# 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#?

##### 1 Answer

The height data provided is confusing: Is the oil level 2 ft high or 2 ft from the rim? The answer assumes y= 2ft high.

#### Explanation:

Work is defined as:

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

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

where

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

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: