Stokes' Theorem parameterization

  • #1
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Homework Statement


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Homework Equations



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The Attempt at a Solution


I only know that they gave the parameterization of the circle: r(t) = <cost, sint, 2>.
My problem is, did they already give the curl of F in the line integral? I don't understand why dx, dy, and dz are separated like that.
 

Answers and Replies

  • #2
vela
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If ##\vec{F} = F_x\,\hat{i}+F_y\,\hat{j}+F_z\,\hat{k}## and ##d\vec{r} = dx\,\hat{i} + dy \,\hat{j} + dz\,\hat{k}##, what's ##\vec{F}\cdot d\vec{r}## equal to?
 
  • #3
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If ##\vec{F} = F_x\,\hat{i}+F_y\,\hat{j}+F_z\,\hat{k}## and ##d\vec{r} = dx\,\hat{i} + dy \,\hat{j} + dz\,\hat{k}##, what's ##\vec{F}\cdot d\vec{r}## equal to?
It would equal Fxdx + Fydy + Fzdz, but wouldn't it be simplified to just F?
 
  • #4
vela
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What do you mean "simplified to just F"?
 
  • #5
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What do you mean "simplified to just F"?
For example, (F/dx)(dx), so the dx would cancel out. That happens to each. Actually, wouldn't it be 3F?
 
  • #6
vela
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##\vec{F}## is a vector-valued function. ##F_x## is the x-component of ##\vec{F}##, not the derivative of ##F## with respect to ##x##. Besides, the expression ##F/dx## is meaningless.
 
  • #7
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##\vec{F}## is a vector-valued function. ##F_x## is the x-component of ##\vec{F}##, not the derivative of ##F## with respect to ##x##. Besides, the expression ##F/dx## is meaningless.
Oh right, so do I just take the derivative of the parameterization which will give me values equal to dx, dy, dz and plug that into the original integral? Do I also plug in the parameterization into F?
 
  • #8
vela
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Sounds like a plan.
 
  • #9
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Sounds like a plan.
How would you go about simplifying the integral though? Some parts are fine but how about e^(-cos^2(t)/2), for example?
 
  • #10
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You don't. The problem is asking you to evaluate the integral using Stoke's theorem.
 
  • #11
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You don't. The problem is asking you to evaluate the integral using Stoke's theorem.
So is the answer just:
t=0 to t=2pi (-e-cos2t/2sint+2sin2t+e-sin2t/2cost+4cos2t)dt
 
  • #12
LCKurtz
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No. That isn't using Stoke's theorem. The other side of Stoke's theorem is$$
\iint_S \nabla \times \vec F \cdot d\vec S$$Do that.
 
  • #13
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No. That isn't using Stoke's theorem. The other side of Stoke's theorem is$$
\iint_S \nabla \times \vec F \cdot d\vec S$$Do that.
So do I find the curl of F first, plug in the parameterization into the curl, then multiply it with the normal vector using dot product?
I'm having a hard time relating these things I think.
 
  • #14
LCKurtz
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So do I find the curl of F first,

Yes

plug in the parameterization into the curl,

Not sure what you mean by "the" parameterization. You are given the parametrization of a curve enclosing an area. You are going to need to parameterize that area and use that. You will need two parameters for a surface.

then multiply it with the normal vector using dot product?
I'm having a hard time relating these things I think.

You will have to show us what you do before we can tell if you are doing it correctly.
 
  • #15
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Yes



Not sure what you mean by "the" parameterization. You are given the parametrization of a curve enclosing an area. You are going to need to parameterize that area and use that. You will need two parameters for a surface.



You will have to show us what you do before we can tell if you are doing it correctly.
So far, I found the curl to be <-x, -y, 2z+2>. Do I find the normal vector by using the circle's parameterization?
 
  • #16
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So far, I found the curl to be <-x, -y, 2z+2>. Do I find the normal vector by using the circle's parameterization?

Let's call that vector you got ##\vec V = \langle -x,-y,2z+2\rangle##, which looks correct for the curl. So now you need to integrate ##\vec V\cdot d\vec S## over the described area up in the ##z=2## plane. Your next issue is to figure out ##d\vec S##. How do you find the area vector? If you look at the geometry for this problem, it should be easy to figure out a unit normal vector and an area element. You can worry about the integral after you figure out the integrand.
 
  • #17
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Let's call that vector you got ##\vec V = \langle -x,-y,2z+2\rangle##, which looks correct for the curl. So now you need to integrate ##\vec V\cdot d\vec S## over the described area up in the ##z=2## plane. Your next issue is to figure out ##d\vec S##. How do you find the area vector? If you look at the geometry for this problem, it should be easy to figure out a unit normal vector and an area element. You can worry about the integral after you figure out the integrand.
I think what I am confused about is do you use F or the circle to find the unit normal vector?
 
  • #18
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I think what I am confused about is do you use F or the circle to find the unit normal vector?
You are given a curve which encloses a surface. Surfaces have normal vectors. It doesn't have anything to do with the vector ##\vec F##. I suggest you draw a picture of your surface.
 
  • #19
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You are given a curve which encloses a surface. Surfaces have normal vectors. It doesn't have anything to do with the vector ##\vec F##. I suggest you draw a picture of your surface.
Is the unit normal vector just <cost, sint, 2>?
 
  • #20
LCKurtz
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Is the unit normal vector just <cost, sint, 2>?
No. That isn't even a unit vector. It is a parametric equation of your circular boundary. Surely your text tells you how to calculate a normal vector to a surface. Or you could just draw a picture of the surface for this problem and look at it.
 
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  • #21
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No. That isn't even a unit vector. It is a parametric equation of your circular boundary. Surely your text tells you how to calculate a normal vector to a surface. Or you could just draw a picture of the surface for this problem and look at it.
Okay, I have two answers I am unsure about for the normal vector. What would you say about <0,0,r>?
 
  • #22
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Okay, I have two answers I am unsure about for the normal vector. What would you say about <0,0,r>?
I would be confused about the use of ##r## because you are using it for the curve so I'm not sure what you are thinking.
 
  • #23
vela
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Do you have a textbook with worked out examples?
 
  • #24
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I would be confused about the use of ##r## because you are using it for the curve so I'm not sure what you are thinking.
How about <sint, -cost, 0>? Sorry, I came up with these with my classmates.
Do you have a textbook with worked out examples?
No, we just depend on MyMathLab and the examples our professor has done in class, but we have not done anything like this.
 
  • #25
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I would be confused about the use of ##r## because you are using it for the curve so I'm not sure what you are thinking.
He also did not really explain well how to find the unit normal vector...
 
  • #26
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1. Do you know how to find a unit normal vector to the ##xy## plane?

2. Where is the surface we are talking about located?
 
  • #27
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1. Do you know how to find a unit normal vector to the ##xy## plane?

2. Where is the surface we are talking about located?
The gradient of F divided by the magnitude of the gradient.
Is at located at z=2?
Sorry, I feel like I know nothing now...
In class, we discussed about how to find this normal vector. We did:
<a,b,c>⋅<cost,sint,2>=0 (since the normal vector should be orthogonal to the surface)
a=sint
b=-cost
c=0
Therefore, the normal vector = <sint,-cost,0>.
 
  • #28
vela
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No, we just depend on MyMathLab and the examples our professor has done in class, but we have not done anything like this.
It would be a good idea to get your hands on a textbook. There's a free one at OpenStax, and I'm sure there are others available for free as well. It'll help fill in the blanks that your professor may have skipped or glossed over in class.
 
  • #29
vela
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<cos t, sin t, 2> is the boundary of the surface S. You can use any surface that has that curve as its boundary. The simplest one to use is a flat surface. Have you sketched it or at least have a picture of the surface in your mind? Tell us in words which way a vector perpendicular to that surface points. You don't need to use any dot products, etc., here; just reason it out geometrically.
 
  • #30
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<cos t, sin t, 2> is the boundary of the surface S. You can use any surface that has that curve as its boundary. The simplest one to use is a flat surface. Have you sketched it or at least have a picture of the surface in your mind? Tell us in words which way a vector perpendicular to that surface points. You don't need to use any dot products, etc., here; just reason it out geometrically.
Isn't it a unit circle centered at (0,0,2)? So the vector perpendicular to it will either point upwards or downwards directly from the surface... Am I even thinking correctly?
 
  • #31
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The gradient of F divided by the magnitude of the gradient.
Is at located at z=2?
Sorry, I feel like I know nothing now...
In class, we discussed about how to find this normal vector. We did:
<a,b,c>⋅<cost,sint,2>=0 (since the normal vector should be orthogonal to the surface)
a=sint
b=-cost
c=0
Therefore, the normal vector = <sint,-cost,0>.

You are apparently talking about a normal vector to the curve, not the normal vector to the enclosed surface.

Yes, the surface lies in the plane z=2. You shouldn't need any calculus to tell me what a unit normal to that plane is. Have you drawn a picture? And it is easy to write down a standard area element in that plane. I think Vela's suggestion for you to read some examples is a good one, and I have to be gone the rest of the afternoon.
 
  • #32
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Isn't it a unit circle centered at (0,0,2)? So the vector perpendicular to it will either point upwards or downwards directly from the surface... Am I even thinking correctly?
Yes, that's right! To choose between the two vectors depends on the way you go around the contour. Have you heard of the right hand rule?

For now, just take it on faith the correct vector is the one that points upward. Can you finish evaluating the integral?
 
  • #33
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Yes, that's right! To choose between the two vectors depends on the way you go around the contour. Have you heard of the right hand rule?

For now, just take it on faith the correct vector is the one that points upward. Can you finish evaluating the integral?
So the unit normal vector would be <0,0,2>?
All I have to do is take the curl and multiply it with the normal vector using the dot product, but I am confused about the bounds now. Do I also have to make the curl = <-cost, -sint, 6> (plugging in what they gave me for the circle)? Then are the bounds just from t=0 to t=2pi?
 
  • #34
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So the unit normal vector would be <0,0,2>?
No. "Unit" means the length of the vector is equal to 1.

All I have to do is take the curl and multiply it with the normal vector using the dot product, but I am confused about the bounds now. Do I also have to make the curl = <-cost, -sint, 6> (plugging in what they gave me for the circle)? Then are the bounds just from t=0 to t=2pi?
It's a surface integral, so you have to integrate over the area bounded by the curve.
 
  • #35
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No. "Unit" means the length of the vector is equal to 1.


It's a surface integral, so you have to integrate over the area bounded by the curve.
So would it be <0,0,sqrt(2)>?
Do I have to use polar coordinates?
 

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