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Curl of a vector field

by Fronzbot
Tags: curl, field, vector
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Fronzbot
#1
Sep19-10, 11:35 PM
P: 62
1. The problem statement, all variables and given/known data
Assume the vector function A = ax(3x[tex]^{2}[/tex]2y[tex]^{2}[/tex])-ax(x[tex]^{3}[/tex]y[tex]^{2}[/tex])
a) Find [tex]\oint[/tex]A[tex]\cdot[/tex]dl around the triangular contour shown in Fig. 2-36 [it is a triangle with base and height of one on the x and y axis. the curl travels so that the normal vector is in the -z direction]
b) Evaluate [tex]\int[/tex]([tex]\nabla[/tex][tex]\times[/tex]A)[tex]\cdot[/tex]ds over the triangular area.


2. Relevant equations
The equation in part a) should be equivalent to that of b) as per Stoke's theorem


3. The attempt at a solution
Part a) is where I'm having trouble. Here is my work for that and I am 100% sure it is incorrect. I have no idea how to set it up, that's my problem. The execution is no big deal, but the set-up part is. Any guidance is appreciated.

A[tex]\cdot[/tex]dl = [ax(3x[tex]^{2}[/tex]2y[tex]^{2}[/tex])-ax(x[tex]^{3}[/tex]y[tex]^{2}[/tex]) [tex]\cdot[/tex] [ax(dx) + ay(dy)]
= 3x[tex]^{2}[/tex]y[tex]^{2}[/tex]dx - x[tex]^{3}[/tex]y[tex]^{2}[/tex]dy
[tex]_{1}[/tex][tex]^{2}[/tex][tex]\int[/tex]3x[tex]^{2}[/tex]y[tex]^{2}[/tex]dx - [tex]_{1}[/tex][tex]^{2}[/tex][tex]\int[/tex]x[tex]^{3}[/tex]y[tex]^{2}[/tex]dy

I then said the area of the triangle is equal to 1/2 (since .5*1*1 = .5) From there I said x = 1/y and y =1/x so I could plug into each equation to get:
[tex]_{1}[/tex][tex]^{2}[/tex][tex]\int[/tex]3x[tex]^{2}[/tex](1/y)[tex]^{2}[/tex]dx - [tex]_{1}[/tex][tex]^{2}[/tex][tex]\int[/tex](1/x)[tex]^{3}[/tex]y[tex]^{2}[/tex]dy
= 3 - ln(2) =2.3068


PART B:
I'm fairly confident in my execution of this, but it would be worth verifying.
[tex]\nabla[/tex][tex]\times[/tex]A = ax([tex]\partial[/tex]/[tex]\partial[/tex]y * Az - [tex]\partial[/tex]/[tex]\partial[/tex]z * Ay) + ay([tex]\partial[/tex]/[tex]\partial[/tex]z * Ax - [tex]\partial[/tex]/[tex]\partial[/tex]x * Az) + az([tex]\partial[/tex]/[tex]\partial[/tex]x * Ay - [tex]\partial[/tex]/[tex]\partial[/tex]y * Ax)
= ax(0-0) + ay(0-0) + az(-6x[tex]^{4}[/tex]y[tex]^{4}[/tex] + 6x[tex]^{5}[/tex]y[tex]^{3}[/tex])

ds = ds[tex]_{z}[/tex] = dxdy

[tex]\int[/tex]([tex]\nabla[/tex][tex]\times[/tex]A)[tex]\cdot[/tex]ds = [tex]_{1}[/tex][tex]^{2}[/tex][tex]\int[/tex][tex]_{1}[/tex][tex]^{2}[/tex][tex]\int[/tex](-6x[tex]^{4}[/tex]y[tex]^{4}[/tex] + 6x[tex]^{5}[/tex]y[tex]^{3}[/tex])dxdy = 5.61

EDIT- for some reason the tex stuff is being really obnoxious for me. the 12 that your seeing is the boundry of the integral (going from 1 to 2) and if you look carefully, there is a dot in between A and dl and others like that. I apologize for it looking crappy!
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Fronzbot
#2
Sep20-10, 12:13 AM
P: 62
I done screwed up. Part b the cross product I did was wrong, stupid mistake. It should be the double integral of -3x^2y^2-6x^2y dxdy from 1 to 2 on both integrals which equals -112/3
vela
#3
Sep20-10, 12:16 AM
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When you write equations with LaTeX, write the whole thing with LaTeX rather than trying to mix it in with plain text.

Your vector field looks screwy, since you have ax twice, which I assume is ax, the unit vector in the x direction. And are the 2 and 3 in the first term really separate numerical factors? It just seems strange they weren't multiplied together already.

In part A, I'm not sure what you're doing with the area and x=1/y and y=1/x stuff. You need to handle each leg of the triangle separately and then add the results. One way to do line integrals is to parameterize the contour so you have x=x(t) and y=y(t) and dx=x'(t) dt and dy=y'(t) dt so you end up with an integral with respect to t. For example, on the leg that goes from the origin to (1,0), you could use x(t)=t, y(t)=0 where t runs from 0 to 1. Then dx=dt and dy=0. Plug everything in and integrate from 0 to 1.

In part B, you calculated the curl incorrectly. You have higher powers of x and y than you do in the original vector field. Differentiation will lower them, if anything. You should figure out what the correct curl is first.

Fronzbot
#4
Sep20-10, 12:26 AM
P: 62
Curl of a vector field

Quote Quote by vela View Post
When you write equations with LaTeX, write the whole thing with LaTeX rather than trying to mix it in with plain text.

Your vector field looks screwy, since you have ax twice, which I assume is ax, the unit vector in the x direction. And are the 2 and 3 in the first term really separate numerical factors? It just seems strange they weren't multiplied together already.

In part A, I'm not sure what you're doing with the area and x=1/y and y=1/x stuff. You need to handle each leg of the triangle separately and then add the results. One way to do line integrals is to parameterize the contour so you have x=x(t) and y=y(t) and dx=x'(t) dt and dy=y'(t) dt so you end up with an integral with respect to t. For example, on the leg that goes from the origin to (1,0), you could use x(t)=t, y(t)=0 where t runs from 0 to 1. Then dx=dt and dy=0. Plug everything in and integrate from 0 to 1.

In part B, you calculated the curl incorrectly. You have higher powers of x and y than you do in the original vector field. Differentiation will lower them, if anything. You should figure out what the correct curl is first.
Yeah, I tried doing everything in Latex but it was giving me errors were there shouldn't have been. It was very odd. As for that 2 and 3 thing- the 2 is a typo. The correct equation is A = ax(3x^2y^2) - ay(x^3y^2), my apologies.

In part A- that's what I wasn't sure about. I wasn't sure if I had to do each line. I'll give that a go.

IN part B- yeah, I realized that soon after I posted that it was stupidly wrong. I think I replied as you were as I addressed that issue already- my new integral is [tex]\int _{1}^{2}\int _{1}^{2}-3x^{2}y^{2}-6x^{2}ydxdy = -112/3[/tex] which makes more sense as the curl has a normal vector in the negative z direction. (At least, that's how it makes sense to me)
vela
#5
Sep20-10, 12:34 AM
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Yeah, you posted while I was writing earlier. For part B, I got 7/60. You're going clockwise around the triangle, right? I inferred that because you said the normal points in the -z direction.

What limits did you use for x and y?
Fronzbot
#6
Sep20-10, 12:55 AM
P: 62
Quote Quote by vela View Post
Yeah, you posted while I was writing earlier. For part B, I got 7/60. You're going clockwise around the triangle, right? I inferred that because you said the normal points in the -z direction.

What limits did you use for x and y?
Yeah, clockwise. And I have limits from 1 to 2 on both (I should've specified that the triangle legs go from (1,1) to (1,2) and (1,1) to (2,1). Sorry 'bout that.

Now my integral I have for part b is [tex]\int_{1}^{2} \int_{1}^{2} (-3x^{2}y^{2} - 6x^{2}y)dxdy [/tex]

I'm assuming this is incorrect as when I input boundaries of 0 to 1 I get -4/3 and not 7/60. I have [tex]\nabla \times A = ax(\partial / \partial z x^{3}y^{2}) + ay(\partial / \partial z 3x^{2}y^{2}) + az(-\partial / \partial x x^{3}y^{2} - \partial / \partial y 3x^{2}y^{2}) = ax(0) + ay(0) +az(-3x^{2}y^{2} - 6x^{2}y)[/tex]

I don't see anything wrong with what I did though.
Fronzbot
#7
Sep20-10, 01:09 AM
P: 62
I re-did part a based on your recommendation (though I didn't see any benefit in parameterizing the curve, am I missing something?)

Here's what I got:
Integral I worked off of:
[tex]\int 3x^{2}y^{2}dx - \int x^{3}y^{2}dy[/tex]

Leg 1 (from (1,1) to (2,1))
[tex]\int_{1}^{2}3x^{2}dx = 7[/tex]

Leg 2 (from (1,1) to (1,2))
[tex]-\int_{1}^{2}8y^{2}dy = -56/3[/tex]

Leg 3 (from (1,1) to (2,2))
[tex]\int_{1}^{2}3x^{2}dx - \int_{1}^{2}8y^{2}dy = -35/3[/tex]

So Leg 1 + Leg 2 + Leg 3 = -70/3

I'm pretty sure I did at least the last leg wrong, but am I getting on the right track?

EDIT- wait, if that first leg is negative, then I get -112/3 which is what I got for part b. But how, mathematically, would I get that first one equal to -7 rather than +7 without needing to change the bounds on all my other functions (which would just give me 70/3 instead)?
vela
#8
Sep20-10, 01:45 AM
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Quote Quote by Fronzbot View Post
I re-did part a based on your recommendation (though I didn't see any benefit in parameterizing the curve, am I missing something?)

Here's what I got:
Integral I worked off of:
[tex]\int 3x^{2}y^{2}dx - \int x^{3}y^{2}dy[/tex]

Leg 1 (from (1,1) to (2,1))
[tex]\int_{1}^{2}3x^{2}dx = 7[/tex]
Wrong sign. The contour goes from (2,1) to (1,1). The direction is important.
Leg 2 (from (1,1) to (1,2))
[tex]-\int_{1}^{2}8y^{2}dy = -56/3[/tex]
You're using the wrong value for x, I think.
Leg 3 (from (1,1) to (2,2))
[tex]\int_{1}^{2}3x^{2}dx - \int_{1}^{2}8y^{2}dy = -35/3[/tex]
The point (2,2) isn't on the contour at all. You can't hold x and y constant in either integral because they both vary as you move along this leg.
So Leg 1 + Leg 2 + Leg 3 = -70/3

I'm pretty sure I did at least the last leg wrong, but am I getting on the right track?

EDIT- wait, if that first leg is negative, then I get -112/3 which is what I got for part b. But how, mathematically, would I get that first one equal to -7 rather than +7 without needing to change the bounds on all my other functions (which would just give me 70/3 instead)?
I redid the integral with the new boundary, and the answer comes out to 709/60.
Fronzbot
#9
Sep20-10, 01:50 AM
P: 62
Quote Quote by vela View Post
Wrong sign. The contour goes from (2,1) to (1,1). The direction is important.
Yeah, I see that I didn't take that into account.

You're using the wrong value for x, I think.
There I'm using x=2 since the line goes from (2,2) down to (2,1) and so x is constant at 2.

The point (2,2) isn't on the contour at all.
Why not though? If the triangle has a line going from (1,1) to (2,1); (2,1) to (2,2) and (1,1) to (2,2) which is the hypotenuse, how is the point (2,2) not on the curve? That doesn't make sense to me.
vela
#10
Sep20-10, 01:55 AM
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Quote Quote by Fronzbot View Post
[it is a triangle with base and height of one on the x and y axis. the curl travels so that the normal vector is in the -z direction]
Quote Quote by Fronzbot View Post
Yeah, clockwise. And I have limits from 1 to 2 on both (I should've specified that the triangle legs go from (1,1) to (1,2) and (1,1) to (2,1). Sorry 'bout that.
Quote Quote by Fronzbot View Post
Why not though? If the triangle has a line going from (1,1) to (2,1); (2,1) to (2,2) and (1,1) to (2,2) which is the hypotenuse, how is the point (2,2) not on the curve? That doesn't make sense to me.
You've described three different contours now!
Fronzbot
#11
Sep20-10, 02:03 AM
P: 62
Quote Quote by vela View Post
You've described three different contours now!
Yeahh... It's late (3am here) and I have an 8am class not to mention I've been doing other homework on top of this. I'm really sorry that it's tough for me to type correctly, I should've double checked that second one I posted (or, even better, been more clear in my opening post). The last one I just wrote is the correct contour.

Sorry, again, I'm just really tired. When I wrote that second one down I seemed to have flipped the y and x axis in my head. It made sense to me, but that's about it haha
vela
#12
Sep20-10, 02:10 AM
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I got a final answer of 593/30 with the last contour you described. Here's what I got for each leg, so you can check your results:

81/10 for (1,1) to (2,2)
56/3 for (2,2) to (2,1)
-7 for (2,1) to (1,1)

So it looks like you have two legs done, except for possible sign problems. On the hypotenuse, neither x nor y is constant, but you know that x=y on that part of the contour.


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