# 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#?

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.

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

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:

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To calculate the work required to pump the oil to the rim of the tank, we first need to determine the volume of oil in the tank. We can use the formula for the volume of a cone, which is V = (1/3)πr^2h, where V is the volume, r is the radius of the base, and h is the height of the cone.

Given that the tank follows the equation y = 2x and is filled to within 2 feet of the top, we can set up an integral to find the volume. The limits of integration for x will be from 0 to the radius of the tank base.

After finding the volume, we can multiply it by the density of the olive oil to find the total weight of the oil. This weight represents the work required to pump the oil to the rim of the tank.

Finally, we can use the formula for work, which is W = Fd, where W is the work, F is the force required to pump the oil, and d is the distance over which the force is applied. In this case, d would be the height of the tank.

By plugging in the calculated weight of the oil and the height of the tank, we can find the work required to pump the oil to the rim of the tank.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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