Richard P. answered 06/24/20
Notice that the end face of the diagram is a rectangle 3 ft wide by 2 ft high.
The origin of coordinates (x,y,z) can be chosen as the lower left corner of the rectangle.
Let z be the vertical coordinate with z = 0 at the lowest point and z = 2 at the highest point. The z axis contains the left edge of the rectangle.
Let y = the coordinate that contains the other edge of the rectangle. 0 < y < 3
The x coordinate is perpendicular to the z and y coordinates. 0 < x < 4.
Of the 5 surfaces of the volume, two are normal to the y axis, one (the top) is normal to the z axis, and one ( the end face) is normal to the x axis. At any point on the fifth surface, it can be shown ( by similar triangles) that x = 2z
The volume can be thought of as a stack of slabs of infinitesimal height dz and area = 3 x = 6 z .
The associated infinitesimal volume is 6 z dz. The (infinitesimal) weight of this infinitesimal volume is
rho 6 z dz. Here the density rho = 62.5. To exit the spout , this weight must be raised up by 2 -z (feet).
This means that the total work is the integral of (2-z) rho 6 z dz, with limits of 0 and 2.
The indefinite integral is easy to carry out = rho (6 z^2 – 2 z^3 )
There is no contribution from the lower limit so the integral is rho (24 -16)
Which works out to be 500 lb-ft