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August 4th, 2017, 05:04 PM   #1
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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?
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August 4th, 2017, 06:24 PM   #2
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You should have got (0, 0) -> (0, 1) and the 4y limit should be 0.
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August 4th, 2017, 06:40 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 View Post
You should have got (0, 0) -> (0, 1) and the 4y limit should be 0.
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August 4th, 2017, 08:37 PM   #4
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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.
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