# Thread: unsolvable in a normal way integral mmn 15 5

1. ## unsolvable in a normal way integral mmn 15 5

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

2. Express

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

where

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

3. 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

4. Originally Posted by transgalactic
how to write R1 and R2

I don't understand.

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

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

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

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

7. 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

8. 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

9. 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

10. how this determinant of what functions??

11. 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.

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

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

14. 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

15. 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 $\displaystyle y=x$ .

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

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