Another quick please tell me if my logic seems correct (change of variables)

  • #1
I'm trying to evaluate the double integral

[tex]\int \int \sqrt{x^2 + y^2} \, dA[/tex]

over the region R = [0,1] x [0,1]
using change of variables.

Well, after fooling around, I've got an answer. I set u = x^2, v =y^2, and then calculated the jacobian of T which was 1. The image transformation limits of integration for u and v turned out to be the same [0,1] x [0,1]

So I did the following calculation (both integrals going from 0 to 1)

[tex]\int \int \sqrt{u + v} * (1) dudv[/tex]


which resulted in a value of roughly 3.238.

Does my logic and answer seem sound here? Thanks in advance.
 

Answers and Replies

  • #2
9
0
Converting to rectangular coordinates would probably be easier
 
  • #3
9
0
That's cylindrical coordinates, sorry
 
  • #4
Answered in Calculus and Analysis.
 
  • #5
I'm totally at a loss here guys. I realized my Jacobian was computed wrong. Can someone please give me a clue as to what would be the most efficient integral setup? I'm completely dumbfounded. :( Thanks


edit: more in-depth post in the calculus forum, thanks
 
Last edited:
  • #7
9
0
Dexter...you need to drop the intensity down a notch. And to the ninja, just convert [tex] x^2 + y^2 [/tex] to [tex] r^2 [/tex] and integrate over the same area in cylindrical.
 
  • #8
dextercioby
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He can't do that,the square [0,1]*[0,1] is not equivalent to the quarter of the circle you're implying...

Daniel.
 

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