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(1, 1) D' C" R (0, 0) (2, 0) A B' (1, –1) x= (u + v) y = (v – u) u = x - y v =x + y C D B A Figure 16.81 2. 2. EXAMPLE 4 Change of variables determined by the integrand Evaluate х — у х+у+1 - dA, where R is the square with vertices (0, 0), (1, –1), (2, 0), and (1, 1) (Figure 16.81).

(1, 1) D' C" R (0, 0) (2, 0) A B' (1, –1) x= (u + v) y = (v – u) u = x - y v =x + y C D B A Figure 16.81 2. 2. EXAMPLE 4 Change of variables determined by the integrand Evaluate х — у х+у+1 - dA, where R is the square with vertices (0, 0), (1, –1), (2, 0), and (1, 1) (Figure 16.81).

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.8: Determinants
Problem 8E
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(1, 1) D' C" R (0, 0) (2, 0) A B' (1, –1) x= (u + v) y = (v – u) u = x - y v =x + y C D B A Figure 16.81 2. 2.
Transcribed Image Text:(1, 1) D' C" R (0, 0) (2, 0) A B' (1, –1) x= (u + v) y = (v – u) u = x - y v =x + y C D B A Figure 16.81 2. 2.
EXAMPLE 4 Change of variables determined by the integrand Evaluate х — у х+у+1 - dA, where R is the square with vertices (0, 0), (1, –1), (2, 0), and (1, 1) (Figure 16.81).
Transcribed Image Text:EXAMPLE 4 Change of variables determined by the integrand Evaluate х — у х+у+1 - dA, where R is the square with vertices (0, 0), (1, –1), (2, 0), and (1, 1) (Figure 16.81).
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