Line integral of a vector field

[yecko]

Homework Statement

[/B]
I would like to ask for Q5b function G & H

Homework Equations

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

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!

 

[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.

 

[yecko]

to be honest, the solution is below, yet I am not good at reading html codes...

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 learned stroke theorem as it is out of our syllabus, and I have learned till divergence theorem right now)

 

[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.

 

[yecko]

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

 

[Delta2]

Why do you take r to be ##<\frac{1}{9}cost,sint>## that's not the r for the unit circle!

if you take ##r=<cost,sint>## and compute the line integral correctly you end up with an answer of -2pi (I did this at wolfram). The final integral is ##\int_0^{2\pi}\frac{-3sin^2(t)-3cos^2(t)}{9cos^2(t)+sin^2(t)}dt=\int_0^{2\pi}\frac{-3}{9cos^2(t)+sin^2(t)}dt##
http://www.wolframalpha.com/input/?i=integral+((-3sin^2(t)-3cos^2(t))/(9cos^2(t)+sin^2(t))dt)+from+0+to+2pi

 

[yecko]

O thanks