Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.7.9

Chapter 6, Problem 6.7.9

9–12. Consider the cylindrical tank in Example 4 that has a height of 10 m and a radius of 5 m. Recall that if the tank is full of water, then ∫₀¹⁰ 25 π ρg(15−y) dy equals the work required to pump all the water out of the tank, through an outflow pipe that is 15 m above the bottom of the tank. Revise this work integral for the following scenarios. (Do not evaluate the integrals.)

The work required to empty the top half of the tank

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Identify the portion of the tank to be emptied: since the tank height is 10 m, the top half corresponds to the water from height 5 m to 10 m.

Set up the integral limits accordingly: the variable of integration \( y \) will go from 5 to 10, representing the vertical position of the water slice being pumped.

Recall the integrand from the original problem: \( 25 \pi \rho g (15 - y) \), where \( 25 \pi \) is the cross-sectional area (since radius \( r = 5 \), area \( = \pi r^2 = 25 \pi \)), \( \rho \) is the density of water, \( g \) is acceleration due to gravity, and \( (15 - y) \) is the distance the water must be lifted to the outflow pipe 15 m above the bottom.

Write the revised integral for the work to pump out the top half of the tank as \( \int_{5}^{10} 25 \pi \rho g (15 - y) \, dy \).

This integral represents the total work required to pump all the water from height 5 m up to 10 m out through the pipe at 15 m.

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Key Concepts

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

Work as an Integral in Fluid Mechanics

Work done to pump fluid is calculated by integrating the force needed to move each layer of fluid times the distance it must be moved. This involves setting up an integral where the integrand represents the weight of a thin slice of water multiplied by the vertical distance to the outflow point.

Setting Limits of Integration for Partial Volumes

When calculating work for only part of the tank, such as the top half, the limits of integration must correspond to the vertical range of that portion. For the top half of a 10 m tank, the integral limits change from 0 to 10 to 5 to 10, reflecting the height interval of the water being pumped.

Distance Function in Work Integrals

The distance each water slice is pumped depends on its vertical position. If the outflow pipe is 15 m above the bottom, the distance function is (15 - y), where y is the height of the slice. This distance varies with y and must be included in the integrand to correctly calculate work.

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Related Practice

Textbook Question

60–63. Equivalent constant velocity Consider the following velocity functions. In each case, complete the sentence: The same distance could have been traveled over the given time period at a constant velocity of ________.

v(t) = t(25−t²)^1/2, for 0≤t≤5

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Textbook Question

9–20. Arc length calculations Find the arc length of the following curves on the given interval.

y = −8x−3 on [−2, 6] (Use calculus.)

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Textbook Question

Leaky Bucket A 1-kg bucket resting on the ground contains 3 kg of water. How much work is required to raise the bucket vertically a distance of 10 m if water leaks out of the bucket at a constant rate of 1/5 kg/m? Assume the weight of the rope used to raise the bucket is negligible. (Hint: Use the definition of work, W = ∫a^bF(y) dy, where F is the variable force required to lift an object along a vertical line from y=a to y=b.)

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Textbook Question

Find the area of the region described in the following exercises.

The region bounded by x=y(y−1) and y=x/3

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Textbook Question

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis.

{Use of Tech} y = √sin^−1x,y = √π/2, and x=0; about the x-axis

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Textbook Question

For the following regions R, determine which is greater—the volume of the solid generated when R is revolved about the x-axis or about the y-axis.

R is bounded by y=1−x^3, the x-axis, and the y-axis.

100

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