# Line integral of a vector field

1. Nov 29, 2017

### yecko

1. The problem statement, all variables and given/known data

I would like to ask for Q5b function G & H

2. Relevant equations
answer: G: -2pi H: 0
by drawing the vector field

3. The attempt at a solution
the solution is like: by drawing the vector field, vector field of function G is always tangential to the circle in clockwise direction, so line integral is -2pi; vector field of function H is always perpendicular to the circle, so line integral is 0.

the definition of flux is the vector field crosses the designated area, "perpendicular" should be crosses while "tangential" is just touching the circle, isn't it?

I am now really confused with how can we visualize the equations with their physical meaning behind in order to obtain the answer... can anyone help me in clarifying the concept?

thank you so much!

2. Nov 29, 2017

### Delta²

Answer for G must be wrong, Curl(G)=0 everywhere in the x-y plane, hence by stokes theorem the line integral around any curve in the x-y plane is zero.

If you haven't been taught stokes theorem yet, then there is another way to see it. If from the answer in 5(a) the potential function for G is well defined in the unit circle of the xy-plane then by using the gradient theorem you can easily see that the answer will be zero. I think all the 3 answers for 5(b) should be zero.

3. Nov 29, 2017

### yecko

to be honest, the solution is below, yet I am not good at reading html codes...
Code (Text):

Context()->texStrings;
SOLUTION(EV3(<<'END_SOLUTION'));
$PAR SOLUTION$PAR

$BBOLD(a)$EBOLD
We see that $$\vec F,\vec G,\vec H$$ are all gradient vector fields, since
$\nabla (f) = \vec F\quad\mbox{for all }x,y,$
$\nabla (g) = \vec G\quad\mbox{except where }y=0,$
and
$\nabla (h) = \vec H\quad\mbox{except at }(0,0).$
Other answers are possible.

$PAR$BBOLD(b)$EBOLD Parameterizing the unit circle, $$C$$, by $$x=\cos t$$, $$y=\sin t$$, $$0\le t\le 2\pi$$, we have $$\vec r'(t) =-\sin t\,\vec i+ \cos t\,\vec j$$, so $\int_C\vec F\cdot d\vec r = a1d\int_0^{2\pi}((\sin t)\vec i+(\cos t)\vec j)\cdot ((-\sin t)\vec i+(\cos t)\vec j)\,dt = a1d\int_0^{2\pi}\cos (2t)\,dt = 0.$ The vector field $$\vec G$$ is tangent to the circle, pointing in the opposite direction to the parameterization, and of length$a2
everywhere.  Thus
$\int_C\vec G\cdot d\vec r= -a2\cdot\hbox{ Length of circle }=-twoa2\pi.$
The vector field $$\vec H$$ points radially outward, so it is
perpendicular to the circle everywhere.  Thus
$\int_C\vec H\cdot d\vec r = 0.$

$PAR${BBOLD}(c)\$EBOLD
Green's Theorem does not apply to the computation of the line
integrals for $$\vec G$$ and $$\vec H$$ because their domains do not
include the origin, which is in the interior, $$R$$, of the circle.
Green's Theorem does apply to $$\vec F=fx\vec i+fy\vec j$$.
$\int_C\vec F\cdot d\vec r = \int_R\left(\frac{\partial}{\partial x} F_2 - \frac{\partial}{\partial y} F_1\right)\,dx\,dy = \int 0\,dx\,dy=0.$

END_SOLUTION

from my notes, the numerical answer have got the similar answer, but I am not sure the reason why...
(i have not learnt stroke theorem as it is out of our syllabus, and I have learnt till divergence theorem right now)

4. Nov 30, 2017

### Delta²

Ok well, seems I am wrong, it says that the line integral of G is as you said 2pi. And it also says that Green's Theorem (which is special case of Stoke's Theorem in 2 dimensions) doesn't apply for G and H (because G and H are not defined at (0,0)).

I guess gradient theorem also doesn't hold for G because it says the gradient of g with $\nabla{g}=G$ isn't defined for y=0.

5. Nov 30, 2017

### yecko

I have tried to evaluate it once more, yet the answer is still wrong... how should i deal with it?

6. Nov 30, 2017

### Delta²

Last edited: Nov 30, 2017
7. Nov 30, 2017

### yecko

O thanks

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