# tough double integral, involves transformation

• November 26th 2008, 08:22 PM
minivan15
tough double integral, involves transformation
use the change of variables x = (u-v)/SQRT(2), y = (u+v)/SQRT(2) to evaluate
[integral from 0 to 1] [integral from 0 to 1] 1/(1-xy) * dxdy
i am given the hint to use the identity (1 - sint)/cost = cost/(1+sint) = tan(pi/4 -t/2)
so i need to prove the identities (have done so...no problem with that so i wont post it), transform the integral into terms of u and v, and then integrate!
the domain in xy is a square in the first quadrant, vertices (0,0) (0,1) (1,0) and (1,1)
let L1 be the bottom edge of the square
y = 0, 0 <= x <=1
i.e. v = -u, 0 <= u <= 1/SQRT(2)
-----------------------------------------------
let L2 be the right edge of the square
0 <= y <= 1, x = 1
i.e. v = u - SQRT(2), 1/SQRT(2) <= u <= SQRT(2)
-----------------------------------------------
let L3 be the top edge of the square
y = 1, 0 <= x <=1
i.e. v = SQRT(2) - u, 1/SQRT(2) <= u <= SQRT(2)
------------------------------------------------
let L4 be the left edge of the square
0 <= y <= 1, x = 0
i.e. v = u, 0 <= u <= 1/SQRT(2)
-------------------------------------------------
so when all is said and done, the transformation from the domain D relating to xy to the domain T relating to uv is a square in the uv plane with vertices (0,0) (1/SQRT(2), 1/SQRT(2)) ((SQRT(2), 0) and (1/SQRT(2), -1/SQRT(2))
also the Jacobian is 1
please if you can double check this, as I'm not positive I got it right
now, from here I can't figure out how to evaluate the integral! Please help if you can, thanks!!
• November 27th 2008, 03:07 PM
Krizalid
I think a better substitution would be $(u,v)=\left( \frac{x+y}{2},\frac{y-x}{2} \right)$ and that will transfer your unit square into a square with coordinates $(0,0),\,(1,0),\,\left( \frac{1}{2},-\frac{1}{2} \right),\,\left( \frac{1}{2},\frac{1}{2} \right).$

And please, try to use some of LaTeX, it's really hard to read a post as you wrote it.