surface integral using substitution I am stuck trying to calculate the following surface integral:
$$\int _{R}\int (x+y)^{2}ds$$
over the the following regions:
$$0\leqslant x+2y\leqslant 2\: \: \wedge \: \: 0\leqslant x-y\leqslant 1$$
My professor suggested us to use the following substitutions:
$$u = x -y \: \: \wedge v = x+2y$$
I have done the following in order to find ds
$$\frac{du}{ds}= \frac{dx}{ds}-\frac{dy}{ds} \: \: \wedge \frac{dv}{ds}= \frac{dx}{ds}-2\frac{dy}{ds}
$$
 but it does not work :(
please help
 A: For any diffeomorphism $\phi \colon A \to B$ on open sets $A,B\subset \mathbb{R}^2$  you get
$$ \int_B f(x,y) \,d(x,y) = \int_A f(\phi(x,y)) \, |det D\phi| \, d(x,y).$$
If you choose 
$$A := \{ (x,y) \mid 0 < x +2y < 2 ,\, 0 < x-y < 1 \} $$
$$ \phi(x,y) := (x+2y,x-y)$$
$$B := (0,2) \times (0,1)$$
$$ f(x,y) := \left(\frac{y+2x}{3}\right)^2$$
and calculate $ |det D\phi| = 3$ you obtain
 $$\int_0^1 \int_0^ 2 \left(\frac{y+2x}{3} \right)^2 \frac{1}{3} \,dx \,dy= \int_A (x+y)^2 \, dxdy$$ 
The only difficulty was to find a function $f$ with $f(\phi(x,y)) = (x+y)^2$.
Note that $A$ and your set $R$ only differ in a zero set.
A: Since you want to use the transformation $u = x-y$ and $v=x+2y$, the problem is reduced to evaluate $$\int_0^1\int_0^2 \left(\frac{2u+v}{3}\right)^2\frac{1}{3}dudv$$
A: $$0\leqslant x+2y\leqslant 2\: \: \wedge \: \: 0\leqslant x-y\leqslant 1 \iff 0\leqslant x\leqslant \frac43\: \: \wedge \: \: \frac{-x}{2}\leqslant y\leqslant 1+\frac{-x}{2}$$
so first integrate along $y$ like $x$ was a constant and then integrate along $x$. 
