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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.7.36
Chapter 8, Problem 8.7.36

[Technology Exercise] When solving Exercises 33-40, you may need to use a calculator or a computer.
Find, to two decimal places, the areas of the surfaces generated by revolving the curves in Exercises 35 and 36 about the x-axis.
y = x²/4, 0 ≤ x ≤ 2

Verified step by step guidance
1
Identify the formula for the surface area of a solid of revolution about the x-axis: \[ S = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]
Given the curve \[ y = \frac{x^2}{4} \], find its derivative with respect to \[ x \]: \[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{x^2}{4} \right) = \frac{x}{2} \]
Substitute \[ y \] and \[ \frac{dy}{dx} \] into the surface area formula: \[ S = \int_0^2 2\pi \left( \frac{x^2}{4} \right) \sqrt{1 + \left( \frac{x}{2} \right)^2} \, dx \]
Simplify the expression inside the square root and the integrand: \[ S = \int_0^2 2\pi \frac{x^2}{4} \sqrt{1 + \frac{x^2}{4}} \, dx = \int_0^2 \frac{\pi x^2}{2} \sqrt{1 + \frac{x^2}{4}} \, dx \]
Evaluate the integral using an appropriate method (such as substitution or numerical integration) to find the surface area, then round the result to two decimal places.

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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 involves integrating 2π times the radius (the y-value) times the arc length element. This concept connects geometry with calculus to measure curved surfaces.
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Example 1: Minimizing Surface Area

Arc Length Element (ds)

The arc length element ds represents a small segment of the curve and is given by √(1 + (dy/dx)²) dx. It accounts for the curve's slope, ensuring the surface area calculation accurately follows the curve's shape rather than just the x-interval.
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Arc Length of Parametric Curves

Definite Integration with Limits

Definite integration calculates the exact accumulated value over an interval, here from x = 0 to x = 2. Applying limits ensures the surface area corresponds precisely to the specified portion of the curve, providing a numerical result to two decimal places.
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Definition of the Definite Integral
Related Practice
Textbook Question

Evaluate the integrals in Exercises 1–24 using integration by parts.

∫ arcsin(y) dy

Textbook Question

Evaluate the integrals in Exercises 51–56 by making a substitution (possibly trigonometric) and then applying a reduction formula.

∫ csc³(√θ) / √θ dθ

Textbook Question

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.

∫ from 2 to ∞ of (dx / √(x² - 1))

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Textbook Question

[Technology Exercise] When solving Exercises 33-40, you may need to use a calculator or a computer.

Find, to two decimal places, the areas of the surfaces generated by revolving the curves in Exercises 35 and 36 about the x-axis.

y = sin x, 0 ≤ x ≤ π

1
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Textbook Question

Use any method to evaluate the integrals in Exercises 15–38. Most will require trigonometric substitutions, but some can be evaluated by other methods.

∫ √(9 - w²) dw / w²

Textbook Question

Evaluate the integrals in Exercises 23–32.

∫₀^(π/6) √(1 + sin(x)) dx

(Hint: Multiply by √((1 - sin(x)) / (1 - sin(x))))