Solving Line Integrals with Vector Cross Products

[Bassalisk]

I posted an actual problem in advanced physics but no answer so i will try to get an math part answer from it.

Suppose I have to solve this integral:

I=\int {\vec{dl} × \vec A }

Where \vec A = -\frac {1}{x} \vec a_{z}

So it has only a z component and I have to find the vector cross of the field with the contour depicted in the picture

http://pokit.org/get/img/7ea5523fdd63cfc5c9374a07f6ab8bb6.jpg First question:

For the part 3:

If I take that the \vec dl = - dy \vec {a_y} will I integrate from b to c or from c to b?

In my intuition if I took the minus into the account in the differential there is no need for flipping the c and b, so we would integrate from c to b?

When you vector cross this all, there shouldn't be net y component am i right?

Link to thread in physics:

https://www.physicsforums.com/showthread.php?t=630013

 

[HallsofIvy]

If I read this correctly you have \vec{A}= -(1/x)\vec{k}. d\vec{r}= dx\vec{i}+ dy\vec{j}+ dz\vec{k} and you should have no trouble finding the cross product of those.

But the exact contour is not clear. Are you saying that the entire contour lies in the xy-plane with z= 0? If so, then yes, d\vec{r}= (1/x)dy and, if a= (x_1, y_x, 0) and b= (x_2, y_2, 0) then we can write the straight line from a to b as x= [(x_2- x_1)/(y_2- y_1)](y- y_1)+ x_1. The integral would be
\int_{y_1}^{y_2}\frac{dy}{\frac{x_2- x_1}{y_2- y_1}(y- y_1)+ x_1}

 

[Bassalisk]

If I read this correctly you have \vec{A}= -(1/x)\vec{k}. d\vec{r}= dx\vec{i}+ dy\vec{j}+ dz\vec{k} and you should have no trouble finding the cross product of those.

But the exact contour is not clear. Are you saying that the entire contour lies in the xy-plane with z= 0? If so, then yes, d\vec{r}= (1/x)dy and, if a= (x_1, y_x, 0) and b= (x_2, y_2, 0) then we can write the straight line from a to b as x= [(x_2- x_1)/(y_2- y_1)](y- y_1)+ x_1. The integral would be
\int_{y_1}^{y_2}\frac{dy}{\frac{x_2- x_1}{y_2- y_1}(y- y_1)+ x_1}

I think I understand.

But can you tell me how can "math" know how I orientated the contour?

Let me show you what I mean.

http://pokit.org/get/img/65e8ba92c1d00bf7fc8be2b178757ed8.jpg If you have a function like in our case for (2) y- \frac{c}{2}=\frac {c}{2b}(x-6b) (a)

How does my equation knows how I orientated the curve? Do I have to manually account for this?

Because (a) is simply a line through 2 points. It does not contain any information about the orientation.

The line integral I have to solve is \vec F = I_{2} \int{\vec {dl} × \vec B}.

And the (3) is not giving me much problems. But (2) is because it has negative slope.

(This is from my EM problem, its a force on a wire problem, check the link provided in first post if you may)

 

[AlephZero]

But can you tell me how can "math" know how I orientated the contour?

It "knows" which way you go round the contour. If you change direction, the sign of the vector product changes.

IMO your diagram is confusing because without any text to explain it, the arrows on the lines suggest the contour is a - b - c - a, but the numbers 1, 2, 3 suggest the reverse direction, a - c - b - a.

 

[Bassalisk]

It "knows" which way you go round the contour. If you change direction, the sign of the vector product changes.

IMO your diagram is confusing because without any text to explain it, the arrows on the lines suggest the contour is a - b - c - a, but the numbers 1, 2, 3 suggest the reverse direction, a - c - b - a.

You have arrow which orients the contour. The 1,2,3 are just there to point out that you should do the problem part by part.