Double integral change of variable

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SUMMARY

The discussion focuses on solving the double integral of xy da over the constraint C defined by the unit circle x^2 + y^2 = 1, using the change of variables x = u^2 - v^2 and y = 2uv. Participants clarify that the transformed equation (u^2 + v^2)^2 = 1 simplifies to u^2 + v^2 = 1, indicating that the domain for u and v remains the unit disk. The conversation highlights the necessity of applying the Jacobian for the transformation, which complicates the integral rather than simplifying it. The substitution's rationale and its implications for the integral's evaluation are also questioned.

PREREQUISITES
  • Understanding of double integrals and their applications
  • Familiarity with change of variables in multivariable calculus
  • Knowledge of Jacobians and their role in transformations
  • Basic concepts of polar coordinates and unit circles
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  • Explore the properties of polar coordinates and their transformations
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Students and educators in calculus, particularly those focusing on multivariable calculus and integral transformations, as well as mathematicians interested in the geometric interpretations of variable changes.

philnow
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Homework Statement



Hey all. The problem is to solve the double integral xy da where the constraints C is x^2 + y^2 = 1, with the change of variables x = u^2 - v^2 and y = 2uv

The problem is applying the change of variables to the constraint unit circle. After the algebra I end up with (u2+v2)^2 = 1. What shape does this represent? How can I find the constraints for U and V given this... odd... equation?
 
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I suppose you mean the domain to by x^2 + y^2 \leq 1, the interior of the circle. That substitution seems strange to me for this problem too. Your equation (u2+v2)2 = 1 is equivalent to (u2+v2) = 1, and the interior of the xy circle maps to the interior of the uv circle. Throw in the Jacobian and it seems like the transformed integral is worse than the original. I am a bit curious where the problem came from and why that substitution is suggested.
 
If (u^2+v^2)^2=1 then u^2+v^2=1. It looks to me like the u,v domain is still the unit disk. In terms of complex numbers it's just the mapping f(z)=z^2. But you don't have to know that.
 

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