Use the transformation T : (u, v) → (x, y) with 1 1 3(u + v), y = 3(v – 2u), to evaluate the integral I = I| (3x +2y) dxdy when D is the region bounded by the lines y = x , Y = x – 2 and y + 2x = 0, y + 2x = 3. 1. I = 4

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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14
2. I =
3
3. I = 5
13
4. I =
3
11
5. I
3
||
Transcribed Image Text:14 2. I = 3 3. I = 5 13 4. I = 3 11 5. I 3 ||
Use the transformation T: (u, v) → (x, y)
with
1
1
3(u + v),
y = 3(v – 2u),
to evaluate the integral
I =
(Зх + 2у) dxdy
D
when D is the region bounded by the lines
y = x ,
y = x – 2
and
y + 2x = 0,
y + 2x = 3.
|
1. I = 4
Transcribed Image Text:Use the transformation T: (u, v) → (x, y) with 1 1 3(u + v), y = 3(v – 2u), to evaluate the integral I = (Зх + 2у) dxdy D when D is the region bounded by the lines y = x , y = x – 2 and y + 2x = 0, y + 2x = 3. | 1. I = 4
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