Line integral of a vector field.

In summary, the problem is that the person is confused about how to do line integrals and is looking for help. They say that they could do part a ok, but part b has them stumped. They are dyslexic, so their handwriting is bad. They try to do the problem using a technique that they found on the internet, but they do not understand why it works. They eventually figure out that the path is not important, and that the only thing that matters is the curl being zero.
  • #1
Jack_O
65
0
Hi all, I'm new to the forums so if i do something stupid don't hesitate to tell me.

Anyway I'm struggling with this problem:

mathsproblem-1.jpg


I could do part a ok, but part b has me stumped, I am in the second year of a physics degree and this is a from a maths problem sheet, i haven't done line integrals before now and they have me a bit confused, my textbook has a few examples but none of them include vectors and http://en.wikipedia.org/wiki/Line_integral" has me even more confused.

Here is my attempt so far:

mathsproblemattempt.jpg

(please excuse bad handwriting, i am dyslexic)

I basically don't know where to begin with it, any help much appreciated.
 
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  • #2
Use that along your path y=x^3. So r=x*i+x^3*j. (i=x hat and j=y hat). What is the vector dr? Change the y's in V to x^3's as well and integrate dx.
 
  • #3
I think dr is just a small difference in the postion vector r, so dr = dx*i+dy*j ? r isn't mentioned anywhere else in the question.
 
  • #4
Jack_O said:
I think dr is just a small difference in the postion vector r, so dr = dx*i+dy*j ? r isn't mentioned anywhere else in the question.

Sure it is. But now let's decide to integrate dx. Replace y with x^3. Now what's dr?
 
  • #5
dr=x*i +(x^3)j

I also have V(r)=(2x-x^6)i + [(6x^6)-(2x^4)]j

But i am a bit unsure about V.dr, is it {[ (2x^2)-(x^7) ]i + [ (6x^9)-(2x^7) ]j}dx ?
 
  • #6
What happened to the 'd's on the right side of dr? You've got y*j, shouldn't it be d(y)*j? Same for i. Let's get dr right before going to the dot product.
 
  • #7
OK, i jumped the gun a bit with dr, is it d(r) = d(x)*i + d(y)*j = d(x)*i + d(x^3)*j ?
 
  • #8
Right. And d(x^3)=3*x^2*dx, right? So dr=(i+3x^2*j)dx.Now do the dot product and you'll get a single integral dx.
 
  • #9
Ok, V.dr=[(2x-x^6)+(18x^8)-(6x^6)]dx

When i integrate this with the limits x1=0 and x2=1 i get 2, does this sound about right?
 
  • #10
Jack_O said:
Ok, V.dr=[(2x-x^6)+(18x^8)-(6x^6)]dx

When i integrate this with the limits x1=0 and x2=1 i get 2, does this sound about right?

That's what I got. Not so hard, hmm?
 
  • #11
Dick said:
That's what I got. Not so hard, hmm?

Yeah thanks that's been a great help.

Part c seems trivial now but part d seems a bit daunting still.
 
  • #12
Don't be daunted. You know dphi/dx=2x-y^2 (partial derivative). What does that tell you about phi?
 
  • #13
Hmm, after much head scratching i think what you are getting at is that [tex]\nabla[/tex](phi(x,y))=(dphi/dx)i+(dphi/dy)j, which means:

phi(x,y)=[int(2x-y^2)dx] + [int((6y^2)-2xy)]

which gives:

phi=x2 + 2y3 -xy2

Which does seem to give 2 when i do phi(1,1) - phi(0,0)
 
  • #14
The answer is right. Since dphi/dx=2x-y^2, that means phi=x^2-x*y^2+f(y). Yeah, that is 'integral dx' with f(y) being the constant of integration. That gives dphi/dy=-2xy+f'(y). Since you are supposed to get 6y^2-2xy, you can figure that the f'(y) part must be the 6y^2, so if you put it all together, phi=x^2-xy^2+2y^3.
 
  • #15
Jack_O said:
Hmm, after much head scratching i think what you are getting at is that [tex]\nabla[/tex](phi(x,y))=(dphi/dx)i+(dphi/dy)j, which means:

phi(x,y)=[int(2x-y^2)dx] + [int((6y^2)-2xy)]

which gives:

phi=x2 + 2y3 -xy2

Which does seem to give 2 when i do phi(1,1) - phi(0,0)

I'm not taking your int()+int() formula literally. If I did I would get a -xy^2 term from each one of those. For a total of -2xy^2, which is not right.
 
  • #16
Well I've managed to do the problem, and i could probably do others now using that technique, but I'm not sure i understand why it works, i mean in questions b, c and d the only constant is x1=0, y1=0, x2=1 and y2=1, why doesn't the chosen path have an effect, is it something to do with the curl being 0?
 
  • #17
It has everything to do with the curl being zero. It's one of the conditions you need for V to be conservative. 'Conservative' means that the line integral between two points is independent of the path chosen. If they hadn't given you a conservative V, then the answer would depend on the path.
 
  • #18
Ok i get it now! Thanks for your help, i wouldn't have been able to do this much without it.
 

FAQ: Line integral of a vector field.

What is a line integral of a vector field?

A line integral of a vector field is a mathematical concept used to calculate the work done by a vector field along a given path. It involves integrating the dot product of the vector field and the direction of the path over the length of the path.

What is the significance of line integrals in science?

Line integrals are used in various fields of science and engineering to calculate quantities such as work, force, and energy. They are particularly useful in fields such as physics, electromagnetism, and fluid mechanics.

How do you calculate a line integral of a vector field?

To calculate a line integral of a vector field, the path is divided into small segments and the dot product of the vector field and the direction of each segment is calculated. These dot products are then summed up over the entire length of the path to get the final value of the line integral.

What is the difference between a line integral and a surface integral?

A line integral is calculated along a one-dimensional path, while a surface integral is calculated over a two-dimensional surface. Additionally, a line integral involves the dot product of the vector field and the direction of the path, while a surface integral involves the cross product of the vector field and the surface normal vector.

What are some applications of line integrals in real-world problems?

Line integrals have many applications in real-world problems such as calculating the work done by a force, finding the electric field strength along a path, and determining the fluid flow through a given path. They are also used in fields such as computer graphics, where they are used to calculate the shading of 3D objects.

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