Explanation Given information: Integral is ∫ − π π ∫ 0 1 − cos θ r d r d θ . Formula used: Integration formulas are, 1. ∫ a b f x d x = F b − F a , where F x is anti - derivative of f x . 2. ∫ x n d x = x n + 1 n + 1 + c , where c is integration constant. 3. ∫ cos a x d x = sin a x a + c , where c is integration constant. Calculation: Consider the integral ∫ − π π ∫ 0 1 − cos θ r d r d θ . Integrate with respect to r , ∫ − π π ∫ 0 1 − cos θ r d r d θ = ∫ − π π r 2 2 0 1 − cos θ d θ = 1 2 ∫ − π π 1 − cos θ 2 − 0 d θ = 1 2 ∫ − π π 1 − cos θ 2 d θ = 1 2 ∫ − π π 1 + cos 2 θ − 2 cos θ d θ Since, cos 2 θ = 2 cos 2 θ − 1 . This implies that cos 2 θ = cos 2 θ + 1 2

Calculus: Early Transcendentals, Enhanced Etext

12th Edition

ISBN: 9781119777984

Author: Howard Anton; Irl C. Bivens; Stephen Davis

Publisher: Wiley Global Education US

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Chapter 14.3, Problem 2ES

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To calculate: The value of the iterated integral ∫−ππ∫01−cosθrdrdθ .

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Chapter 14.3, Problem 1ES

Chapter 14.3, Problem 3ES

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Chapter 14 Solutions

Calculus: Early Transcendentals, Enhanced Etext

Ch. 14.1 - Prob. 1QCE Ch. 14.1 - The iterated integral 1524fx,ydxdy integrates f...Ch. 14.1 - Supply the missing integrand and limits of...Ch. 14.1 - Prob. 4QCE Ch. 14.1 - Evaluate the iterated integrals. 0102x+3dydx Ch. 14.1 - Evaluate the iterated integrals. 13112x4ydydx Ch. 14.1 - Evaluate the iterated integrals. 2401x2ydxdy Ch. 14.1 - Evaluate the iterated integrals. 2012x2+y2dxdy Ch. 14.1 - Evaluate the iterated integrals. 0ln30ln2ex+ydydx Ch. 14.1 - Prob. 6ES

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The value of the iterated integral ∫ − π π ∫ 0 1 − cos θ r d r d θ .

Solution Summary: The author calculates the value of the iterated integral displaystyle, in the form of a formula.

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To calculate: The value of the iterated integral ∫−ππ∫01−cosθrdrdθ .

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Chapter 14 Solutions