Textbook Question
9–20. Arc length calculations Find the arc length of the following curves on the given interval.
x = y⁴/4 + 1/8y², for 1≤y≤2
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Ch. 6 - Applications of Integration
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 6, Problem 6.6.31
Find the area of the surface generated when the given curve is revolved about the given axis.
y=x^3/2−x^1/2 / 3, for 1≤x≤2; about the x-axis
Verified step by step guidance1
Identify the given function and the interval: The curve is given by \(y = \frac{x^{3/2} - x^{1/2}}{3}\) for \(1 \leq x \leq 2\), and it is revolved about the x-axis.
Recall the formula for the surface area of a curve revolved about the x-axis:
\(S = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\),
where \(y\) is the function and \(\frac{dy}{dx}\) is its derivative.
Compute the derivative \(\frac{dy}{dx}\) of the function:
First, rewrite \(y\) as \(y = \frac{1}{3} x^{3/2} - \frac{1}{3} x^{1/2}\).
Then, differentiate term-by-term using the power rule.
Substitute \(y\) and \(\frac{dy}{dx}\) into the surface area integral formula:
\(S = \int_1^2 2\pi \left( \frac{x^{3/2} - x^{1/2}}{3} \right) \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx\).
Set up the integral for evaluation or numerical approximation as needed to find the surface area.
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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 formed by revolving a curve around an axis is found using an integral formula. For revolution about the x-axis, the formula is S = ∫ 2πy √(1 + (dy/dx)²) dx over the given interval. This calculates the total area of the curved surface generated.
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Example 1: Minimizing Surface Area
Derivative of the Function
To apply the surface area formula, you need the derivative dy/dx of the given function y(x). The derivative measures the slope of the curve at each point, which is essential for computing the arc length element √(1 + (dy/dx)²) in the integral.
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06:30Derivatives of Other Trig Functions
Integration over the Given Interval
After setting up the integral for surface area, you must evaluate it over the specified interval [1, 2]. This involves integrating the function 2πy √(1 + (dy/dx)²) with respect to x from 1 to 2 to find the exact surface area.
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11:11Improper Integrals: Infinite Intervals
Related Practice
Textbook Question
Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given line.
y=2 sin x and y=0 on [0,π]; about y=−2
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Textbook Question
Find the area of the surface generated when the given curve is revolved about the given axis.
x=√12y−y^2, for 2≤y≤10; about the y-axis
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Textbook Question
Explain how to find the mass of a one-dimensional object with a variable density ρ.
Textbook Question
What is the pressure on a horizontal surface with an area of 2 m² that is 4 m underwater?
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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=4−2x, the x-axis, and the y-axis.