ok so if we change the variables
$\displaystyle y=x^2$ is now $\displaystyle u=v^2$
and
$\displaystyle y^2=x$ is now $\displaystyle u^2=v$
this is R2
and in the integral of R2 the yacobian is -1
so the integral R1 equals minus the integral over R2
but why the sum of the integrals over R2 and R1 is the integral over R
?
because when we go from one coordinate systen to another out area is described in a totlay new axes
u ,v axes
so we cant do this thing
$\displaystyle R_1=R\cap\{y\geq x\},\;R_2=R\cap\{y< x\}$
because R1 and R2 are on a diiferent axes system
Let $\displaystyle f(x,y)=\frac{sin(x)-sin(y)}{xy+1}.$
We can first set up the double integral like this:
$\displaystyle \int\int_{R} f(x,y)dA=\int^{1}_{0}\int^{sqrt(x)}_{x^2} f(x,y)dydx.$
By changing the order of integration, we get:
$\displaystyle \int\int_{R} f(x,y)dA=\int^{1}_{0}\int^{sqrt(y)}_{y^2} f(x,y)dxdy.$
Now, if we change $\displaystyle y$ to $\displaystyle x$ and vice versa on the right hand side, we will still get the same thing:
$\displaystyle \int\int_{R} f(x,y)dA=\int^{1}_{0}\int^{sqrt(x)}_{x^2}f(y,x)dyd x,$
$\displaystyle \int\int_{R} f(x,y)dA=\int\int_{R}f(y,x)dA.$
Thus, $\displaystyle \int\int_{R} f(x,y)dA-\int\int_{R}f(y,x)dA=0.$
Since $\displaystyle f(y,x)=-f(x,y)$, we get
$\displaystyle \int\int_{R} f(x,y)dA+\int\int_{R}f(x,y)dA=0.$
Hence $\displaystyle \int\int_{R} f(x,y)dA=0.$
Hope this helps.