how to use the clue that f(x,y)=-f(x,y)
Using the substitution $\displaystyle x=v,\;y=u$ the corresponding Jacobian is $\displaystyle \frac{{\partial (x,y)}}{{\partial (u,v)}}=-1$ so,
$\displaystyle \iint_{R_1}f(x,y)\;dxdy=\iint_{R_2}f(v,u)|-1|\;dudv=$
$\displaystyle -\iint_{R_2}f(u,v)\;dudv=-\iint_{R_2}f(x,y)\;dxdy$
i cant understand this expression $\displaystyle \frac{{\partial (x,y)}}{{\partial (u,v)}}=-1$
i know for example that:
$\displaystyle \frac{\partial f(x,y)}{\partial x}$
means the partial derivetive of "f" by x
but here the is no clear structure of what is the function
and by what variable we derive it by
This is the Jacobian matrix
Jacobian matrix and determinant - Wikipedia, the free encyclopedia
$\displaystyle \det \dfrac{{\partial (x,y)}}{{\partial (u,v)}}=\det \begin{bmatrix}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}\end{bmatrix}=\det \begin{bmatrix}{0}&{1}\\{1}&{0}\end{bmatrix}=...=-1$
The domain $\displaystyle R$ is symmetric with respect to $\displaystyle y=x$ .
and in the end why you go back to x y coordinates withoud multiplying by the yacobian
Try to find the Jacobian of $\displaystyle u=x,v=y$ .
and you didnt use the riven R its between y^2=x and x^2=y
If you mean the given $\displaystyle R$ , yes I did : $\displaystyle R=R_1\cup R_2$ .