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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
Chapter 6, Problem 6.PE.25

Areas of Surfaces of Revolution
In Exercises 23–26, find the areas of the surfaces generated by revolving the curves about the given axes.
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y = √4y ― y² , 1 ≤ y ≤ 2 ; y-axis

Verified step by step guidance
1
First, recognize that the problem involves finding the surface area generated by revolving a curve around the y-axis. The formula for the surface area when revolving around the y-axis is: \[ S = \int_{a}^{b} 2\pi x(y) \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy \] where \(x\) is expressed as a function of \(y\).
Given the curve \(y = \sqrt{4y - y^2}\), the first step is to rewrite the curve to express \(x\) as a function of \(y\). Notice that the equation looks like it might represent \(x\) in terms of \(y\), so isolate \(x\) accordingly. Since the curve is given as \(y = \sqrt{4y - y^2}\), check if this is a typo or if it should be \(x = \sqrt{4y - y^2}\) to revolve around the y-axis.
Assuming the curve is \(x = \sqrt{4y - y^2}\) for \(1 \leq y \leq 2\), compute the derivative \(\frac{dx}{dy}\) using the chain rule: \[ \frac{dx}{dy} = \frac{d}{dy} \left( \sqrt{4y - y^2} \right) = \frac{1}{2\sqrt{4y - y^2}} (4 - 2y) \]
Substitute \(x(y)\) and \(\frac{dx}{dy}\) into the surface area integral formula: \[ S = \int_{1}^{2} 2\pi \sqrt{4y - y^2} \sqrt{1 + \left( \frac{4 - 2y}{2\sqrt{4y - y^2}} \right)^2} \, dy \]
Simplify the expression inside the square root and the integral as much as possible to make the integral easier to evaluate. Then, set up the integral for evaluation either by hand or using a computational tool 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 by integrating the circumference of circular slices along the curve. The formula involves the radius (distance from the axis) and the arc length element, typically expressed as an integral involving the function and its derivative.
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Example 1: Minimizing Surface Area

Parametrization and Variable of Integration

Choosing the correct variable of integration and expressing the function accordingly is crucial. Since the curve is given as y in terms of y, and the revolution is about the y-axis, the function must be expressed as x in terms of y, and integration is performed with respect to y over the given interval.
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Parameterizing Equations

Arc Length Element (ds)

The arc length element ds represents a small segment of the curve and is calculated using the derivative of the function. For a function x = f(y), ds = sqrt(1 + (dx/dy)^2) dy, which accounts for the curve's slope and is essential in computing the surface area integral.
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Arc Length of Parametric Curves