A right circular cone (with the vertex at the bottom) has a height of 6m, and aradius at the base of 3m. The cone is full of water. Recall that the density of water isp = 1000 kg/m³ and gravity is g = 9.8 m/s².Set up an integral which computes the amount of work performed in pumping waterover the top of the cone until 2m of water remain in the cone. Show a diagram as partof your work. Identify on your diagram: the y-coordinate at the top of the cone, the y-coordinate at the bottom of the cone, and the volume and thickness of an arbitrary"slice" of water.

Aim: To find the work done in pumping out water from a cone-shaped container (schematic diagram shown below). The initial height, h = 6m The final height, h = 2m The base radius,r = 3m A pump is just at the top of the container. In generalWork Done, W (in Joules)W = mghwherem = mass of the fluid to be pumped out (kg)g = acceleration due to gravity (m/s2) h = height to which the fluid to be lifted from its present height(m) Specific to given problemConsider a strip of water of thickness 'dh' at height 'h' from the bottom point. The volume 'dV' of this cylindrical strip of radius 'r' and height 'dh'. dV = πr2dh (1)Calculate 'r' in terms of h. △ABC and △ADE are similar (AA similarity rule)Hence, the ratio of corresponding sides is also equal.⇒BCAC = DEAErh = 36 r = h2 (2)HencedV = π4h2dh (3)dV = Volume of the strip = π4h2dh (m3)ρ = density of the fluid = 1000 (kgm3)dm = mass of the strip (kg) = ρ*dVdm = 1000π4h2dh (4)The fluid strip is to be lifted from 'h'm to '6'm.So the net height to be lifted is = (6-h)mdW = work done in carrying up this strip of mass dm to height (6-h)m dW = dm*g*(6-h)dW = 1000π4h2dh *g((6-h)dW = 250πg(6h-h2)dh (5)W = Total work done to empty the the tank upto depth of 2m.W = ∫dW = ∫62250πg6h-h2dhW = 250πg3h2-h3362 (Given g = 9.8m/s2)W = 250*9.8π322-62-1323-63W = 250*9.8π-803W =-205250.72J (-ve sign shows that the work is done opposite to the force of gravity )So the work done in pumping out water=|W|=205250.72J