unsolvable in a normal way integral mmn 15 5

**mr fantastic** · May 28th 2011, 01:53 AM

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

You were given good advice in post #11. Follow it.

**transgalactic** · May 28th 2011, 11:51 AM

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

**transgalactic** · May 29th 2011, 08:28 AM

at the start i cant understand this transition

**FernandoRevilla** · May 29th 2011, 08:37 AM

Originally Posted by transgalactic

at the start i cant understand this transition

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

**transgalactic** · June 7th 2011, 01:18 PM

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

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
$R_1=R\cap\{y\geq x\},\;R_2=R\cap\{y< x\}$
because R1 and R2 are on a diiferent axes system

**sakodo** · June 12th 2011, 07:14 AM

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

We can first set up the double integral like this:

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

By changing the order of integration, we get:

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

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

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

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

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

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

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

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

Hope this helps.

**transgalactic** · June 15th 2011, 02:44 PM

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

**transgalactic** · June 15th 2011, 11:38 PM

i got it thanks

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