Show that Integral K (x^2)dxdy=2, K being parallelogram...

[Zan4]

Whole instruction is:
Show that Integral_K (x^2)dxdy = 2, when K is parallelogram which has its angles (peaks) in (0,0), (-1,-2), (1,-1) and (2,1).
Make substitution x= 2u-v and y= u-2v.

I managed to draw this parallelogram. I have problems with substitution and constructing limits for integration, that is why I cant go on.
Please help me, if you happen to be master of this subject.

 

[JohnZ]

Whole instruction is:
Show that IntegralK (x^2)dxdy = 2, when K is parallelogram which has its angles (peaks) in (0,0), (-1,-2), (1,-1) and (2,1).
Make substitution x= 2u-v and y= u-2v.

I managed to draw this parallelogram. I have problems with substitution and constructing limits for integration, that is why I cant go on.
Please help me, if you happen to be master of this subject.

you want to change variables x, y to u, v.
formula is:
IntegralK (x^2)dxdy = IntegralR (2u-v)^2 |D(x,y)/D(u,v)| dudv
where D(x,y)/D(u,v) is the jacobian of the transformation.

 

[JohnZ]

R is region in uv-plane: R = T (K).
T is the inverse transformation of T1: x = 2u-v, y = u-2v