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

1. Originally Posted by transgalactic
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

2. ok so we got the area between those to curves
and x=y between them
i cant understand this transition
why you change the area from R1 to R2 ,on what basis
$\displaystyle \iint_{R_1}f(x,y)\;dxdy=\iint_{R_2}f(v,u)|-1|\;dudv$

3. at the start i cant understand this transition

4. Originally Posted by transgalactic
at the start i cant understand this transition

http://www.ucl.ac.uk/~ucahjva/changeint.pdf

5. ok so if we change the variables
$\displaystyle y=x^2$ is now $\displaystyle u=v^2$
and
$\displaystyle y^2=x$ is now $\displaystyle u^2=v$

this is R2

and in the integral of R2 the yacobian is -1
so the integral R1 equals minus the integral over R2

but why the sum of the integrals over R2 and R1 is the integral over R
?

because when we go from one coordinate systen to another out area is described in a totlay new axes
u ,v axes
so we cant do this thing
$\displaystyle R_1=R\cap\{y\geq x\},\;R_2=R\cap\{y< x\}$
because R1 and R2 are on a diiferent axes system

6. Let $\displaystyle f(x,y)=\frac{sin(x)-sin(y)}{xy+1}.$

We can first set up the double integral like this:

$\displaystyle \int\int_{R} f(x,y)dA=\int^{1}_{0}\int^{sqrt(x)}_{x^2} f(x,y)dydx.$

By changing the order of integration, we get:

$\displaystyle \int\int_{R} f(x,y)dA=\int^{1}_{0}\int^{sqrt(y)}_{y^2} f(x,y)dxdy.$

Now, if we change $\displaystyle y$ to $\displaystyle x$ and vice versa on the right hand side, we will still get the same thing:

$\displaystyle \int\int_{R} f(x,y)dA=\int^{1}_{0}\int^{sqrt(x)}_{x^2}f(y,x)dyd x,$

$\displaystyle \int\int_{R} f(x,y)dA=\int\int_{R}f(y,x)dA.$

Thus, $\displaystyle \int\int_{R} f(x,y)dA-\int\int_{R}f(y,x)dA=0.$

Since $\displaystyle f(y,x)=-f(x,y)$, we get

$\displaystyle \int\int_{R} f(x,y)dA+\int\int_{R}f(x,y)dA=0.$

Hence $\displaystyle \int\int_{R} f(x,y)dA=0.$

Hope this helps.

7. ## Re: unsolvable in a normal way integral mmn 15 5

i understand the transtion from x,y to v,u
nut i cant undertand how you changed the untervals for these new variables
sso you got R2 area

8. ## Re: unsolvable in a normal way integral mmn 15 5

i got it thanks

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