00:01
The solution is u is equals to x plus y comma v is equals to y minus x.
00:07
So there is integration 0 to 2, integration y, 4 minus y, x plus y is power y minus x dx dy.
00:21
So we can write this equation as x equals to u minus v upon 2 and y is equals to u plus v upon 2.
00:32
So there is the jacobian j is equals to del x comma y upon del u comma v which is equals to, determine it, del x upon del u, del x upon del v, del y upon del u and del y upon del v which is equals to 1 upon 2 minus 1 upon 2, 1 upon 2, 1 upon 2.
01:00
So which is equals to 1 upon 4 plus 1 upon 4 which is equals to 1 upon 2.
01:09
So then there is dx dy is equals to, determine it, jacobian, there is bar, there is mod, dx dy is equals to mod j du dv which is equals to 1 upon 2 du dv.
01:29
So also x equals to 0, so there is u equals to v and v is equals to u.
01:39
There is y equals to 0, so there is u equals to minus v and v is equals to minus u.
01:46
So then we have x equals to, x equals to 4 minus y, so there is x plus y is equals to 4 and u equals to 0, so there is u equals to 4.
02:00
Therefore, integral becomes double integration r u e's power v 1 upon 2 dv du...