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 December 29th, 2016, 09:00 AM #1 Member   Joined: Dec 2015 From: malaysia Posts: 66 Thanks: 1 Evaluate the line integral vector filed (curve) Evaluate the line integral vector filed ∫F dr, where F(x,y,z) = (2xyi + ((x^2) + (z^2))j + 2zyk) and C is the piecewise smooth curve from (1,1,0) to (0,2,3). I am having trouble of finding the r here; since this is a curve, I can't use equation of y = mx + c, so how should I find the r? Thanks from zylo Last edited by skipjack; December 30th, 2016 at 08:47 PM.
 December 29th, 2016, 09:12 AM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 1,238 Thanks: 637 the fact that they don't specify a curve should give you a hint about the nature of your force field.
 December 29th, 2016, 12:45 PM #3 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,056 Thanks: 85 1) Ck that $\displaystyle \int f\cdot dr$ is independent of path. 2) Evaluuate $\displaystyle \int f\cdot dr$ for simplest path. $\displaystyle \int f\cdot dr$ = Xdx+Ydy+Zdz is independent of path if: $\displaystyle f=\triangledown u$ so that $\displaystyle \int f\cdot dr = \int \triangledown u\cdot dr=\int du$ f=$\displaystyle \triangledown u$ if $\displaystyle \frac{\partial Z}{\partial y}=\frac{\partial Y}{\partial z}$, $\displaystyle \frac{\partial X}{\partial z}=\frac{\partial Z}{\partial x}$, $\displaystyle \frac{\partial Y}{\partial x}=\frac{\partial X}{\partial y}$
 December 29th, 2016, 02:52 PM #4 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,056 Thanks: 85 Actually, by inspection, u=$\displaystyle (x^{2}+z^{2})y$. Then $\displaystyle \int f\cdot dr = \oint_{1}^{2} du$
December 29th, 2016, 06:46 PM   #5
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 Originally Posted by xl5899 Evaluate the line integral vector filed ∫F dr, where F(x,y,z) = (2xyi + ((x^2) + (z^2))j + 2zyk) and C is the piecewise smooth curve from (1,1,0) to (0,2,3). I am having trouble of finding the r here; since this is a curve, I can't use equation of y = mx + c, so how should I find the r?
As we've gone through before when a force field is the gradient of a scalar potential function then the line integral through this field is path independent and depends only on the start and end points.

They don't give you a specific curve so your first thought should be that the line integral is path independent and thus the force field must be the gradient of some potential function.

So let's figure out what that potential function is.

$\dfrac{\partial \Phi}{\partial x} = 2 x y$

$\dfrac{\partial \Phi}{\partial y} = x^2 + z^2$

$\dfrac{\partial \Phi}{\partial z} = 2 y z$

by inspection $\Phi(x,y,z) = (x^2 + z^2)y$

and thus

$\displaystyle{\int_C}~f \cdot dr = \Phi(0,2,3) - \Phi(1,1,0)=18-1=17$

Last edited by skipjack; December 30th, 2016 at 08:48 PM.

 December 29th, 2016, 07:22 PM #6 Member   Joined: Dec 2015 From: malaysia Posts: 66 Thanks: 1 Yes, I know that the line integral is conservative, but is it possible to use the traditional method to do this question, which is to find the r first, then get the dr/dt? Last edited by skipjack; December 30th, 2016 at 08:50 PM.
 December 30th, 2016, 05:21 AM #7 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,489 Thanks: 630 If, by "traditional method", you mean writing x, y, and z in terms of parametric equations for the curve (which I would have called "the hard way), you have already been told that the integral is independent of the path. So choose whatever path you want! You can choose the path to be the straight line from (1,1,0) to (0,2,3). Or use a broken line- the line from (1, 1, 0) to (0, 1, 0), then from (0, 1, 0) to (0, 2, 0), then the line from (0, 2, 0) to (0, 2, 3). Thanks from xl5899
December 30th, 2016, 06:18 AM   #8
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Quote:
 Originally Posted by Country Boy If, by "traditional method", you mean writing x, y, and z in terms of parametric equations for the curve (which I would have called "the hard way), you have already been told that the integral is independent of the path. So choose whatever path you want! You can choose the path to be the straight line from (1,1,0) to (0,2,3). Or use a broken line- the line from (1, 1, 0) to (0, 1, 0), then from (0, 1, 0) to (0, 2, 0), then the line from (0, 2, 0) to (0, 2, 3).
From where ? I was told that the field is independent of path ?

December 30th, 2016, 11:31 AM   #9
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Quote:
 Originally Posted by xl5899 From where ? I was told that the field is independent of path ?
you weren't exactly told this but we were able to work out a potential function corresponding to your force field.

If a potential function exists then the line integral is independent of path.

Which you would know if you actually read these responses.

December 30th, 2016, 07:41 PM   #10
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 Originally Posted by romsek you weren't exactly told this but we were able to work out a potential function corresponding to your force field. If a potential function exists then the line integral is independent of path. Which you would know if you actually read these responses.
If we don't want to carry out the test to check whether it's conservative or not, we still can use the traditional method, which is using ∫Fdr, right?

Last edited by skipjack; December 30th, 2016 at 08:51 PM.

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