How much work does it take to pump the oil to the rim of the tank if the conical tank from #y=2x# is filled to within 2 feet of the top of the olive oil weighing #57(lb)/(ft)^3#?

Work is defined as: #W = intvecf*dvecs#

The minimal work done to lift a fluid particle with mass #deltam# from the bottom of the cone to height y is:

#deltaW = int_0^y(deltam) g dy = deltam g y#

Hence the work to bring a layer of fluid at level y in the cone should be:

#deltaW =deltam g y=(rhodv)gy=rhogy(pix^2)dy#
where
#rho=# density of the fluid; #dv=Ady = pix^2dy #
#x =y/2 # cross-section radius of a fluid layer in the cone

#W= int_0^y 1/4pirhogy^3dy = 1/16pirhog(y^4)_0^y=1/16pirhogy^4 #

Let y = 2ft fills when the fluid fills the cone up to the rim

#W= 1/16pirhog (2)^4 =pirhog#

#W = pi (57 (lb)/(ft^3))(32 (ft)/sec^2)(ft^4)=5730 lb*ft#

Basically the work is equal to potential energy of the cone filled with oil. Applying the potential energy concept (mgh) yields the same result.

Note: the data on the fluid level is confusing. Is the fluid 2 ft high or 2 ft from the rim? If the later is the case, the height of the cone is not given. The solution if assumes the general form is:

#W= int_(y_1)^(y_2) 1/4pirhogy^3dy = 1/16pirhog(y^4)_(y_1)^(y_2)=1/16pirhog(y_2^4-y_1^4) #