calculusmultivariable-calculus
Use the change of variables formula and an appropriate transformation to evaluate $\int \int_R xy \ dA$ , where R is the square with vertices $(0,0), (1,1),(2,0)$ and $(1,-1)$.
The answer is 0.
Can I do this ? Why ?
$$ \int_0 ^2 \int_{-1} ^1 xy \ dy \ dx = 0$$
I am adding some points for the first one so you can find the limits easily. Try to do the rest by yourself. :)
If we consider the 4 points above and mark them on $xy$ plane, we have the area as you see below:
Now, as you probably know, we find $x$ and $y$ respect to $u$ and $v$ as we are indicated. So $$u=y-x,v=x+y,~~~\to y=\frac{u+v}2,~~x=\frac{v-u}2~~~~~~~(I)$$ For finding the limits for $u$ and $v$ in $uv$ plane do as follows and by $I$: $$(1,0)\longrightarrow u=-1,v=1\\\\\ (2,0)\longrightarrow u=-2,v=2\\\\\ (0,1)\longrightarrow u=1,v=1\\\\\ (0,2)\longrightarrow u=2,v=2\\\\\ $$ This means that you have a new region like:
Obviously $$-1\le v\le +1, ~~~-v\le u\le v$$
Let the variable changes (equivalent to 45-degree rotation),
$$x = \frac 1{\sqrt2} (u-v),\>\>\>\>\>y=\frac 1{\sqrt2}(u+v)$$
The integrand becomes,
$$\frac{1}{4}(y^2-x^2)=\frac12 uv$$
and the three vertexes are instead $(0,0)$, $(2\sqrt2,0)$ and $(2\sqrt2,\sqrt2)$. The corresponding area in the $uv$-plane is confined by $v=0$, $u=2\sqrt2$ and the line $v= \frac12 u$. The area integral becomes,
$$\frac12\int_0^{2\sqrt2}\int_0^{\frac12u}uv\>dvdu=1$$
Best Answer
You are not integrating the region $R$. The idea of the exercise is to deform $R$ into a square with sides parallel to the axis in the plane by a counterclockwise rotation of an angle equal to $\frac{\pi}{4}$. The required transformation is then
$$(x,y)\mapsto (u,v)=\frac{\sqrt{2}}{2}(x-y,x+y).$$
To finish the proof you need the formula for the change of variables in multiple integrals recalling that the original integrand
$$f(x,y)=xy$$
in the new coordinates reads (I inverted the coordinate transformation given above)
$$\tilde{f}(u,v)=(u+v)(v-u)$$
Computing the determinant of the Jacobian of the coord. transofrmation the exercise is finished.