Problem 52 - Calculus Single Variable

Question

Are concerned with a tank filled to the top with water. Its shape is obtained by rotating about the \(y\) -axis the region that is bounded above by the horizontal line \(y=10\) \(\left(1-1 / e^{2}\right),\) on the left by the \(y\) -axis, and on the right by the graph of \(y=10\left(1-\exp \left(-x^{2} / 2\right)\right) .\) Both \(x\) and \(y\) are measured in feet. A pump floats on the surface of the water and pumps water to the top, at which point the water runs off. How much work is done in pumping out half the water in the tank?

Step-by-Step Solution

Answer

The work done is the evaluation of the integral using these steps.

Step 1: Understanding the Problem

We need to calculate the work done in pumping the water from a tank. The tank has a shape formed by rotating a specific curve around the y-axis. We aim to find the work done to pump half of the water to the top.

Step 2: Determine the Shape and Volume of the Tank

The shape of the solid is generated by rotating the area between the curve \( y = 10(1 - e^{-x^2/2}) \) and the y-axis around the y-axis. We need to find the volume of the tank and then determine half of it, as we are interested in half of the water.

Step 3: Identifying the Limits of Integration

The region of integration is bounded from \( y = 0 \) to \( y = 10(1 - 1/e^2) \). It's necessary to set up the integral in terms of \( y \) since the tank's rotation revolves around the y-axis and work calculations require the integration over y.

Step 4: Compute the Differential Volume Element

Consider a horizontal slice at height \( y \) with a small thickness \( dy \). The area of this slice is \( A(y) = \pi r^2(y) \), where \( r(y) \) is the inverse function representing the radius at height \( y \). Since \( x = \sqrt{-2 \, \ln(1 - y/10)} \) for the given function, we have \( r(y) = \sqrt{-2 \, \ln(1 - y/10)} \).

Step 5: Calculate Work for the Entire Volume

The work required to lift a slice of water to the top (10 feet) is \( \, dW = \rho g A(y) (10 - y) dy \), where \( \rho \) is the density of water (62.4 lbs/ft³), \( g \) is the gravitational force, and \( (10 - y) \) is the height it must be lifted. Setting up the integral of \( dW \) over \( y \) from 0 to half the volume limits.

Step 6: Evaluate the Integral for Work

Compute the definite integral \( W = \int_0^{0.5 \times 10(1 - 1/e^2)} \rho g \pi (-2 \ln(1 - y/10)) (10 - y) \, dy \). Simplifying and evaluating this integral will provide the work done in pumping out half the volume.

Key Concepts

Volume of Revolution

In many mathematical and physical problems, the concept of "Volume of Revolution" is vital. When we talk about the volume of revolution, we are referring to the volume of a three-dimensional object created by rotating a two-dimensional shape around an axis. This method is often used when dealing with calculus problems that require finding volumes of oddly-shaped objects.

For instance, in the exercise, the tank's shape is formed by rotating a region defined by a function about the y-axis. Here's how you can visualize it: imagine tracing out the boundary of the shape on a piece of paper. Then, rotate that paper shape around the y-axis. The shape you've created in this way is what we call the "Volume of Revolution."

To calculate such volumes, calculus is often employed using integration methods. Recognizing the specific function and setting appropriate limits for your integration are key steps in determining the volume.

• The area between the curve and the axis forms the core of calculation.
• In this problem, the boundaries are defined by specific curves and lines.
• Rotating the curve about the y-axis requires understanding the inverse relationship between x and y.

Integration in Calculus

Integration in calculus is a fundamental concept used to calculate things like area under curves and volume of solids, including those of revolution. In our exercise, integration plays a critical role in determining the work needed to pump water from the tank.

By setting up the integral correctly, we can find the exact volume of water or, in this case, compute the work needed to pump part of that water. When dealing with shapes revolved around an axis, integration allows us to account for every tiny slice of the volume, adding it all together to get a total.

• Integration helps in finding the total work done by summing infinitesimal work elements (dW) across the volume.
• Here, the integral incorporates the area's formula minus the small, finite slices throughout the volume.
• The limits of integration, from 0 to half the "final height," are vital to accurately representing the water being pumped out.

Using these principles, the integral calculates work by considering force and distance for each slice, ultimately offering a precise solution to our problem.

Physics of Fluids

The physics of fluids encompasses how liquids behave under various conditions and forces. In our problem, this specifically involves understanding the work needed to move water in a gravitational field, such as when pumping water out of a tank.

Work, in physical terms, is defined as the force applied over a distance. When you're lifting water, this force is the weight of the water, determined by its density and volume, and the distance is how far it has to be lifted to reach the top of the tank.

• The formula for work, integrating weight over distance, is central in solving this problem.
• Water's density (62.4 lbs/ft³) is used to calculate the weight of each differential volume element.
• Gravitational forces are constant, requiring the integration of the product of mass, gravity, and height.

By applying these principles, combining mathematics with physical understanding, we effectively determine the total amount of work required to pump out half the water from the tank.