Change of variables double integral

[eibon]

Homework Statement

Use the transformation [itex]x= \sqrt{v- u}[/itex], y = u + v to evaluate the double integral of [itex]f(x, y) = \frac{x}{(x^2 + y)}[/itex]
over the smaller region bounded by y = x^2, y = 4 − x^2, x = 1.

Homework Equations

The Attempt at a Solution

d:={ (x,y)| [itex]-\sqrt{2}<x<1[/itex] , x^2<y< 4-x^2}

using the jacobian the integral becomes

[tex]\int\int f(x,y)\\da, =\int\int frac{1}{2v}\\dudv[/tex]

[tex]u=\frac{ y-x^2}{2}[/tex]
[tex]v=\frac{y+x^2}{2}[/tex]



but now i am unsure of how to find the limit of integration now

 

[Disinterred]

Hi eibon,

I believe your latex code is not working, since the equations are not showing up nicely for me. I think you should forget about the latex code for now and just write the fractions like x/(x^2 +y), might make it clearer, but maybe my browser is just retarded.

But let me attempt to answer your question. I believe what you need to do here is

1) compute the Jacobian
2) find the new differential area element in the u, v coordinates

both of these I believe you did, but then again the equations are not showing up nicely for me.

Now to get the limits, find what the region bounded by the equations y = x^2, y = 4 − x^2, x = 1. would look like in the u,v coordinate system and then figure out the limits accordingly for u and v. Maybe it is easier than that, perhaps you can find the limits in the x,y coordinate system first and it will then be easy to figure out the new limits in u,v coordinate system by solving the transformation equations.I believe this is how you would go about solving this question. But these are just some ideas to help before someone more knowledgeable answers your question.

 

[HallsofIvy]

The "smaller region" is bounded on the left by x= 1, above by [itex]y= 4- x^2[/itex] and below by [itex]y= x^2[/itex] which cross, of course, at [itex]x= \sqrt{2}[/itex].

With [itex]x= \sqrt{v- u}[/itex] and [itex]y= u+ v[/itex], [itex]y= x^2[/itex] becomes [itex]u+ v= v- u[/itex] which reduces to u= 0. Similarly, [itex]y= 4- x^2[/itex] becomes [itex]u+ v= 4- v+ u[/itex] which reduces to v= 2. Finally, x= 1, which is the same as [itex]x^2= 1[/itex], becomes v- u= 1 or v= u+ 1 which intersects u= 0 at (0, 1) and v= 2 at (1, 2). Your region of integration is the triangle with vertices (0, 1), (1, 2) and (0, 2) in the uv-plane.