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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 6, Problem 6.6.5a

A surface is generated by revolving the line f(x)=2−x, for 0≤x≤2, about the x-axis. Find the area of the resulting surface in the following ways.


a. Using calculus

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Step 1: Recall the formula for the surface area of a solid of revolution about the x-axis. The formula is: A = 2π ∫[a,b] f(x)√(1 + (f'(x))²) dx, where f(x) is the function being revolved, and f'(x) is its derivative.
Step 2: Identify the function f(x) = 2 - x and its domain [0, 2]. Compute the derivative of f(x): f'(x) = -1.
Step 3: Substitute f(x) and f'(x) into the formula. The integrand becomes: 2π ∫[0,2] (2 - x)√(1 + (-1)²) dx. Simplify the square root term: √(1 + 1) = √2, so the integrand is 2π√2 ∫[0,2] (2 - x) dx.
Step 4: Break down the integral ∫[0,2] (2 - x) dx into two simpler integrals: ∫[0,2] 2 dx - ∫[0,2] x dx. Compute each integral separately using the power rule for integration.
Step 5: Multiply the result of the integral by 2π√2 to find the surface area. This completes the calculation process.

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Key Concepts

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

Surface Area of Revolution

The surface area of a solid of revolution is calculated by revolving a curve around an axis. For a function f(x) revolved around the x-axis, the formula is given by A = 2π ∫[a to b] f(x) √(1 + (f'(x))^2) dx, where [a, b] is the interval of x. This formula accounts for both the height of the function and the slope, providing the total surface area generated.
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Definite Integral

A definite integral represents the accumulation of quantities, such as area under a curve, over a specific interval [a, b]. It is denoted as ∫[a to b] f(x) dx and provides a numerical value that corresponds to the total area between the curve f(x) and the x-axis from x = a to x = b. Understanding how to evaluate definite integrals is crucial for calculating surface areas.
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Derivative and Its Role

The derivative of a function, denoted as f'(x), measures the rate of change of the function with respect to x. In the context of surface area, the derivative is used to find the slope of the curve, which is essential for the formula involving √(1 + (f'(x))^2). This term adjusts the length of the curve segment being revolved, ensuring accurate surface area calculations.
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Related Practice
Textbook Question

2–3. Displacement, distance, and position Consider an object moving along a line with the following velocities and initial positions. Assume time t is measured in seconds and velocities have units of m/s.


d. Determine the position function s(t) using the Fundamental Theorem of Calculus (Theorem 6.1). Check your answer by finding the position function using the antiderivative method.


v(t) = 12t²-30t+12, for 0 ≤ t ≤ 3; s(0)=1

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Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

c. If water flows into a tank at a constant rate (for example, 6 gal/min), the volume of water in the tank increases according to a linear function of time.

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Textbook Question

Oscillating growth rates Some species have growth rates that oscillate with an (approximately) constant period P. Consider the growth rate function N'(t) = r+A sin 2πt/P, where A and r are constants with units of individuals/yr, and t is measured in years. A species becomes extinct if its population ever reaches 0 after t=0.


a. Suppose P=10, A=20, and r=0. If the initial population is N(0)=10, does the population ever become extinct? Explain.

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Textbook Question

Filling a tank A 2000-liter cistern is empty when water begins flowing into it (at t=0 at a rate (in L/min) given by Q′(t) = 3√t, where t is measured in minutes.


a. How much water flows into the cistern in 1 hour?

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Textbook Question

Work in a gravitational field For large distances from the surface of Earth, the gravitational force is given by F(x) = GMm / (x+R)², where G = 6.7×10^−11 N m²/kg² is the gravitational constant, M = 6×10^24 kg is the mass of Earth, m is the mass of the object in the gravitational field, R = 6.378×10⁶ m is the radius of Earth, and x≥0 is the distance above the surface of Earth (in meters).


a. How much work is required to launch a rocket with a mass of 500 kg in a vertical flight path to a height of 2500 km (from Earth’s surface)?

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Textbook Question

70–72. Variable density in one dimension Find the mass of the following thin bars.


A bar on the interval 0≤x≤6 with a density ρ(x) = {1 if 0 ≤ x < 2

2 if 2 ≤ x < 4

4 if 4 ≤ x ≤ 6

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