August 4th, 2017, 04:04 PM  #1 
Member Joined: Jan 2017 From: Toronto Posts: 97 Thanks: 2  Changing of variables issues
Evaluate $\displaystyle \int_{0}^{4} \int_{y}^{4} \sqrt{ x^{2} + y^{2} } dx dy $ over the triangle with corners (0, 0), (4, 4), and (4, 0) using x = u, y = uv. Answer with direct integral: 24.4862629 Here is my solution for changing the variables to u and v (x, y) => (u, v)  (0, 0) => (0, 0) (4, 4) => (4, 1) (4, 0) => (4, 0) u = x v = y / x $\displaystyle \int_{0}^{1} \int_{4y}^{4} u^{2} \sqrt{ 1 + v^{2} } du dv $ But the above answer is 17.6192, which is not correct. Any idea why I got it wrong? 
August 4th, 2017, 05:24 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 17,725 Thanks: 1359 
You should have got (0, 0) > (0, 1) and the 4y limit should be 0.

August 4th, 2017, 05:40 PM  #3 
Member Joined: Jan 2017 From: Toronto Posts: 97 Thanks: 2  
August 4th, 2017, 07:37 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 17,725 Thanks: 1359 
Each point (0, v) (in the uv plane) where 0 < v < 1 corresponds to (x, y) = (0, 0), so the original triangle has become a rectangle. I should have pointed this out instead of just stating (0, 0) > (0, 1). Your slip was understandable (though even given your approach, using 4y doesn't make sense). It can be easy to get the new limits wrong  that's why you were given a slightly tricky example. 

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