A circular swimming pool has a diameter of 12 meters. The sides are 3 meters high and the depth of the water is 2.5 meters. How much work is required to pump all of the water over the side?
Let the x-axis point upward, with x=0 at ground level. Consider a thin horizontal increment of water, of thickness dx, located at some x (0 ≤ x ≤ 2.5). This water has mass dm given by dm = ρA dx where ρ is the mass density of water (in kg/m?) and A is the area of this increment (A = 36π for all increments here). This increment must be lifted (3-x) meters. The increment of work dW is given by dW = gh dm where g is the acceleration due to gravity and h is the distance that the increment of mass must be lifted (which we've seen is equal to 3-x meters). The total work W is given by 2.5 ?∫ g (3-x) 36π ρ dx 0
the quantity of the water interior the trough is, V(h) the place h is the peak of the water interior the trough. The one million/2 perspective on the backside of the trough is B tan B = h /(w/2) the place w is the area around the trough tan B = 4 /(12/2) = 4/6 = 2/3 2/3 = h /(w/2) (w/2) 2/3 = h w = 3 h the component to the bypass-component to the trough A, is then A = h w /2 = h(3h)/2 = 3/2 h^2 the quantity is then V(h) = A * L = (3/2 h^2) 12 = 18 h^2 dV/dt = 36 h dh/dt dh/dt = dV/dt /(36 h) dV/dt = 9 ft^3/min dh/dt = 9 /(36 * 12) = one million/forty 8 ft /min
