What is the approach to calculating line integrals in a vector field?

[pakkanen]

First I want to greet everyone because I am new here.

I have attended to applied electromagnetic course which seems to be pretty hard to understand and issues came up at very first time after I went at calculations.

I try to explain this as good as possible.

1. Vectorfield F(x,y,z) = (y-2x)u_x + (y²-x²)u_x. Calculate the fields line integral among the parabola y=3x², from origin to poin r (r=u_x-3u_y).

Does it mean that every point of the slope y=3x² can be expressed in terms of vector field F?

I have tried the following
I put F=r to express the amound of x and y is needed in vector field F to reach point r. It gives me the following:
y-2x = 1
y²-x² = 3
Does this make sense? I know that vector field´s line integral is int(F-dot-d)
Next I have tried to make this happen and had
int(F_x,y,z-dot-d_x,y)
int[-u_x+u_y(2y-2x)].

How I implement the y=3x² in this integral? Shouldn´t the integral be
int(3x²) from point 0 to r?

I also have tried to put the function y=3x² to F(x,y,z) and tried int[(3x²u_x-2xu_x+9x⁴u_y-x²u_y)*(du_x+du_y)
After this I had 2x²+9x⁴-2x and tried to put value for X which i got when I did F = r but it doesn´t seem to work.

Can anyone give me a clue how should I THINK or understant this? How do I make the calculation?

The answe should be 15/2

 

[rock.freak667]

ok so we want to calculate


[tex]\int_C \vec{F} . d\vec{r} [/tex]

Where C is curve y=3x² from (0,0) to i+3j (i is the same as u_x and j is the same as u_y). So it is basically from (0,0) to (1,3) .

I think it should be i+3j or (1,3) as (1,-3) does not lie on the curve. So you can easily parameterize the curve by letting x= some function of t.

so now find [itex]\vec{r}[/itex] for the curve which is in the form [itex] \vec{r} = x(t)\hat{i} + y(t) \hat{j}[/itex].

So what is your range of values for t on the vector r?

When you have this. You can easily parameterize F (make x in terms of t and y in terms of t) and you can find dr

 

[pakkanen]

Ok, I have tried this but something is still going wrong.

I replace the y = 3x^2 to in F(x,y,z) and I get F(x,y,z) = (2x²-2x)u_x + (9x⁴-x²)u_y

So I know the y in terms of x in the vector field.

I tried to somehow force r in terms of x and y and got r = 1u_x and 720u_y but I think this is not going to work..

What it means that I have to take [itex]\vec{r} = x(t)\hat{i} + y(t) \hat{j}[/itex] in terms of t?
How I express r in values of x and y? Do I need to replace u_x and u_y or how I erase them

 

[LCKurtz]

You can always parameterize an equation y = f(x) by letting x = t, y = f(t). In your case you would get:

r(t) = t i + 3t²j

 

[rock.freak667]

How I express r in values of x and y? Do I need to replace u_x and u_y or how I erase them

Well you need to do what LCKurtz suggested. I don't think it matters whether you use u_x for i or u_y for j.

When you find [itex]\vec{r}[/itex]. You need to also get [itex]F(\vec{r(t)})}[/itex]

 

[pakkanen]

Ok, thank you by far. this is makes sense now but I still can´t get the right answer.

I got got that the following,
r = ti + 9t²j => dr = d(ti) + d(9t²j)

And the F would be F = (3t²-2t)i + (9t⁴-t²)j

Can I now take dr = (i + 9tj)dt? Ok, I have done this and int(F.dt):

I had int(3t²-2t+81t⁵-9t³) with only t:s

How do I get t:s? i tried to put F(t) = r(t) and got 27t⁴-2t-10 = 0. i mean I put 3t²-2t = 1 and 9t⁴-t² = 3 and got very nonsense with t = about -/+ 0,809... I tried to integrate from 0 to t and got some ********.

 

[LCKurtz]

Your r should be ti + 3t²j. When t = 0 the point is at (0,0) and when t = 1 the point is at (1,3) on your parabola. So the t variable in the integral should go from 0 to 1.

 

[pakkanen]

hmmm. of course. int(3t²-2t+81t⁵-9t³) that is
/(t³ - t² + 13,5t⁶ - 2,25t⁴) right?
Now from 0 to 1 gives me 11,25 and the answer should be 7,5. Somewhere is still error but I can´t find it.