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1. Let T be the triangular region enclosed by the lines y = 0, y = 2x, x=1. Evaluate the double integral [[(x+y)dA using an iterated integral with: a) y-integration first (Type I); b) x-integration first (Type II)

1. Let T be the triangular region enclosed by the lines y = 0, y = 2x, x=1. Evaluate the double integral [[(x+y)dA using an iterated integral with: a) y-integration first (Type I); b) x-integration first (Type II)

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 39RE
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### Double Integral Evaluation in a Triangular Region **Problem Statement:** 1. Let \( T \) be the triangular region enclosed by the lines \( y = 0 \), \( y = 2x \), and \( x = 1 \). Evaluate the double integral \[ \iint_{T} (x + y) \, dA \] using an iterated integral with: - a) \( y \)-integration first (Type I) - b) \( x \)-integration first (Type II) **1. Type I Region (y-integration first):** For Type I regions, where \( y \) is integrated first, \( y \) typically varies between two functions of \( x \), and \( x \) varies between two constants. In this case: - The region \( T \) is delineated by the lines \( y = 0 \), \( y = 2x \), and \( x = 1 \). - The \( y \)-limits go from \( y = 0 \) to \( y = 2x \). - The \( x \)-limits go from \( x = 0 \) to \( x = 1 \). Thus, the double integral with \( y \)-integration first is formulated as: \[ \iint_{T} (x + y) \, dA = \int_{0}^{1} \left( \int_{0}^{2x} (x + y) \, dy \right) dx \] **2. Type II Region (x-integration first):** For Type II regions, where \( x \) is integrated first, \( x \) typically varies between two functions of \( y \), and \( y \) varies between two constants. In this case: - The region \( T \) is delineated by the lines \( x = \frac{y}{2} \), \( x = 1 \), and \( y = 2x \). - The \( x \)-limits go from \( x = \frac{y}{2} \) to \( x = 1 \). - The \( y \)-limits go from \( y = 0 \) to \( y = 2 \). Thus, the double integral with \( x \)-integration first is formulated as: \[ \iint_{T} (x + y) \, dA = \
Transcribed Image Text:### Double Integral Evaluation in a Triangular Region **Problem Statement:** 1. Let \( T \) be the triangular region enclosed by the lines \( y = 0 \), \( y = 2x \), and \( x = 1 \). Evaluate the double integral \[ \iint_{T} (x + y) \, dA \] using an iterated integral with: - a) \( y \)-integration first (Type I) - b) \( x \)-integration first (Type II) **1. Type I Region (y-integration first):** For Type I regions, where \( y \) is integrated first, \( y \) typically varies between two functions of \( x \), and \( x \) varies between two constants. In this case: - The region \( T \) is delineated by the lines \( y = 0 \), \( y = 2x \), and \( x = 1 \). - The \( y \)-limits go from \( y = 0 \) to \( y = 2x \). - The \( x \)-limits go from \( x = 0 \) to \( x = 1 \). Thus, the double integral with \( y \)-integration first is formulated as: \[ \iint_{T} (x + y) \, dA = \int_{0}^{1} \left( \int_{0}^{2x} (x + y) \, dy \right) dx \] **2. Type II Region (x-integration first):** For Type II regions, where \( x \) is integrated first, \( x \) typically varies between two functions of \( y \), and \( y \) varies between two constants. In this case: - The region \( T \) is delineated by the lines \( x = \frac{y}{2} \), \( x = 1 \), and \( y = 2x \). - The \( x \)-limits go from \( x = \frac{y}{2} \) to \( x = 1 \). - The \( y \)-limits go from \( y = 0 \) to \( y = 2 \). Thus, the double integral with \( x \)-integration first is formulated as: \[ \iint_{T} (x + y) \, dA = \
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