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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 6 - Applications of IntegrationProblema 6.6.35
Capitolo 6, Problema 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

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1
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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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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Differentiation 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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Arc Length of Parametric Curves
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