Question

Filling a spherical tank A spherical water tank with an inner radius of 8 m has its lowest point 2 m above the ground. It is filled by a pipe that feeds the tank at its lowest point (see figure). Neglecting the volume of the inflow pipe, how much work is required to fill the tank if it is initially empty?

Accepted Answer

Identify the physical setup: The spherical tank has a radius of 8 m, and its lowest point is 2 m above the ground. The tank is filled from the bottom through a pipe located at this lowest point. Set up a coordinate system: Let the vertical axis y measure height from the bottom of the tank (where the pipe is). Thus, y ranges from 0 to 16 m (since the diameter is 16 m). Express the volume of a thin horizontal slice of water at height y: The cross-sectional area of the slice is a circle with radius r(y) given by the sphere equation. Using the sphere centered at y=8, the radius of the slice is $$ r(y) = \sqrt{8^2 - (y - 8)^2} . $$The area of the slice is then $$ A(y) = \pi r(y)^2 = \pi (64 - (y - 8)^2) . $$The volume of a thin slice of thickness $$ dy  is$$ $$ dV = A(y) dy . $$Determine the work to lift the slice of water to the top of the tank: The water at height y must be lifted to the top of the tank, which is at height 16 m. The distance the slice must be lifted is $$ 16 - y . $$The weight of the slice is $$ \rho g dV $$, where $$ \rho  is $$the density of water and $$ g  is $$acceleration due to gravity. The work to lift this slice is $$ dW = \rho g (16 - y) A(y) dy . $$Set up the integral for total work: Integrate the expression for $$ dW $$ from $$ y = 0  to$$ $$ y = 16  to $$find the total work required to fill the tank: $$ W = \int_0^{16} \rho g (16 - y) \pi (64 - (y - 8)^2) dy .$$

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