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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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Joined: Jan 2017
From: Toronto

Posts: 209
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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 Tags changing, issues, variables Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zollen Calculus 5 August 2nd, 2017 08:46 PM fixxit Algebra 4 September 10th, 2014 11:29 AM CherryPi Calculus 10 April 21st, 2012 10:23 PM Brightstar Geometry 3 January 19th, 2012 12:39 PM isaace Calculus 6 November 28th, 2008 12:59 PM

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