We need to get to the mathematical application behind this problem.
Work=force times distance. The distance d is clearly 2 feet because the depth of the water is already 4 feet and the radius of the tank is 6 feet. So what is the force? Force, F=mass times acceleration, and in this case the acceleration is gravity g, so F=mg downwards. Therefore the work required=mgd will involve the force needed to overcome gravity. What about the mass? ρ=m/v, where m is mass and, because the liquid is water, the density ρ is 1g/cm3=1g/mL=1kg/L, so m=v.
Unfortunately, this means we have to convert feet into cms or metres. 2 feet=60.96cm=0.6096m approximately.
So we need to find the volume of water between 4ft and 6ft in the hemispherical tank.
x2+(y-R)2=R2, x2=R2-(y-R)2, where R is the radius of the tank, is a circle centre (0,R), the lower half representing the cross-section of the tank. y=R is a line representing the top of the tank, and y=r is a line representing the top surface of the water already in the tank. These lines plus the circle are the boundaries of the volume of revolution about the y-axis. So πx2dy is the volume of a disc of radius x, thickness dy. The limits of integration are r and R for y, so πr∫R(R2-(y-R2))dy should give the volume of water to be added to the tank. R2-(y-R2)=(R-y+R)(R+y-R)=2Ry-y2.
The result of integrating is π[Ry2-y3/3]rR=π[(R3-R3/3)-(r3-r3/3)=⅔π(R3-r3) (a result which could be deduced from geometry). m=⅔π(R3-r3), and W, the work done=⅔π(R3-r3)g(R-r).
R=182.88cm=1.8288m, r=121.92cm=1.2192, so d=R-r=60.96cm=0.6096m; g=9.81m/s2.
W=53.91 Newton-metres approx.