Physical meaning of vector line integration

[A Dhingra]

hello..

I understand what line integral of a real function mean (the area) as well explained in https://www.physicsforums.com/showthread.php?t=115057
(Mentioning this link because there line integral of a real function has been explained.)

I was trying to understand what would the vector line integral graphically mean.

Like evaluate ∫F.dr where F is = x² i+xy j over c which is a square in the z plane bounded by lines x=0,x=a,y=0,y=a in the counterclockwise direction.

The result would be a³/2.

What would this mean graphically??

Please explain

 

[HallsofIvy]

hello..

I understand what line integral of a real function mean (the area) as well explained in https://www.physicsforums.com/showthread.php?t=115057
(Mentioning this link because there line integral of a real function has been explained.)

Area is one common application of the integral. It is NOT the "meaning".

I was trying to understand what would the vector line integral graphically mean.

Like evaluate ∫F.dr where F is = x² i+xy j over c which is a square in the z plane bounded by lines x=0,x=a,y=0,y=a in the counterclockwise direction.

The result would be a³/2.

No, it isn't. It is a²/2. Was that "3" a typo?

What would this mean graphically??

"Graphically", one could interpret it as the area of the region swept out by the vector function as it moves along the curve. In your title, you say "Physical meaning". The most common application is interpreting the vector being integrated as a force vector, calculating the total work done by the force along the path.

Please explain

But I want to emphasize, again, that these are possible "interpretations" or "applications" of the path integral. Mathematical methods do not, in general, have specific graphical or physical "meanings".

 

[A Dhingra]

"Graphically", one could interpret it as the area of the region swept out by the vector function as it moves along the curve.

What about a Vector A = x i + y j and its line integral over the curve c, which is a circle of say radius r centered at the origin? Here (A.dr) will be zero and hence the integral will also be zero. Then interpreting it as the area of the region swept by the vector function moving along the curve does not satisfy this example..

Can you please explain this method of interpretation using a simple example?
(Because i am not able to do it myself..)

 

[Muphrid]

Yeah, to find the area of a curve, you'd have to integrate $r \times dr$, not a dot product.

 

[A Dhingra]

So what would line integration be interpreted as? Just a line...may be

But if i consider a vector A= -y i + x j, around a circle c of radius a centered at the origin, the line integral would yield 2 times the area of the circle...
It seems for different vector functions the line integration yields different interpretation of line integration !

 

[Muphrid]

This should emphasize that integrals are a way of doing a calculation; the "physical interpretation" of an integral depends on the geometry that drives the calculation in the first place. Integrals themselves have no intrinsic interpretation. The idea that 1D integrals calculate areas under curves is the way it's often taught, yes, but only because this is a geometry that requires that integral.

 

[A Dhingra]

So it can't be interpreted? Then what do we want when we are solving a line integral, What should be expected from its outcome?

 

[Muphrid]

Without any other information about what the quantities signify and what geometry is being represented, all that you should expect to get from an integral is a number.

 

[A Dhingra]

I happen to find a demonstration of line integral at the link:
http://demonstrations.wolfram.com/IntegratingAVectorFieldAlongACurve/

And it appears to me as if the graphical aspect or interpretation of line integral of a vector is the area under the boundary of the vector function at all required points associated with the curve c or path a to b..
The area may be negative when the vectors are pointing in the negative direction or may be zero when they cancel out..

So Please have a look at the demonstration and please tell me if i am headed on the right track or am i making some mistake in interpreting it..

 

[A Dhingra]

Am i making some silly error in trying to graphically interpret this?
Someone please tell me whether this is right or not?