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EXAMPLE 3 Evaluate xy dA, where D is the region bounded by the line y = x - 1 and the parabola -3 y2 = 2x + 6. SOLUTION The region D is shown in the figure. Again D is both type I and type II, but the description of D as a type I region is more complicated because the lower boundary consists of two parts. Therefore we prefer to express D as a type II region: -1 {ex.v) |-2 5 y s 4, - 3 sxsy+1}. -6 -2 2 Then this equation gives ху xy dx dy J-2J1/2y² - 3 Video Example ) y +1 dy - ly•ar -( + 1)2 dy + 4y3 + 2y2 - 8y) dy +*+ - ay", If we had expressed D as a type I region, then we would have obtained 2x + 6 / 2x + 6 xy dA = xy dy dx + xy dy dx -V 2x + 6 but this would have involved more work than the other method. Need Help? Talk to a Tutor Read It

EXAMPLE 3 Evaluate xy dA, where D is the region bounded by the line y = x - 1 and the parabola -3 y2 = 2x + 6. SOLUTION The region D is shown in the figure. Again D is both type I and type II, but the description of D as a type I region is more complicated because the lower boundary consists of two parts. Therefore we prefer to express D as a type II region: -1 {ex.v) |-2 5 y s 4, - 3 sxsy+1}. -6 -2 2 Then this equation gives ху xy dx dy J-2J1/2y² - 3 Video Example ) y +1 dy - ly•ar -( + 1)2 dy + 4y3 + 2y2 - 8y) dy +*+ - ay", If we had expressed D as a type I region, then we would have obtained 2x + 6 / 2x + 6 xy dA = xy dy dx + xy dy dx -V 2x + 6 but this would have involved more work than the other method. Need Help? Talk to a Tutor Read It

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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EXAMPLE 3 Evaluate xy dA, where D is the region bounded by the line y = x - 1 and the parabola -3 y2 = 2x + 6. SOLUTION The region D is shown in the figure. Again D is both type I and type II, but the description of D as a type I region is more complicated because the lower boundary consists of two parts. Therefore we prefer to express D as a type II region: -1 {ex.v) |-2 5 y s 4, - 3 sxsy+1}. -6 -2 2 Then this equation gives ху xy dx dy J-2J1/2y² - 3 Video Example ) y +1 dy - ly•ar -( + 1)2 dy + 4y3 + 2y2 - 8y) dy +*+ - ay", If we had expressed D as a type I region, then we would have obtained 2x + 6 / 2x + 6 xy dA = xy dy dx + xy dy dx -V 2x + 6 but this would have involved more work than the other method. Need Help? Talk to a Tutor Read It
Transcribed Image Text:EXAMPLE 3 Evaluate xy dA, where D is the region bounded by the line y = x - 1 and the parabola -3 y2 = 2x + 6. SOLUTION The region D is shown in the figure. Again D is both type I and type II, but the description of D as a type I region is more complicated because the lower boundary consists of two parts. Therefore we prefer to express D as a type II region: -1 {ex.v) |-2 5 y s 4, - 3 sxsy+1}. -6 -2 2 Then this equation gives ху xy dx dy J-2J1/2y² - 3 Video Example ) y +1 dy - ly•ar -( + 1)2 dy + 4y3 + 2y2 - 8y) dy +*+ - ay", If we had expressed D as a type I region, then we would have obtained 2x + 6 / 2x + 6 xy dA = xy dy dx + xy dy dx -V 2x + 6 but this would have involved more work than the other method. Need Help? Talk to a Tutor Read It
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