Double Integral Over a General Region Evaluate the $\iint_R \sin\left(\frac{(x+y)}{2}\right)\cos\left(\frac{(x-y)}{2}\right) dA$ where R is a triangle with vertices $(0,0), (2,0), (1,1)$ using $u = \frac{(x+y)}{2}$ and $v = \frac{(x-y)}{2}$
I had trouble performing the substitution because I was not sure how to find $du$ or $dv$ or if I even needed to to perform the substitution.
Sorry for my poor formatting and any help would be appreciated
 A: Here is how to find the equations for the new region.
The segment connecting the points $(0,0)$ and $(1,1)$ has equation: $y = x$. So this gives:
$x - y = 0$ and so: $v = \dfrac{x-y}{2} = 0$.
The segment connecting the points $(1,1)$ and $(2,0)$ has equation: $x + y = 2$. So:
$u = \dfrac{x+y}{2} = 1$,
and finally the segment connecting the points $(0,0)$ and $(2,0)$ has equation: $y = 0$.
So: $u - v = \dfrac{x+y}{2} - \dfrac{x-y}{2} = y = 0$. In summary your new region is defined in the $u$-$v$ coordinates are:
$v = 0$, $u = 1$, and $u = v$.
A: The technique you need to use isn't as directly related to $u$-substitution as you may be thinking.  Instead, you are supposed to calculate the Jacobian of the change of variables.
Specifically, you want to compute:
$$J = \det\pmatrix{\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}}$$
Then, your integral becomes
$$\iint_G \sin\left(u\right)\cos\left(v\right) |J|dA$$
(Where $G$ is the appropriate transformation of $R$ under the new variables -- see $8\pi r$'s answer for more information regarding that.)
Let me know if you need more guidance.
