'simplify' integration by transformation: Parallelogram

Question

I have to calculate the volume (?) of the function $f(x,y) = 3\,x-2\,y$ through the area that is spanned by the points $(1,0)\quad (0,1) \quad (-1,0) \quad (0,-1)$

A simple parallelogram. I'd solve it by integrating over one of the 4 rectangles of which it consist: $\displaystyle{\int_0^1\int_0^{1-x}}(3\,x-2\,y)\,\mathrm{dydx} = \frac{2}{12}$. Multiplying by $4$ should give the resultant: $\frac{8}{12}$.

However I wonder if you could also approach this problem by rotating the parallelogram by $45^\circ$ and afterwards integrating. The thing is, I have no clue how to use the integral transformation in order to do so: $\int\int_Rf(x,y)\,\mathrm{dx dy} = \int\int_Sf(g(u,v))\vert\det g'\vert\,\mathrm{du dv}$

I assume it has something to do with x = $r\,\cos(\varphi), \quad y = r\,\sin(\varphi)$, where $\varphi\in[0,\pi/2], \quad r\in[0,1]$, but this seems just like integrating over a quarter of a circle. So, how to apply the formula correctly?

Answer 1

The integral of $(3x-2y)$ over the given square (the region is a square) is zero due to symmetry across x-axis and y-axis and both $x$ and $y$ being odd functions.

You can absolutely rotate and I am sharing a diagram that may help.

Say we rotate the coordinate axes by $45^0$ counterclockwise $x \to X, y \to Y$.

Take example of line $x+y = 1$, which is one of the sides of the square. In new coordinate system, it becomes $Y = \frac{1}{\sqrt2}$ as you can see in the diagram. So you can arrive at the transformation without any additional work.

But in the way you started with polar coordinates,

$x = r \cos \theta, y = r \sin\theta$

Say, $X = r \cos (\frac{\pi}{4} + \theta), Y = \sin (\frac{\pi}{4} + \theta)$

$X = \frac{1}{\sqrt2} (x - y), Y = \frac{1}{\sqrt2} (x + y)$

$x = \frac{X+Y}{\sqrt2}, y = \frac{Y-X}{\sqrt2}$

And the square region transforms to $ - \frac{1}{\sqrt2} \leq X \leq \frac{1}{\sqrt2}$ and $ - \frac{1}{\sqrt2} \leq Y \leq \frac{1}{\sqrt2}$.

Answer 2

Rotating the parallelogram corresponds to the variable changes $x=\frac1{\sqrt2}(u+v) $ and $y=\frac1{\sqrt2}(u-v) $, with $u\in (-\frac1{\sqrt2}, \frac1{\sqrt2})$ and $v\in (-\frac1{\sqrt2}, \frac1{\sqrt2})$. Then, the integral becomes

$$\frac1{\sqrt2}\int_{-\frac1{\sqrt2} }^{\frac1{\sqrt2}}\int_{-\frac1{\sqrt2} }^{\frac1{\sqrt2}}(u+5v)dudv=0 $$