Using an appropriate change of variables evaluate double int

Using an appropriate change of variables evaluate double integral_T 9y-3x+6/32 dA where T is the triangle with vertices (1,1), (4,2) and (2,4). Draw the region before and after the transformation.

Solution

u = 3 (x-1) -(y-1)

v = (x-1)-3(y-1)

initial points (1,1),(4,2),(2,4),

after transformation

points (0,0),(8,0),(0,-8)

x = 1/8 (3u -v +8)

y = 1/8 (u-3v +8)

|J| = x u y v x v y u = 1/8

integral become (9 (1/8 (3u -v +8)) -3(1/8 (u-3v +8)) +6 )/32 *1/8 = (3(4-v))/256 as jacobian is 1/8

now integration can be done

u varies from 0 to v-8

v varies from 0 to 8