Francisco P. answered 10/16/14
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Rigorous Physics Tutoring
Let x = 2u and y = 2v.
Then (1/4) ∫∫ √(x2 + y2 + 1) dx dy is your transformed integral. Looking at this you might think of polar coordinates.
Then using polar coordinates:
x2 + y2 = r2 , dx dy = r dr dθ.
(1/4) ∫∫ √(r2 + 1) r dr dθ
This looks a lot better.
Now, transform your limits using x = r cos(θ) and y = r sin(θ). Good luck.
Paul S.
Thanks for the help. I follow your thought process in changing the integrand to polar coordinates, but I still am having an issue in properly changing the limits of integration. How would I go about doing so correctly?
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10/16/14
Francisco P.
Paul,
You will have to break the area integral into two integrals because the area of integration is a rectangle.
The limits for one integral would be from 0 to tan-1(2) for θ, 0 to 2sec(θ) for r.
The limits for the other integral would be from tan-1(2) to π/2 for θ, 0 to 4csc(θ) for r.
Yohan,
I will post what I get this evening.
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10/17/14
Francisco P.
1/24 [16√21 -7 ln(5)+14 ln(4+√21)-2 tan-1(8/√21)+76 sinh-1(2/√17)]
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10/17/14
Yohan C.
10/16/14