• Support PF! Buy your school textbooks, materials and every day products Here!

Change of Variables

  • Thread starter bodensee9
  • Start date
178
0
I am wondering if someone could help me with the following? I am supposed to show that ∫∫f(x+y)dA evaluated from the triangular region with the vertices (0,0), (1,0) and (0,1) is equal to ∫∫uf(u)du.

This triangular region has the equations, x = 0, x = 1, and y = -x + 1. If I set x+y = u, then I know that u must satisfy the equations u = 1, u = 0. But, I’m not sure what to do after that? If someone could give me some hints, that be great. Thanks!!
 

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
41,732
893
I am wondering if someone could help me with the following? I am supposed to show that ∫∫f(x+y)dA evaluated from the triangular region with the vertices (0,0), (1,0) and (0,1) is equal to ∫∫uf(u)du.

This triangular region has the equations, x = 0, x = 1, and y = -x + 1. If I set x+y = u, then I know that u must satisfy the equations u = 1, u = 0. But, I’m not sure what to do after that? If someone could give me some hints, that be great. Thanks!!
No, the triangular region has boundaries x= 0, y= 0, and x+ y= 1. The upper line, x+ y= 1 corresponds to u= 1 but neither x=0 nor y= 0 corresponds to u= 0. In any case, your problem is incomplete. ∫∫f(x+y)dA is a double integral. In ∫∫uf(u)du du is not an area differential. Did you mean show that it is equal to ∫∫uf(u)dudx? If so then the limits of integration are irrelevant. You need only show that dA= dydx= ududx. for some choice of u.
 

Related Threads for: Change of Variables

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
795
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
1
Views
3K
Top