# Double Integral Substitution

#### Imo

The question is Evaluate the double integral over the region R of the function f(x,y)=(x/y -y/x), where R is in the first quadrant, bounded by the curves xy=1, xy=3, x^2 -y^2 =1, x^2-y^2 =4.

Now it seems that a substitution would be the best bet. What I've done is make u=xy, and v=x^2 -y^2. From this, I calculate the Jacobian and get |J|=2(x^2 +y^2). The problem is, I can't then find J in terms of u,v. I've tried many other substitutions, but finding the limits of integration just become much more difficult.

Can anyone either suggest another substitution (right or wrong doesn't matter, I'll check anyways) or give me a hint as to how to find J in terms of u,v? I've been doing this question for about 3-5 hours and I just keep going in circles.

Thank you very much

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#### xanthym

Imo said:
The question is Evaluate the double integral over the region R of the function f(x,y)=(x/y -y/x), where R is in the first quadrant, bounded by the curves xy=1, xy=3, x^2 -y^2 =1, x^2-y^2 =4.
Have you considered applying Green's Theorem (shown below). The Line Integrals are probably easier to compute in this case. (Remember to traverse "C" in the positive sense, that is, counter-clockwise.)
P = y2/(2x)
Q = x2/(2y)
(∂P/∂y) = y/x
(∂Q/∂x) = x/y
∫∫R f(x,y) dx dy = ∫∫R (x/y) - (y/x) dx dy = ∫∫R (∂Q/∂x) - (∂P/∂y) dx dy =
= ∫C P dx + Q dy = ∫C y2/(2x) dx + ∫C x2/(2y) dy

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#### Imo

I don't think Green's Theorem is even needed, considering I haven't learnt it in that class yet. I'm pretty sure there's an appropriate substitution.

#### Imo

thanks xanthym for answering, i figured out the answer a little while ago. If anyone is interested: You just make the substitution I said above, and you can square and take square root x^2 +y^2 and then complete the square to get a value of the Jacobian as a function of u and v. You need an integral table (or a good memory) to do the final integral.

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