unsolvable in a normal way integral mmn 15 5

**transgalactic** · May 24th 2011, 03:24 PM

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

**FernandoRevilla** · May 24th 2011, 10:42 PM

Express

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

where

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

**transgalactic** · May 25th 2011, 02:11 AM

Originally Posted by FernandoRevilla

Express

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

where

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

how to write R1 and R2

**FernandoRevilla** · May 25th 2011, 02:18 AM

Originally Posted by transgalactic

how to write R1 and R2

I don't understand.

**transgalactic** · May 25th 2011, 07:30 AM

i dont know how to continue with what you hve said
dont know how to write R1 and R2 integrals

**FernandoRevilla** · May 25th 2011, 07:58 AM

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

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

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

**transgalactic** · May 25th 2011, 09:11 AM

i cant understand this expression $\frac{{\partial (x,y)}}{{\partial (u,v)}}=-1$

i know for example that:
$\frac{\partial f(x,y)}{\partial x}$
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

**TheEmptySet** · May 25th 2011, 09:13 AM

Originally Posted by transgalactic

i cant understand this expression $\frac{{\partial (x,y)}}{{\partial (u,v)}}=-1$

i know for example that:
$\frac{\partial f(x,y)}{\partial x}$
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

**transgalactic** · May 25th 2011, 09:33 AM

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

**transgalactic** · May 25th 2011, 01:42 PM

how this determinant of what functions??

**FernandoRevilla** · May 25th 2011, 04:01 PM

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.

**transgalactic** · May 25th 2011, 09:36 PM

what this $\frac{{\partial (x,y)}}{{\partial (u,v)}}=-1$
means
what function what determinants
?

**FernandoRevilla** · May 25th 2011, 11:25 PM

$\det \dfrac{{\partial (x,y)}}{{\partial (u,v)}}=\det \begin{bmatrix}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}\end{bmatrix}=\det \begin{bmatrix}{0}&{1}\\{1}&{0}\end{bmatrix}=...=-1$

**transgalactic** · May 27th 2011, 11:33 AM

Originally Posted by FernandoRevilla

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

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

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

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

**FernandoRevilla** · May 27th 2011, 11:51 PM

Originally Posted by transgalactic

on what basis you change the area from R1 to R2 in the integral??

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

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

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

and you didnt use the riven R its between y^2=x and x^2=y

If you mean the given $R$ , yes I did : $R=R_1\cup R_2$ .

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