Change of variables in a double integral

tjackson3
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Homework Statement



Find the mass of the plane region R in the first quadrant of the xy plane that is bounded by the hyperbolas [itex]xy=1, xy=2, x^2-y^2 = 3, x^2-y^2 = 5[/itex] where the density at the point x,y is [itex]\rho(x,y) = x^2 + y^2.[/itex]

Homework Equations





The Attempt at a Solution



The region of integration lends itself to the change of variables [itex]u = xy, v = x^2-y^2.[/itex] However, if I make this change of variables, it seems impossible to solve for x and y. Is there a better change of variables to make?
 
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tjackson3 said:

Homework Statement



Find the mass of the plane region R in the first quadrant of the xy plane that is bounded by the hyperbolas [itex]xy=1, xy=2, x^2-y^2 = 3, x^2-y^2 = 5[/itex] where the density at the point x,y is [itex]\rho(x,y) = x^2 + y^2.[/itex]

Homework Equations





The Attempt at a Solution



The region of integration lends itself to the change of variables [itex]u = xy, v = x^2-y^2.[/itex] However, if I make this change of variables, it seems impossible to solve for x and y. Is there a better change of variables to make?
Why do you want to solve for x & y ?
 
At the very least, I need to solve for [itex]x^2+y^2[/itex]
 
tjackson3 said:
At the very least, I need to solve for [itex]x^2+y^2[/itex]

Square v, that gives you x4 - 2x2y2 + y4

If you add 4x2y2 to that you will have x4 + 2x2y2 + y4 .

Does that help ?
 
Very much. Thank you!
 
By the way:

If you use the change of variables u=2xy, v=x2−y2 , then u2 + v2 = (x2 + y2)2 , which is a bit nicer.

The only reason I was able to help so quickly, was that I recently helped with a problem having a similar change of variable.
 
Ah that is much nicer. Haha don't be modest now! Thanks again!
 

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