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? Last edited by skipjack; December 30th, 2016 at 08:47 PM. 
December 29th, 2016, 09:12 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,653 Thanks: 839 
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: 1,194 Thanks: 91 
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: 1,194 Thanks: 91 
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  
Senior Member Joined: Sep 2015 From: USA Posts: 1,653 Thanks: 839  Quote:
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 $\displaystyle{\int_C}~f \cdot dr = \Phi(0,2,3)  \Phi(1,1,0)=181=17$ 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: 2,875 Thanks: 766 
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). 
December 30th, 2016, 06:18 AM  #8  
Member Joined: Dec 2015 From: malaysia Posts: 66 Thanks: 1  Quote:
 
December 30th, 2016, 11:31 AM  #9 
Senior Member Joined: Sep 2015 From: USA Posts: 1,653 Thanks: 839  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 
Member Joined: Dec 2015 From: malaysia Posts: 66 Thanks: 1  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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curve, evaluate, filed, integral, line, vector 
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