Explanation Given information: Integral is ∬ R x y d A Here, R is the region enclosed by y = cos − 1 x , y = 0 and x = − 1 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. ∫ sin a x d x = − cos a x a + c , where c is integration constant. 4. ∫ cos a x d x = sin a x a + c , where c is integration constant. 5. ∫ u x × v x d x = u × ∫ v d x − ∫ d d x u × ∫ v d x d x Calculation: Consider the integral ∬ R x y d A Region is given by R which is enclosed by y = cos − 1 x , y = 0 and x = − 1 The region can be drawn as, At y = 0 , 0 = cos − 1 x x = cos 0 = 1 Hence, the limits of y are, y = 0 to y = cos − 1 x And limits of x are, x = − 1 to x = 1 Then the integral becomes, ∬ R x y d A = ∫ − 1 1 ∫ 0 cos − 1 x x y d y d x Integrate with respect to y , ∬ R x y d A = ∫ − 1 1 ∫ 0 cos − 1 x x y d y d x = ∫ − 1 1 x y 2 2 0 cos − 1 x d x = 1 2 ∫ − 1 1 x cos − 1 x 2 − 0 d x = 1 2 ∫ − 1 1 x cos − 1 x 2 d x Let, cos − 1 x = t this implies that x = cos t

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 14.2, Problem 22ES

To determine

To calculate: The double integral ∬RxydA , here R is the region enclosed by y=cos−1x , y=0 and x=−1

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Chapter 14.2, Problem 21ES

Chapter 14.2, Problem 23ES

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 double integral ∬ R x y d A , here R is the region enclosed by y = cos − 1 x , y = 0 and x = − 1

Solution Summary: The author calculates the value of the double integral displaystyle 'underset' Riint's xydA, where R is the region enclosed by y=mathrm

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To calculate: The double integral ∬RxydA , here R is the region enclosed by y=cos−1x , y=0 and x=−1

Want to see the full answer?

Chapter 14 Solutions