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L sinty?) dy dx. EXAMPLE 5 Evaluate the iterated integral SOLUTION If we try to evaluate the integral as it stands, we are faced with the task of first evaluating sin(y?) dy. But it's impossible to do so in finite terms since sin(y?) dy is not an elementary function. So we must change the order of integration. This is accomplished by first expressing the given iterated integral as a double integral. Using this equation backward, we have sin(y?) dy dx = sin(y?) dA where Video Example ) D = y) |0 We sketch this region D in the figure. Then we see that an alternate description of D is D = {(x, v) I 0 s y s 6, 0 sx sy}. This enables us to use this equation to express the double integral as an iterated integral in the reverse order: sin(y?) dy dx = sin(y?) dA = L7 sin(y?) dx dy dy %3D - Lvsin(,2) = 0.20

L sinty?) dy dx. EXAMPLE 5 Evaluate the iterated integral SOLUTION If we try to evaluate the integral as it stands, we are faced with the task of first evaluating sin(y?) dy. But it's impossible to do so in finite terms since sin(y?) dy is not an elementary function. So we must change the order of integration. This is accomplished by first expressing the given iterated integral as a double integral. Using this equation backward, we have sin(y?) dy dx = sin(y?) dA where Video Example ) D = y) |0 We sketch this region D in the figure. Then we see that an alternate description of D is D = {(x, v) I 0 s y s 6, 0 sx sy}. This enables us to use this equation to express the double integral as an iterated integral in the reverse order: sin(y?) dy dx = sin(y?) dA = L7 sin(y?) dx dy dy %3D - Lvsin(,2) = 0.20

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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6 EXAMPLE 5 Evaluate the iterated integral sin(y?) dy dx. If we try to evaluate the integral as it stands, we are faced with the task of first evaluating | sin(y?) dy. SOLUTION But it's impossible to do so in finite terms since sin(y2) dy is not an elementary function. So we must change the order of integration. This is accomplished by first expressing the given iterated integral as a double integral. Using this equation backward, we have 9. 9. sin(y?) dy dx = sin(y?) dA where Video Example {(x, v) | 0s x < 6, x < y s 6}. D = We sketch this region D in the figure. Then we see that an alternate description of D is {(x, y) I 0s ys 6, 0 s x s y. D = This enables us to use this equation to express the double integral as an iterated integral in the reverse order: 9. 9. sin(y?) dy dx sin(y?) dA %3D Jo Jx -T sinty*) dx dy 9. 1 sin (2)x X = y dy x = 0 9. b (y sin(y?) dy (- cos(,2) 1 COS 6. 0.20
Transcribed Image Text:6 EXAMPLE 5 Evaluate the iterated integral sin(y?) dy dx. If we try to evaluate the integral as it stands, we are faced with the task of first evaluating | sin(y?) dy. SOLUTION But it's impossible to do so in finite terms since sin(y2) dy is not an elementary function. So we must change the order of integration. This is accomplished by first expressing the given iterated integral as a double integral. Using this equation backward, we have 9. 9. sin(y?) dy dx = sin(y?) dA where Video Example {(x, v) | 0s x < 6, x < y s 6}. D = We sketch this region D in the figure. Then we see that an alternate description of D is {(x, y) I 0s ys 6, 0 s x s y. D = This enables us to use this equation to express the double integral as an iterated integral in the reverse order: 9. 9. sin(y?) dy dx sin(y?) dA %3D Jo Jx -T sinty*) dx dy 9. 1 sin (2)x X = y dy x = 0 9. b (y sin(y?) dy (- cos(,2) 1 COS 6. 0.20
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