Changing Variables in Double Integrals?

Question

I'm trying to solve a problem, but I can't get the right answer. (Hint. It's -3)

Below is the problem.

Double Integral (x-3y)dA, where R is the triangular region with vertices (0,0), (2,1), (1,2); x= 2u +v, y = u+2v

Please help. I'm trying to study for a midterm.

Answer 1

(x,y) = (0,0) maps to (u,v) = (0,0); (x,y) = (2,1) maps to (u,v) = (1,0); (x,y) = (1,2) maps to (u,v) = (1,0).

Since the mapping (x,y) to (u,v) is linear, lines in xy-space map to lines in uv-space. So the region in uv-space is the triangle with vertices (0,0), (1,0), (0,1).

Or:

The boundaries of the region are the lines y=x/2, y=2x, and y=3-x. Using the transformation, y=x/2 maps to u+2v=u + v/2, i.e., v=0. Similarly, y=2x maps to u=0, and y=3-x maps to u+v=1.

The integrand x - 3y becomes (2u + v) - 3(u + 2v) = -u -5v

Finally, you need the Jacobian of the transformation:

| ∂x/∂u ∂x/∂v |
| ∂y/∂u ∂y/∂v |

which is 3, so dxdy = 3dudv

So your integral becomes

integral v=0 to 1 (integral u=0 to (1-v)) of (-u - 5v) 3dudv