Question

Find the work done in pumping all the oil (density \(\delta=50\) pounds per cubic foot) over the edge of a cylindrical tank that stands on one of its bases. Assume that the radius of the base is 4 feet, the height is 10 feet, and the tank is full of oil.

Short answer

The work done is \( 40000\pi \) foot-pounds.

Step-by-step solution

Identify the Problem

We need to compute the work done to pump all the oil out of a cylindrical tank over its edge.

Set Up the Situation

The tank is a cylinder, with a radius of 4 feet and height of 10 feet. The oil has a density of \(\delta = 50\) pounds per cubic foot.

Calculate Volume of a Slice

Consider a thin horizontal slice of oil at a height \(y\) from the bottom with a thickness \(dy\). The volume of this slice is \( \pi \times 4^2 \times dy = 16\pi \ dy \) cubic feet.

Calculate Weight of the Slice

The weight of this oil slice is its volume multiplied by its density: \( 16\pi \delta \ dy = 800\pi \ dy \) pounds.

Calculate Work for a Slice

The work required to lift this slice to the top of the tank is the weight of the slice times the distance it needs to be lifted, which is \(10-y\). Therefore, the work for this slice is \( 800\pi \ (10-y) \ dy \).

Integrate to Find Total Work

Integrate from \(y = 0\) to \(y = 10\):\[ \text{Work} = \int_0^{10} 800\pi (10-y)\, dy = 800\pi \left[ 10y - \frac{y^2}{2} \right]_0^{10} \]

Compute the Integral

Evaluate the integral:\[ = 800\pi \left( 10(10) - \frac{10^2}{2} \right) = 800\pi (100 - 50) = 800\pi \times 50 \]

Calculate the Final Answer

The total work done to pump all the oil over the top is: \( 40000\pi \) foot-pounds.

Key concepts

Cylindrical Tank

A cylindrical tank is a common geometric shape used in various practical applications, including storing liquids like oil or water. It is characterized by its circular base and constant height. In this exercise, we are dealing with a tank where the radius of the base is 4 feet, and the height is 10 feet.

Understanding the geometry of the tank is crucial here because it influences how we calculate both the volume of the tank and the work required to move oil out of it. The tank stands upright, meaning the circular base is parallel to the ground and the height runs vertically.

• The radius of the cylinder is the distance from the center of the base to its edge, which is 4 feet in this case.
• The height is the vertical length of the tank, given here as 10 feet.
• The formula for the volume of a cylinder is given by: \[ V = ext{base area} imes ext{height} = \ \ \ = \ \ \ \pi r^2 h \]
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When calculating the work, we only move the oil above the edge of the tank, so understanding its shape and dimensions helps us determine the path and amount of oil being lifted. This scenario demonstrates that by breaking down a larger problem, the geometry helps solve for more intricate components of the solution.

Integration

Integration is a fundamental concept in calculus which allows us to find the total amount of a quantity that is accumulated continuously over an interval. In this problem, we use integration to determine the total work done in moving the oil from the bottom of the tank to the top.

Here's how integration is applied in the context of this exercise:

• First, we consider a slice of oil at a height \(y\) with an infinitesimally small thickness \(dy\).
• This slice has a specific weight, and moving it to the top requires a certain amount of work based on its distance from the top of the tank (\(10 - y\)).
• We then set up an integral to sum up the work of moving each slice from \(y = 0\) (the bottom) to \(y = 10\) (the top):\[ \int_0^{10} 800\pi (10-y) dy\ \, \, \,\]

By calculating this integral, you're effectively adding up every tiny bit of work done for each slice to get the total work done in moving all the oil. Integrals are powerful because they help compute sums of continuous, infinitely small contributions over an interval, which is ideal for real-world applications like fluid dynamics.

Density

Density is a concept representing how much mass is contained in a given volume. For this problem, the density of the oil in the tank is given as \(\delta = 50\) pounds per cubic foot. This parameter is crucial for calculating both the weight of the oil and the work necessary to move it.

To find out why density matters here, consider the following:

• Density helps us determine the weight of a slice of oil, which is needed to calculate the work done in moving that slice.
• Using the formula \(\text{Weight} = \text{Density} \times \text{Volume}\), density allows us to convert a volume (in cubic feet) into weight (in pounds). For each slice, the weight is obtained by multiplying the volume by \(50\).
• This problem highlights that without density, figuring out work would be impossible, as you would lack the weight component required to calculate energy expense.

In everyday life, knowing the density of a substance assists in logistics and planning, such as understanding the weight limits of containers or calculating shipping requirements when transporting fluids or solids. In terms of physics, density is a bridge between volume and weight, essential for various calculations in science and engineering.