Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.7.42a

Chapter 6, Problem 6.7.42a

Emptying a water trough A water trough has a semicircular cross section with a radius of 0.25 m and a length of 3 m (see figure).
a. How much work is required to pump the water out of the trough (to the level of the top of the trough) when it is full?

Verified step by step guidance

Identify the shape and dimensions of the trough: it has a semicircular cross section with radius \(r = 0.25\) m and length \(L = 3\) m.

Set up a coordinate system for the vertical direction, letting \(y\) represent the depth of a thin horizontal slice of water measured from the bottom of the trough (where \(y=0\)) up to the top (where \(y=0.25\) m).

Express the width of the water slice at height \(y\). Since the cross section is semicircular, the width \(w(y)\) is twice the horizontal distance from the center to the edge, which can be found using the circle equation: \(w(y) = 2 \sqrt{r^2 - (r - y)^2}\).

Calculate the volume of a thin slice of water at height \(y\) with thickness \(dy\): \(dV = w(y) \times L \times dy\).

Determine the work to pump this slice to the top of the trough. The distance the water must be lifted is \((r - y)\), and the weight density of water is \(\rho g\) (where \(\rho\) is the density of water and \(g\) is acceleration due to gravity). The work for the slice is \(dW = \rho g \times dV \times (r - y)\). Integrate \(dW\) from \(y=0\) to \(y=r\) to find the total work.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Work Done by a Variable Force

Work is the integral of force over distance. When pumping water, the force varies with the weight of each water slice and the distance it must be moved. Calculating work involves integrating the product of force and distance over the volume of water.

Volume and Cross-Sectional Area of a Semicircle

The trough has a semicircular cross section, so the area of a horizontal slice depends on the radius and the height of the water at that slice. Understanding how to express the area of a semicircle segment as a function of depth is essential for setting up the integral.

Density and Weight of Water

The force exerted by the water is its weight, which depends on the volume, density, and gravitational acceleration. Knowing the density of water allows conversion from volume to weight, which is necessary to calculate the force in the work integral.

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