In this problem we use the change of variables x = 3s + t, y = s t to compute the integral √(x + y) dA, where R is the parallelogram with vertices (x, y) = (0, 0), (6, 2), (8, 0), and (2,-2). First find the magnitude of the Jacobian, Then, with a = C = √(x + y) dA= fb fd ( , and d = d(x,y) d(s,t) ,b= S+ = t+ ) dt ds =
In this problem we use the change of variables x = 3s + t, y = s t to compute the integral √(x + y) dA, where R is the parallelogram with vertices (x, y) = (0, 0), (6, 2), (8, 0), and (2,-2). First find the magnitude of the Jacobian, Then, with a = C = √(x + y) dA= fb fd ( , and d = d(x,y) d(s,t) ,b= S+ = t+ ) dt ds =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:In this problem we use the change of variables x = = 3s + t, y = s − t to compute the integral √(x + y) dA, where R is the parallelogram
with vertices (x, y) = (0, 0), (6, 2), (8, 0), and (2, −2).
First find the magnitude of the Jacobian,
Then, with a =
C =
√(x + y) dA = √₂ fac
"
and d =
d(x,y)
d(s, t)
,b=
S+
=
t+
) dt ds =
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