6. A hemispherical tank, placed so that the top is a circular region of radius 6ft. is filled with water to a depth of 4ft. Find the work done in pumping the water to the top of the tank.

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Answer 1

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 π_r∫^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]_r^R=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.