10. a right cylindrical tank is filled with sea water. the tank has a radius of 2 feet and a height of 8…

Explanation:

Step1: Set up the integral for work

The formula for work (W=\int_{a}^{b}F(x)dx), where (F(x)) is the force and (dx) is the distance. For a cylindrical tank, we use the disk method. The volume of a thin disk of water at height (y) (from the bottom of the tank) with thickness (\Delta y) is (V = \pi r^{2}\Delta y). Here (r = 2), so (V=\pi(2)^{2}\Delta y=4\pi\Delta y). The force (F) (weight) of the disk is (F=\text{density}\times V). Given density (\rho=64\ lb/ft^{3}), so (F = 64\times4\pi\Delta y=256\pi\Delta y). The distance the disk of water at height (y) needs to be pumped is (d=(8 - y)) (since the total height of the tank is (8) feet).

Step2: Write the integral

We integrate from (y = 0) (bottom of the tank) to (y=8) (top of the tank). The work (W=\int_{0}^{8}256\pi(8 - y)dy).

Step3: Evaluate the integral

First, expand the integrand: (256\pi(8 - y)=2048\pi-256\pi y). Then (\int_{0}^{8}(2048\pi-256\pi y)dy=\int_{0}^{8}2048\pi dy-\int_{0}^{8}256\pi y dy).

• For (\int_{0}^{8}2048\pi dy), using the power rule (\int k dy=ky + C) ((k = 2048\pi)), we get (2048\pi y\big|_{0}^{8}=2048\pi\times(8 - 0)=16384\pi).
• For (\int_{0}^{8}256\pi y dy), using the power rule (\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C) ((n = 1)), we get (256\pi\times\frac{y^{2}}{2}\big|{0}^{8}=128\pi y^{2}\big|{0}^{8}=128\pi\times(8^{2}-0^{2})=128\pi\times64 = 8192\pi).

Then (W=16384\pi-8192\pi).

Step4: Simplify the result

(W = 8192\pi\approx8192\times3.14 = 25722.88) lb - ft.

Answer:

(25722.88) lb - ft