An rectangular tank 5 m long, 2 m wide, and 8 m high is filled to a depth of 5 m of water. Set up the integral that would compute the work needed to pump the water out of a spout located at the top of the tank. Use the location of the origin and the direction of the positive axis as specified in the different parts. Note: Do not input values for the constant p9 (A) The origin is at the top of the tank with with positive y-values going down (B) The origin is at the bottom of the tank with positive y-values going up. (C) The origin is at the water level of the tank with positive y-values going down.

W = integral ( p_w*98.1*(8 - y ) ) . dy ,        limits = 0 to 8

Step-by-step explanation:

Given:

Dimension of the tank = 5 x 2 x 8

Find:

Set up the integral that would compute the work needed to pump the water out of a spout located at the top of the tank

Solution:

- Make a differential volume (slab) at a depth y with thickness dy and rest dimension the same. We calculate the differential volume as:

                                  dV = 5*2*dy m^3

- Next compute the weight of the differential volume of water:

                                    F_g = p_w*V*g

Where,

p_w : The density of water

g: gravitational constant = 9.81 m/s^2

- Hence, we have:      

                                    F_g = p_w*dV*g

                                    F_g = p_w*98.1*dy

- The work done(W) to lift the differential slab of water up:

                                    W = integral ( F_g*(8 - y ) )

- Hence, the integral of work done W is :

                                  W = integral ( p_w*98.1*(8 - y ) ) . dy

- The limits of integration are 0 - 8 m.