# Line integral in vector field

1. Sep 30, 2009

### pakkanen

First I wanna 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)ux + (y2-x2)ux. Calculate the fields line integral among the parabola y=3x2, from origin to poin r (r=ux-3uy).

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(Fx,y,z-dot-dx,y)
int[-ux+uy(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[(3x2ux-2xux+9x4uy-x2uy)*(dux+duy)
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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 30, 2009

### rock.freak667

ok so we want to calculate

$$\int_C \vec{F} . d\vec{r}$$

Where C is curve y=3x2 from (0,0) to i+3j (i is the same as ux and j is the same as uy). 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 $\vec{r}$ for the curve which is in the form $\vec{r} = x(t)\hat{i} + y(t) \hat{j}$.

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

Last edited: Sep 30, 2009
3. Oct 4, 2009

### 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)ux + (9x4-x2)uy

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 = 1ux and 720uy but I think this is not gonna work..

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

4. Oct 4, 2009

### 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

5. Oct 4, 2009

### rock.freak667

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

When you find $\vec{r}$. You need to also get $F(\vec{r(t)})}$

6. Oct 6, 2009

### 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 ********.

Last edited: Oct 6, 2009
7. Oct 6, 2009

### 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.

8. Oct 8, 2009

### 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.