Chapter 10, Problem 112RE

Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347

Chapter
Section

Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
Textbook Problem

Finding the Area of a Surface of Revolution In Exercises 111 and 112, find the area of the surface formed by revolving the polar equation over the given interval about the given line.Polar Equation Interval Axis of Revolution r = 2 sin θ 0 ≤ θ ≤ π 2 θ = π 2

To determine

To calculate: The area of the surface formed by revolving the polar equation of a curve given as, r=2sinθ over the interval 0θπ2 about the angle given as, θ=π2.

Explanation

Given:

The polar equation is r=2sinÎ¸ and interval is 0â‰¤Î¸â‰¤Ï€2. The axis of revolution is Î¸=Ï€2.

Formula used:

The area of the surface formed by revolving the curve given as, r=f(Î¸) in the interval Î±â‰¤Î¸â‰¤Î², about the polar axis is given by;

S=2Ï€âˆ«Î±Î²f(Î¸)sinÎ¸[f(Î¸)]2+[fâ€²(Î¸)]2dÎ¸

The derivative of ddÎ¸(sinÎ¸)=cosÎ¸.

The trigonometric identity sin2x+cos2x=1, sin2Î¸=1âˆ’cos2Î¸2.

Calculation:

The polar equation is;

r=2sinÎ¸

The derivative of the polar equation with respect to Î¸ is;

drdÎ¸=ddÎ¸(2sinÎ¸)=2cosÎ¸

Then, the surface area formed is;

S=2Ï€âˆ«0Ï€2(2sinÎ¸)cosÎ¸[2sinÎ¸]2+[2cosÎ¸]2dÎ¸=2Ï€âˆ«0Ï€2(2cosÎ¸sinÎ¸)4sin2Î¸+4cos2Î¸dÎ¸=4Ï€âˆ«0Ï€2(cos<

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