Double Integral: Change of Variables for dx/(x+y) = 3

[cos(e)]

Homework Statement

Let D be the region bounded by x=0, y=0, x+y=1, x+y=4. Using the change on variables x=u-uv, y=uv and the jacobian, evaluate the double integral
double integral of dxdy/(x+y)

Homework Equations

answer is 3

The Attempt at a Solution

i drew the graph and found the boundaries x=1,y=0,x+y=1,x+y=4 and solved u and v for each boundary

for x=0: 0=u-uv
v=1

for y=0: 0=uv
so either v=0 or v=0 or both u and v are 0

for x+y=1: u-uv+uv=1
u=1

for x+y=4: u-uv+uv=4
u=4

therefore i have the region in the uv plane [1,4]X[0,1]

now dx=(1-v)du
dy=udv

therefore i get double integral(i-v)dvdu dudv with v varying from 0 to1 and u varying from 1 to 4 and i get 3/2?

im not sure I am doing the right method, i have a feeling I am wrong when i let y=0 and assume v=0?

 

[cos(e)]

ok i forgot bout the jacobian so i tried J=
|delx/delu delx/delv |
|dely/delu dely/delv |
and i got J=u, which gives the right answer, yet I am not sure if I am doing this right i feel i fluked the answer