Textbook Question
Find the area of the surface generated when the given curve is revolved about the given axis.
y=3x+4, for 0≤x≤6; about the x-axis
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 6.6.35
Find the area of the surface generated when the given curve is revolved about the given axis.
x=4y^3/2−y^1/2 / 12, for 1≤y≤4; about the y-axis
Verified step by step guidance1
Identify the formula for the surface area generated by revolving a curve around the y-axis. For a function \( x = f(y) \), the surface area \( S \) is given by:
\[
S = 2\pi \int_{a}^{b} x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy
\]
where \( a \) and \( b \) are the bounds for \( y \).
Write down the given function explicitly:
\[
x = \frac{4y^{3/2} - y^{1/2}}{12}
\]
and the interval for \( y \) is \( 1 \leq y \leq 4 \).
Compute the derivative \( \frac{dx}{dy} \). Differentiate \( x \) with respect to \( y \) using the power rule for each term inside the numerator, then divide by 12.
Substitute \( x \) and \( \frac{dx}{dy} \) into the surface area integral formula:
\[
S = 2\pi \int_{1}^{4} x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy
\]
Set up the integral with the expressions found and prepare to evaluate it either by hand (if possible) or using a computational tool. This integral will give the surface area of the surface generated by revolving the curve about the y-axis.
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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 y-axis, the formula involves integrating 2π times the radius (distance from the axis) times the arc length element along y. This concept connects geometry with calculus to measure curved surfaces.
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Example 1: Minimizing Surface Area
Parametric Representation and Differentiation
When the curve is given as x in terms of y, we treat y as the independent variable and differentiate x with respect to y. This derivative is essential to compute the arc length element, which appears inside the surface area integral. Understanding how to differentiate and manipulate these expressions is crucial.
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06:49Differentiation of Parametric Curves
Arc Length Element in Terms of y
The arc length element ds for a curve defined as x = f(y) is given by ds = sqrt(1 + (dx/dy)^2) dy. This expression measures the infinitesimal length along the curve in the y-direction and is used in the surface area integral to account for the curve's shape when revolved.
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06:29Arc Length of Parametric Curves
Related Practice
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Textbook Question
3–6. Setting up arc length integrals Write and simplify, but do not evaluate, an integral with respect to x that gives the length of the following curves on the given interval.
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