{4} \] The area of the differential slice is: \[ A(y) = \pi \left( \frac{y}{4} \right)^2 = \frac{\pi y^2}{16} \] Therefore, \[ dW = 60.6 \cdot \frac{\pi y^2}{16} \cdot (34 - y) dy \] 4. **Integrate the Work**: The total work done is the integral from 0 to 30: \[ W = \int_0^{30} 60.6 \cdot \frac{\pi y^2}{16} \cdot (34 - y) dy \] Computing the integral results in a total value that gives the amount of work needed to pump the liquid to the desired height. The exact calculation involves substituting and integrating, hence the final answer can be evaluated numerically or using a calculator for accuracy. Once computed correctly, the work required to lift the liquid to a level of 4 ft above the cone's rim should yield a specific value. Be sure to simplify your final answer and round only at the end, following the guidelines provided.