spliting integral question

**transgalactic** · Aug 13th 2009, 03:36 AM

$ x=\sqrt{2}\cos t\\ $
$ y=\sqrt{2}\sin t \\ $
$ z=\frac{4t}{\pi}\\ $
$ 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 \\ $
$ 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+ $
$ +\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

**NonCommAlg** · Aug 13th 2009, 05:36 AM

Originally Posted by transgalactic

$ 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 $dz$ is not $\frac{xy}{1+x^2y^2z^2}$ instead?

**transgalactic** · Aug 13th 2009, 05:52 AM

Originally Posted by NonCommAlg

check for a possible typo: are you sure that the coefficient of $dz$ is not $\frac{xy}{1+x^2y^2z^2}$ 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?

**NonCommAlg** · Aug 13th 2009, 06:04 AM

Originally Posted by transgalactic

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:

$A=\int_{(0,0,0)}^{(1,1,1)} \frac{d(xyz)}{1+x^2y^2z^2}=\int_{(0,0,0)}^{(1,1,1) } d(\tan^{-1}(xyz))=\tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4}.$

**transgalactic** · Aug 13th 2009, 06:17 AM

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
$ A=\int_{(0,0,0)}^{(1,1,1)} \frac{d(xyz)}{1+x^2y^2z^2}= $
from a splitted 3 integrals
??

**NonCommAlg** · Aug 13th 2009, 06:31 AM

Originally Posted by transgalactic

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 $\int F \cdot dr.$

2.how you made it into one integral
$ A=\int_{(0,0,0)}^{(1,1,1)} \frac{d(xyz)}{1+x^2y^2z^2}= $
from a splitted 3 integrals
??

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

**transgalactic** · Aug 13th 2009, 06:38 AM

Originally Posted by NonCommAlg

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 $\int F \cdot dr.$

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

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

**NonCommAlg** · Aug 13th 2009, 06:41 AM

Originally Posted by transgalactic

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: $d(xyz)=yzdx+xd(yz) = yzdx + x(zdy + ydz)=yzdx + xzdy + xy dz.$

**transgalactic** · Aug 13th 2009, 06:46 AM

Originally Posted by NonCommAlg

it's just the product rule used a couple of times: $d(xyz)=yzdx+xd(yz) = yzdx + x(zdy + ydz)=yzdx + xzdy + xy dz.$

i thought its because of the definition of a deferential

we find the differential of xyz

**NonCommAlg** · Aug 13th 2009, 06:50 AM

Originally Posted by transgalactic

i thought its because of the definition of a deferential

we find the differential of xyz

that is correct too.

**transgalactic** · Aug 13th 2009, 06:55 AM

thanks

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