Substitution in double integral

[DWill]

Homework Statement

Use the transformation u = 3x + 2y and v = x + 4y to evaluate:

The double integral of (3x^2 + 14xy + 8y^2) dx dy for the region R in the first quadrant bounded by the lines y = -(3/2)x + 1, y = -(3/2)x + 3, y = -(1/4)x, and y = (-1/4)x + 1.

Homework Equations

The Attempt at a Solution

I first wrote the equations for x and y in terms of u and v and got:
x = (2u - v)/5, y = (3v - u)/10

Then I solved for the Jacobian and got 1/10. Next I saw that the region R has the following boundaries:
x = 2/3 - (2/3)y
x = 2 - (2/3)y
y = 0

Plugging in the equations I got earlier for x and y, I get the equivalent in terms of u and v:
u = 2
u = 6
v = u/3

So I set up my integral now like this:

Double integral of (u * v) * (1/10) dv du, with 2 <= u <= 6, and 0 <= v <= u/3. Solving this I get the answer 16/9, while the correct answer is 64/5. Which part(s) did I do wrong here? thanks

 

[Defennder]

I didn't check through all your working but it looks like you didn't express x and y in terms of u,v properly.

 

[Nick89]

You don't have to express x and y in terms of u and v.

You can calculate the reciprocal of the 'inverse jacobian' (don't know if it's called that):

\left( \frac{ \partial(u,v)}{\partial(x,y)} \right)^{-1} = \frac{ \partial(x,y)}{\partial(u,v)}

So if you got an equation for u, and one for v, you can calculate \frac{ \partial(u,v)}{\partial(x,y)} and then take it's reciprocal to find the jacobian \frac{ \partial(x,y)}{\partial(u,v)}