Calculating Stokes's Theorem on a Triangular Contour

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In summary: Thanks for the reply. I have 2 questions:1) Why is the Stokes theorem useful?2) Why is there a need to do 3 integrations?
  • #1
EugP
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Homework Statement


For the vector field [tex]\bold{E} = \bold{ \hat x} (xy) - \bold{ \hat y} (x^2 + 2y^2)[/tex], calculate the following:

[tex]\oint \bold{E} \cdot d\bold{l}[/tex] around the triangular contour shown.

I don't have a scanner at the moment so I will explain the drawing. The picture is a right triangle. They show an x and y axis. From (0, 0), there is a line going to (1, 0), then from there it goes up to (1, 1), then a diagonal back to (0, 0).


Homework Equations





The Attempt at a Solution



I know how to approach it but I seem to be stuck. This is what I have so far:

I gave each coordinate a name: a (0, 0), b (1, 0), and c (1, 1).

[tex]\oint \bold{E} \cdot d\bold{l} = \oint_a^b \bold{E}_{ab} \cdot d\bold{l} + \oint_b^c \bold{E}_{bc} \cdot d\bold{l} + \oint_c^a \bold{E}_{ca} \cdot d\bold{l}[/tex]

At this point I'm stuck. Could someone please point me in the right direction?
 
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  • #2
Your confusion probably lies in that you don't know what [tex]\vec{dl}[/tex] is. For example, for the integral from a to b, [tex]\vec{dl_{1}}=\hat{x}dl[/tex], whereas b to c it becomes [tex]\vect{dl_{2}}=\hat{y}dl[/tex]. Just work out the dot products, which are especially easy for a to b and b to c since either x or y is constant.
 
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  • #3
Thanks for the quick reply. I have 2 questions:

1) Shouldn't the integral from a to b, be [tex]\oint_a^b \bold{E}_{ab} \cdot \hat x d\bold{l}[/tex] since ab is horizontal?

2) I'm not sure what to put for [tex]\bold{E}_{ab}[/tex]. Should it be the component that acts only in the direction of ab, in other words:

[tex]\oint_a^b \bold{ \hat x} (xy) \cdot \hat x d\bold{l}[/tex]

Am I in the right direction?
 
  • #4
Yes whoops you are right about 1). I will change it. And 2) yes you are right. Don't forget that y is a constant.
 
  • #5
Awsome. So now that I set it up, I have [tex]\oint_a^b xy d\bold{l}[/tex], but I don't know l, or am I missing something obvious?
 
  • #6
Remember that a is (0,0) and b is (1,0). [tex]dl[/tex] is simply dx in that particular case. The only tricky part is going to be c to a, where you're going to have to express all components of dl in terms of one variable to integrate over.
 
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  • #7
EugP said:
Awsome. So now that I set it up, I have [tex]\oint_a^b xy d\bold{l}[/tex], but I don't know l, or am I missing something obvious?

If the path is along the positive x axis, you must use [tex] d \vec{l} = dx \vec{i} [/tex]

and so on.

In general, [tex] d\vec{l} = dx \vec{i} + dy \vec{j} + dz \vec{k} [/tex]


If you have a parametrized curve, you replace all the variables in terms of the parameter and the integral is only over one variable,
 
  • #8
Alright I got it. Thanks for all the help guys.
 
  • #9
Since you titled this "Stokes Theorem", are you also going to integrate -2x-2y over the triangular region and show that they are the same?
 
  • #10
HallsofIvy said:
Since you titled this "Stokes Theorem", are you also going to integrate -2x-2y over the triangular region and show that they are the same?

Actually yes, and I'm stuck on that too.

Also, I thought I figured out how to finish the first part, but it's not working out.

Here's what I'm getting:

[tex]\oint \bold{E} \cdot d\bold{l} = \oint_0^1 xy dx + \oint_0^1 -x^2-2y^2 dy + \oint_c^a \bold{E}_{ca} \cdot d\bold{l}[/tex]
but I don't know what to do with the last term. And when I integrate the first two terms, I get
[tex]\frac{y}{2}-x^2-\frac{2}{3}[/tex]

which doesn't make sense, because the answer for the whole is -1. What should I do with the last term? Shouldn't it be:

[tex]\oint_0^1 (xy - x^2 +2y^2) dxdy[/tex]
 
  • #11
I got -1 for both methods. Did you get -1 for the line integral?
 
  • #12
this is a perfect question to use Green"s theorem. And the answer is -1. Stokes' theorem is just a generalized version of Green's th. and I don't see how that would be useful in this case. Also there is no need to do 3 integrations. Just convert the whole thing to a double integral using Green's theorem, integrate over the triangle and get -1
 

Related to Calculating Stokes's Theorem on a Triangular Contour

What is Stokes's Theorem?

Stokes's Theorem is a mathematical tool used in vector calculus to relate surface integrals to line integrals. It states that the integral of a vector field over a closed surface is equal to the line integral of the vector field over the boundary of the surface.

What is a Triangular Contour?

A triangular contour is a two-dimensional figure made up of three straight lines connected to form a triangle. In the context of Stokes's Theorem, a triangular contour is used as the boundary of a surface over which the theorem is applied.

How do you calculate Stokes's Theorem on a Triangular Contour?

To calculate Stokes's Theorem on a triangular contour, you first need to parameterize the contour by defining a vector function that describes each line segment. Then, you need to calculate the line integral of the vector field over each line segment. Finally, you can use the line integrals to calculate the surface integral using Stokes's Theorem.

What are the applications of Calculating Stokes's Theorem on a Triangular Contour?

Calculating Stokes's Theorem on a triangular contour has various applications in physics and engineering, such as in fluid dynamics, electromagnetism, and heat transfer. It is also used in real-world problem-solving, such as calculating the circulation of a fluid around a triangular object or the flow of electricity around a triangular circuit.

What are the limitations of Calculating Stokes's Theorem on a Triangular Contour?

One limitation of calculating Stokes's Theorem on a triangular contour is that it can only be applied to closed surfaces. Additionally, the vector field and the contour must satisfy certain conditions, such as being continuously differentiable, for the theorem to be valid. Furthermore, the calculation can become more complex and challenging for more complex shapes and vector fields.

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