how to use the clue that f(x,y)=-f(x,y)

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- May 24th 2011, 03:24 PMtransgalacticunsolvable in a normal way integral mmn 15 5
how to use the clue that f(x,y)=-f(x,y)

http://i56.tinypic.com/1zpsmm8.jpg - May 24th 2011, 10:42 PMFernandoRevilla
Express

$\displaystyle \iint_Rf(x,y)\;dxdy=\iint_{R_1}f(x,y)\;dxdy+\iint_ {R_2}f(x,y)\;dxdy$

where

$\displaystyle R_1=R\cap\{y\geq x\},\;R_2=R\cap\{y< x\}$ - May 25th 2011, 02:11 AMtransgalactic
- May 25th 2011, 02:18 AMFernandoRevilla
- May 25th 2011, 07:30 AMtransgalactic
i dont know how to continue with what you hve said

dont know how to write R1 and R2 integrals - May 25th 2011, 07:58 AMFernandoRevilla
Using the substitution $\displaystyle x=v,\;y=u$ the corresponding Jacobian is $\displaystyle \frac{{\partial (x,y)}}{{\partial (u,v)}}=-1$ so,

$\displaystyle \iint_{R_1}f(x,y)\;dxdy=\iint_{R_2}f(v,u)|-1|\;dudv=$

$\displaystyle -\iint_{R_2}f(u,v)\;dudv=-\iint_{R_2}f(x,y)\;dxdy$ - May 25th 2011, 09:11 AMtransgalactic
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 - May 25th 2011, 09:13 AMTheEmptySet
This is the Jacobian matrix

Jacobian matrix and determinant - Wikipedia, the free encyclopedia - May 25th 2011, 09:33 AMtransgalactic
i saw the link and i know what is jacobian

but this is not a determinant

its a fracture of some undefined stuff

and in the link i didnt see it - May 25th 2011, 01:42 PMtransgalactic
how this determinant of what functions??

- May 25th 2011, 04:01 PMFernandoRevilla
- May 25th 2011, 09:36 PMtransgalactic
what this $\displaystyle \frac{{\partial (x,y)}}{{\partial (u,v)}}=-1$

means

what function what determinants

? - May 25th 2011, 11:25 PMFernandoRevilla
$\displaystyle \det \dfrac{{\partial (x,y)}}{{\partial (u,v)}}=\det \begin{bmatrix}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}\end{bmatrix}=\det \begin{bmatrix}{0}&{1}\\{1}&{0}\end{bmatrix}=...=-1$

- May 27th 2011, 11:33 AMtransgalactic
- May 27th 2011, 11:51 PMFernandoRevilla

The domain $\displaystyle R$ is symmetric with respect to $\displaystyle y=x$ .

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and in the end why you go back to x y coordinates withoud multiplying by the yacobian

Try to find the Jacobian of $\displaystyle u=x,v=y$ .

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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 : $\displaystyle R=R_1\cup R_2$ .