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Ch. 6 - Applications of Definite Integrals
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 6 - Applications of Definite IntegralsProblem 6.PE.23
Chapter 6, Problem 6.PE.23

Areas of Surfaces of Revolution
In Exercises 23–26, find the areas of the surfaces generated by revolving the curves about the given axes.
_____
y = √2x + 1 , 0 ≤ x ≤ 3 ; x-axis"

Verified step by step guidance
1
Identify the curve and the interval: The curve is given by \(y = \sqrt{2x + 1}\), and the interval for \(x\) is \(0 \leq x \leq 3\).
Recall the formula for the surface area of a curve revolved about the x-axis: \(S = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\)
Compute the derivative \(\frac{dy}{dx}\) of the function \(y = \sqrt{2x + 1}\). Use the chain rule to find this derivative.
Substitute \(y\) and \(\frac{dy}{dx}\) into the surface area formula, and simplify the expression inside the square root as much as possible.
Set up the definite integral with limits from \(x=0\) to \(x=3\) and prepare to evaluate or approximate the integral 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 a curve y = f(x) revolved about the x-axis, the surface area is given by S = ∫ 2πy √(1 + (dy/dx)²) dx over the interval. This formula accounts for the circumference of circular slices and the curve's slope.
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Example 1: Minimizing Surface Area

Derivative and Slope of the Curve

Calculating the derivative dy/dx is essential to determine the slope of the curve at each point. The slope affects the length element of the curve, which is part of the surface area integral. For y = √(2x + 1), use the chain rule to find dy/dx accurately.
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Setting up Definite Integrals with Limits

The problem specifies the interval 0 ≤ x ≤ 3, which defines the limits of integration. Properly setting these limits ensures the surface area corresponds exactly to the portion of the curve being revolved. Understanding how to apply these bounds in the integral is crucial for an accurate solution.
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Related Practice
Textbook Question

Volumes

Find the volumes of the solids in Exercises 1–18.

The solid lies between planes perpendicular to the x-axis at x = 0 and x = 4. The cross-sections of the solid perpendicular to the x-axis between these planes are circular disks whose diameters run from the curve x² = 4y to the curve y² = 4x.

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Textbook Question

Centers of Mass and Centroids

Find the centroid of a thin, flat plate covering the region enclosed by the parabolas 𝔂 = 2𝓍² and 𝔂 = 3 ― 𝓍² .

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Textbook Question

Volumes

Find the volumes of the solids in Exercises 1–18.

The solid lies between planes perpendicular to the x-axis at x = 0 and x = 1. The cross-sections perpendicular to the x-axis between these planes are circular disks whose diameters run from the parabola y = x² to the parabola y = √x.

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Textbook Question

Work

Pumping a conical tank A right-circular conical tank, point down, with top radius 5 ft and height 10 ft, is filled with a liquid whose weight-density is 60lb/ft³. How much work does it take to pump the liquid to a point 2 ft above the tank? If the pump is driven by a motor rated at 275ft-lb/sec (1/2 hp), how long will it take to empty the tank? 

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Textbook Question

Centers of Mass and Centroids

Find the center of mass of a thin, flat plate covering the region enclosed by the parabola 𝔂² = 𝓍 and the line 𝓍 = 2𝔂 if the density function is δ(𝔂) = 1 + 𝔂. (Use horizontal strips.)

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Textbook Question

Find the lengths of the curves in Exercises 19–22.

y = x¹/² ― (1/3) x³/² , 1 ≤ x ≤ 4

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