# Math Help - change of variables, triangular region

1. ## change of variables, triangular region

so i have the double integral (x-3y)da, and R is a triangular region with vertices (0,0) (1,2) (2,1). transformations are x=2u+v and y=u+2v

So i drew out the triangle and found equations for the 3 lines.
I got:
y=.5x
y=2x
y=-x+3.

then i plugged in my transformations for each of these three equations.
the first equation left me with v=0
The second left me with u=0
the third with u=1-v

So i drew out this new triangle shape.
I want to make sure my bounds are correct, if i'm integrating wrong, i can figure that out.
so integrating with respect to u first, my bounds should be
0<u<1-v
0<v<1
right?

then plug in the transformations for x-3y, and integrate away.
i came out with an answer of -6. but it's supposed to be -3. again, if i integrated wrong, i can figure that out. i've done it a few times and am still not getting -3, so i'm wondering if my bounds are wrong.

2. Hi

I am getting -3
Could you show your work ?
Maybe your problem comes from the Jacobian matrix
Jacobian matrix and determinant - Wikipedia, the free encyclopedia

3. yeah, i got that the matrix came out to be 3. is that correct?

alright, so i must be integrating wrong and i still can't figure it out

so bounds of
0<u<1-v
0<v<1

and plugging in the transformations of x=2u +v and y = u +2v into x-3y, i get
2u+v-3u-6v which is -u -5v.

so integrate that with respect to u, i get
-.5u^2 - 5vu.
use the top bound of 1-v, i get
-.5(1-2v+v^2) - (5v-5v^2)
so i multiplied everything by 2 to get rid of the .5. (and i remember there is a 3 outside the integral from the jacobian matrix)

so i would get 6 outside the integral (bounds of 0 and 1)
and then inside is 1-2v+v^2 - 10v-10v^2
simplify to
1-12v-9v^2.
integrate for v
v-6v^2-3v^3. plug in top bound of 1

1-6-3 = -8. and times the 6 from outside, -48. so not even close... i don't know what i am doing wrong

4. Originally Posted by isuckatcalc
yeah, i got that the matrix came out to be 3. is that correct?

alright, so i must be integrating wrong and i still can't figure it out

so bounds of
0<u<1-v
0<v<1

and plugging in the transformations of x=2u +v and y = u +2v into x-3y, i get
2u+v-3u-6v which is -u -5v.

so integrate that with respect to u, i get
-.5u^2 - 5vu.
use the top bound of 1-v, i get
-.5(1-2v+v^2) - (5v-5v^2)
so i multiplied everything by 2 to get rid of the .5. (and i remember there is a 3 outside the integral from the jacobian matrix)
It's OK up to this point

-.5(1-2v+v^2) - (5v-5v^2) = -.5-4v+4.5v^2 that you integrate from 0 to 1 and multiply the result by 3