How to Solve Integrals with Change of Variables and Jacobians?

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The discussion focuses on solving the integral ∫[0 to 2/3]∫[y to 2-2y] (x+2y)e^(y-x) dx dy using the change of variables defined by u=x+2y and v=x-y. The user identifies the triangular region bounded by the lines x=y, y=0, and y=1-(x/2). Despite creating systems with U and V, they struggle to define a new bounded region in the uv-coordinate system, particularly questioning the validity of the transformation.

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I need to find out how to solve this integral with the indicated changes/transformations.

int.[0 to 2/3]int[y to 2-2y]
(x+2y)e^(y-x)
dxdy

u=x+2y v=x-y

I know that the xy region is x=y y=0 and y=1- (x/2)
which is a triangle

so I created systems with U and V but can't get a new bounded region...

I just get u=v and two other lines that intersect at the origin.
If anyone can tell me what's up with this it would be helpfull.

Also, does it seem weird that my given v doesn't match the problem exactly?
the u and v were given changes.
 
Last edited:
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robierob said:
I need to find out how to solve this integral with the indicated changes/transformations.

int.[0 to 2/3]int[y to 2-2y]
(x+2y)e^(y-x)
dxdy

u=x+2y v=x-y

I know that the xy region is x=y y=0 and y=1- (x/2)
which is a triangle

so I created systems with U and V but can't get a new bounded region...

I just get u=v and two other lines that intersect at the origin.
If anyone can tell me what's up with this it would be helpfull.

Also, does it seem weird that my given v doesn't match the problem exactly?
the u and v were given changes.
What does the region look like in the xy-coordinate system?

One line is y= x. If v= x- y, what is v on that line? Another line is y= 1- (1/2)x or 2y= 2- x so x+ 2y= 2. What is u on that line? Finally, the third line is y= 0. In that case, u= x and v= x so y= 0 corresponds to the line u= v. What does that look like in the uv-coordinate system?
 
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Thanks

Thanks, that put me back on track!
 

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