Question

use the given transformation to evaluate the integral. u=x+y, v=-2x+y; integrateR: 2y dx dy where R is the parallelogram bounded by the lines y=1-x, y=4-x,y=2x+2,y=2x+5.
I have already transformed the region into a square that intersects at (1,2), (4,2), (1,5), (4,5). I need help with setting up the integrals for calculation.

Answer 1

after making the transformations u are asked to do i get this 4 equations corresponding to each of y=mx+b you got:
u=-3/2,u=-12,v=4/3u +2,v=4/3u+5
Can you confirm befor continuing?

Answer 2

I am unsure of how you did that, I haven't had algebra class in quite a few years.

Answer 3

:)
just solve u=x+y, v=-2x+y for x and y and then plug in the results in y=1-x, y=4-x,y=2x+2,y=2x+5
This is no algebra method

Answer 4

I keep getting things like x=y-u and y=v-2x

Answer 5

plug the 1º into the secondand solve for y

Answer 6

so you have y=(y-u)-u

Answer 7

eh??
y = v-2(y-u) = v-2y+2u

Answer 8

but anyway i think this is wrong, becouse i got y=(v-2u)/3, x=(u-v)/3

Answer 9

ok I am with you so far. I still am unsure of why though.

Answer 10

this is for setting comfortably the integration limits

Answer 11

how do I put the other problems into it? would i just use y=2x+2 and put the (y-u) in for x?

Answer 12

now you have this type of region

Answer 13

That is the region I started with.

Answer 14

I know. But that's what your problem whants you to do.
Now you have to find this circled points that intersect horizontal lines with skewed ones. Those will be 3 intervals of integration for u

Answer 15

1º going from left to right should be -21/8.
2º -51/4

Answer 16

i mean values of u

Answer 17

for that points

Answer 18

so new u integration values are from -51/4 to -21/8?

Answer 19

no

Answer 20

yeah i was thinking no.

Answer 21

you need another 2 points