Explanation Let there be a continuous function f ( x , y ) on R = [ a , b ] × [ c , d ] , then the below integrals are referred to as iterated integrals. ∬ R f ( x , y ) d A = ∫ a b ∫ c d f ( x , y ) d y d x = ∫ c d ∫ a b f ( x , y ) d x d y Given: A double integral as ∬ R s i n x d A , R = [ 0 , 2 π ] × [ 0 , 2 π ] Formula Used: In order to compute the double integral ∬ R f ( x ) d A using symmetry, we need to check for following points: 1. If the f ( x ) is an odd function, then ∫ − a a f ( x ) d x = 0 27. If the f ( x ) is even function, then ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x Calculation: The given problem is ∬ R s i n x d A Writing the integral as an iterated integral, we get ∬ R s i n x d A = ∫ 0 2 π ∫ 0 2 π s i n x d y d x Let us first compute the indefinite integral ∫ 0 2 π s i n x d y ∫ 0 2 π s i n x d y = s i n x ∫ 0 2 π 1 d y Applying the constant rule of integration, we get, ∫ 0 2 π s i n x d y = s i n x y | 0 2 π Compute the value, by substituting y = 0 and

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 16.1, Problem 17E

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The value of the double integral $\iint_{R} s i n x d A, R = [0, 2 π] \times [0, 2 π]$ using symmetry.

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Chapter 16.1, Problem 16E

Chapter 16.1, Problem 18E

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Chapter 16 Solutions

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The value of the double integral ∬ R s i n x d A , R = [ 0 , 2 π ] × [ 0 , 2 π ] using symmetry.

Solution Summary: The author explains how to compute the double integral R s i n x d A using symmetry.

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The value of the double integral $\iint_{R} s i n x d A, R = [0, 2 π] \times [0, 2 π]$ using symmetry.

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Chapter 16 Solutions

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