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Thread: spliting integral question

  1. #1
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    spliting integral question

    $\displaystyle
    x=\sqrt{2}\cos t\\
    $
    $\displaystyle
    y=\sqrt{2}\sin t \\
    $
    $\displaystyle
    z=\frac{4t}{\pi}\\
    $
    $\displaystyle
    A=\int^{(1,1,1)}_{(0,0,0)}\frac{yz}{1+x^2y^2z^2}dx +\frac{xz}{1+x^2y^2z^2}dy+\frac{xy}{1+x^2y^2z^2}dz \\
    $
    $\displaystyle
    A=\int^{(1)}_{(0)}\frac{\sqrt{2}\sin tz}{1+(\sqrt{2}\cos t)^2(\sqrt{2}\sin t)^2z^2}(dx)+\frac{xz}{1+x^2(\sqrt{2}\sin t)^2z^2}0+
    $
    $\displaystyle
    +\frac{xy}{1+x^2(\sqrt{2}\sin t)^2z^2}0
    $
    i dont know if i am doing it the right way
    becausei got a very complexed expression and this only one third
    Last edited by transgalactic; Aug 13th 2009 at 05:53 AM.
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    $\displaystyle
    A=\int^{(1,1,1)}_{(0,0,0)}\frac{yz}{1+x^2y^2z^2}dx +\frac{xz}{1+x^2y^2z^2}dy+\frac{xz}{1+x^2y^2z^2}dz \\
    $
    check for a possible typo: are you sure that the coefficient of $\displaystyle dz$ is not $\displaystyle \frac{xy}{1+x^2y^2z^2}$ instead?
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    check for a possible typo: are you sure that the coefficient of $\displaystyle dz$ is not $\displaystyle \frac{xy}{1+x^2y^2z^2}$ instead?
    yeas you are right i change it there

    and it doesnt realy matter
    because i am dealing with walking on the x axes
    and i get this monster integral

    is there a short cut or someting?
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    yeas you are right i change it there

    and it doesnt realy matter
    because i am dealing with walking on the x axes
    and i get this monster integral

    is there a short cut or someting?
    actually it does matter very much! and who said you're working on the x axes?!! anyway, now that you corrected your typo, the problem is quite easy:

    $\displaystyle A=\int_{(0,0,0)}^{(1,1,1)} \frac{d(xyz)}{1+x^2y^2z^2}=\int_{(0,0,0)}^{(1,1,1) } d(\tan^{-1}(xyz))=\tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4}.$
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  5. #5
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    how you done it?
    there is tree integrals
    in each one we go by one axes
    first we got from 0 to 1 by x
    so in his integral dy and dz are 0 because there is no moving on y and z axes
    thats how i was tought to think.
    1.is this a correct way??

    2.how you made it into one integral
    $\displaystyle
    A=\int_{(0,0,0)}^{(1,1,1)} \frac{d(xyz)}{1+x^2y^2z^2}=
    $
    from a splitted 3 integrals
    ??
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    how you done it?
    there is tree integrals
    in each one we go by one axes
    first we got from 0 to 1 by x
    so in his integral dy and dz are 0 because there is no moving on y and z axes
    thats how i was tought to think.
    1.is this a correct way??
    no! not even close! your integral is over a curve which passes through (0,0,0) and (1,1,1). you need to review what you've been taught about integrals in the form $\displaystyle \int F \cdot dr.$


    2.how you made it into one integral
    $\displaystyle
    A=\int_{(0,0,0)}^{(1,1,1)} \frac{d(xyz)}{1+x^2y^2z^2}=
    $
    from a splitted 3 integrals
    ??
    very simple: $\displaystyle yzdx + xzdy + xydz=d(xyz).$
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  7. #7
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    Quote Originally Posted by NonCommAlg View Post
    no! not even close! your integral is over a curve which passes through (0,0,0) and (1,1,1). you need to review what you've been taught about integrals in the form $\displaystyle \int F \cdot dr.$



    very simple: $\displaystyle yzdx + xzdy + xydz=d(xyz).$

    if you do a derivative by x you get yz etc..
    so d(xyz) is the derivative of xyz by x +derivative by y of xyz +derivative by z of xyz
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  8. #8
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    Quote Originally Posted by transgalactic View Post
    if you do a derivative by x you get yz etc..
    so d(xyz) is the derivative of xyz by x +derivative by y of xyz +derivative by z of xyz
    it's just the product rule used a couple of times: $\displaystyle d(xyz)=yzdx+xd(yz) = yzdx + x(zdy + ydz)=yzdx + xzdy + xy dz.$
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  9. #9
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    Quote Originally Posted by NonCommAlg View Post
    it's just the product rule used a couple of times: $\displaystyle d(xyz)=yzdx+xd(yz) = yzdx + x(zdy + ydz)=yzdx + xzdy + xy dz.$
    i thought its because of the definition of a deferential

    we find the differential of xyz
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  10. #10
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    Quote Originally Posted by transgalactic View Post
    i thought its because of the definition of a deferential

    we find the differential of xyz
    that is correct too.
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  11. #11
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    thanks
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