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 August 4th, 2017, 04:04 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 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: 20,931 Thanks: 2207 You should have got (0, 0) -> (0, 1) and the 4y limit should be 0.
August 4th, 2017, 05:40 PM   #3
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(0, 0) => (0, 0)

u = x
v = y / x

Did you mean v = 0/0 = 1?

Quote:
 Originally Posted by skipjack You should have got (0, 0) -> (0, 1) and the 4y limit should be 0.

 August 4th, 2017, 07:37 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2207 Each point (0, v) (in the u-v 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. Thanks from zollen

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