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Math Help - unsolvable in a normal way integral mmn 15 5

  1. #16
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    Quote Originally Posted by transgalactic View Post
    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
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  2. #17
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    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
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  3. #18
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    at the start i cant understand this transition
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  4. #19
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by transgalactic View Post
    at the start i cant understand this transition

    http://www.ucl.ac.uk/~ucahjva/changeint.pdf
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  5. #20
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    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
    Last edited by transgalactic; June 7th 2011 at 01:29 PM.
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  6. #21
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    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.
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  7. #22
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    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
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  8. #23
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    Re: unsolvable in a normal way integral mmn 15 5

    i got it thanks
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