surface integral using substitution

Question

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

Answer 1

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.

Answer 2

$$0\leqslant x+2y\leqslant 2\: \: \wedge \: \: 0\leqslant x-y\leqslant 1 \iff 0\leqslant x\leqslant \frac43\: \: \wedge \: \: \frac{-x}{2}\leqslant y\leqslant 1+\frac{-x}{2}$$ so first integrate along $y$ like $x$ was a constant and then integrate along $x$.

Answer 3

Since you want to use the transformation $u = x-y$ and $v=x+2y$, the problem is reduced to evaluate $$\int_0^1\int_0^2 \left(\frac{2u+v}{3}\right)^2\frac{1}{3}dudv$$