# unsolvable in a normal way integral mmn 15 5

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• May 24th 2011, 03:24 PM
transgalactic
unsolvable 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 PM
FernandoRevilla
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 AM
transgalactic
Quote:

Originally Posted by FernandoRevilla
Express

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

where

$\displaystyle R_1=R\cap\{y\geq x\},\;R_2=R\cap\{y< x\}$

how to write R1 and R2
• May 25th 2011, 02:18 AM
FernandoRevilla
Quote:

Originally Posted by transgalactic
how to write R1 and R2

I don't understand.
• May 25th 2011, 07:30 AM
transgalactic
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 AM
FernandoRevilla
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 AM
transgalactic
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 AM
TheEmptySet
Quote:

Originally Posted by transgalactic
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 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
• May 25th 2011, 09:33 AM
transgalactic
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 PM
transgalactic
how this determinant of what functions??
• May 25th 2011, 04:01 PM
FernandoRevilla
Quote:

Originally Posted by transgalactic
how this determinant of what functions??

You should study the double integrals corresponding theory. It is almost useless to select parts of a theory with the only aim of solving a specific problem.
• May 25th 2011, 09:36 PM
transgalactic
what this $\displaystyle \frac{{\partial (x,y)}}{{\partial (u,v)}}=-1$
means
what function what determinants
?
• May 25th 2011, 11:25 PM
FernandoRevilla
$\displaystyle \det \dfrac{{\partial (x,y)}}{{\partial (u,v)}}=\det \begin{bmatrix}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}\end{bmatrix}=\det \begin{bmatrix}{0}&{1}\\{1}&{0}\end{bmatrix}=...=-1$
• May 27th 2011, 11:33 AM
transgalactic
Quote:

Originally Posted by FernandoRevilla
Using the substitution $\displaystyle x=v,\;y=u$ the corresponding Jacobian is $\displaystyle \frac{{\partial (x,y)}}{{\partial (u,v)}}=-1$ so,

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

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

on what basis you change the area from R1 to R2
in the integral
??
and in the end why you go back to x y coordinates withoud multiplying by the yacobian
?

and you didnt use the riven R
its between y^2=x and x^2=y
• May 27th 2011, 11:51 PM
FernandoRevilla
Quote:

Originally Posted by transgalactic
on what basis you change the area from R1 to R2 in the integral??

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

Quote:

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

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

Quote:

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