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Evaluate the integral (x + 4y) dA, where D is the region bound by the parabolas EXAMPLE 1 y = 3x2 and y = 2 + x2. 3- SOLUTION The parabolas intersect when 3x2 = 2 + x2, that is x? = , so x = ±1. We note that the region D, sketched in the figure, is a type I region but not a type II region and we can write >= {(x, v) 3x? sy s2+ D = Since the lower boundary is y = 3x? and the upper boundary is 2 + x2, this equation gives + x? (x + 4y) dy dx (x + 4y) dy dx = -1 1y = 2 + x2 dx Video Example () y = 3x2 + 2(2 + x²)2 – x(3x2) – 2(3x²)2 dx -16x* - 2x3 + 8x2 + 2x + dx - ([

Evaluate the integral (x + 4y) dA, where D is the region bound by the parabolas EXAMPLE 1 y = 3x2 and y = 2 + x2. 3- SOLUTION The parabolas intersect when 3x2 = 2 + x2, that is x? = , so x = ±1. We note that the region D, sketched in the figure, is a type I region but not a type II region and we can write >= {(x, v) 3x? sy s2+ D = Since the lower boundary is y = 3x? and the upper boundary is 2 + x2, this equation gives + x? (x + 4y) dy dx (x + 4y) dy dx = -1 1y = 2 + x2 dx Video Example () y = 3x2 + 2(2 + x²)2 – x(3x2) – 2(3x²)2 dx -16x* - 2x3 + 8x2 + 2x + dx - ([

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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Evaluate the integral (x + 4y) dA, where D is the region bound by the parabolas EXAMPLE 1 y = 3x2 and y = 2 + x2. 3- SOLUTION The parabolas intersect when 3x2 = 2 + x2, that is x? = , so x = ±1. We note that the region D, sketched in the figure, is a type I region but not a type II region and we can write >= {(x, v) 3x? sy s2+ D = Since the lower boundary is y = 3x? and the upper boundary is 2 + x2, this equation gives + x? (x + 4y) dy dx (x + 4y) dy dx = -1 1y = 2 + x2 dx Video Example () y = 3x2 + 2(2 + x²)2 – x(3x2) – 2(3x²)2 dx -16x* - 2x3 + 8x2 + 2x + dx - ([
Transcribed Image Text:Evaluate the integral (x + 4y) dA, where D is the region bound by the parabolas EXAMPLE 1 y = 3x2 and y = 2 + x2. 3- SOLUTION The parabolas intersect when 3x2 = 2 + x2, that is x? = , so x = ±1. We note that the region D, sketched in the figure, is a type I region but not a type II region and we can write >= {(x, v) 3x? sy s2+ D = Since the lower boundary is y = 3x? and the upper boundary is 2 + x2, this equation gives + x? (x + 4y) dy dx (x + 4y) dy dx = -1 1y = 2 + x2 dx Video Example () y = 3x2 + 2(2 + x²)2 – x(3x2) – 2(3x²)2 dx -16x* - 2x3 + 8x2 + 2x + dx - ([
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