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Checking Stokes' Theorem

  • Thread starter JD_PM
  • Start date
207
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Problem Statement
I want to check Stokes' theorem for the exercise below.
Relevant Equations
##\int_{a} (\vec \nabla \times \vec F) \cdot d \vec a = \oint_{C} \vec F \cdot d \vec l##
I want to check Stokes' theorem for the following exercise:

Consider the vector field ##\vec F = ye^x \hat i + (x^2 + e^x) \hat j + z^2e^z \hat k##.

A closed curve ##C## lies in the plane ##x + y + z = 3##, oriented counterclockwise. The parametric representation of this curve is defined as:

$$C = (1+cost) \hat i + (1+sint) \hat j + (1 - cost - sint) \hat k$$

Where ##t[0, 2\pi]##.

Stokes Theorem states:

$$\int_{a} (\vec \nabla \times \vec F) \cdot d \vec a = \oint_{C} \vec F \cdot d \vec l$$

Evaluating the LHS:

The curl of ##\vec F## :

$$\vec \nabla \times \vec F = 2x \hat k$$

This is an important point: How can I know the projection on the xy plane of the surface enclosed by C?

I am told that it is the circle with radius 1 and center (1, 1) but I do not know why. We can verify it graphically but I am seeking for a numerical method to get it, rather than having to draw the plot.

The unit normal vector to the circle is:

$$\hat n = (x-1) \hat i + (y+1) \hat j$$

Setting up the integral:

$$d \vec a = \hat k dxdy$$

$$\int_{a} (\vec \nabla \times \vec F) \cdot d \vec a = \int_{a} 2x dxdy = 2\int_{0}^{2\pi}\int_{0}^{1} rcos \theta drd \theta = 0$$

BUT I GET 0... What does that imply?

Evaluating the RHS:

$$\oint_{C} \vec F \cdot d \vec l = \oint_{C}[ye^x dx + (x^2 + e^x)dy + z^2e^z dz]$$

But we know how the curve looks like:

$$x = 1 + cost$$
$$y = 1 + sint$$
$$z = 1 - cost - sint$$

So it is a matter of solving the integral:

$$\oint_{C} = C_1 + C_2 + C_3$$

Where:

$$C_1 = \int_{0}^{2\pi} (1+sint) e^{1+cost}(-sint)dt$$

$$C_2 = \int_{0}^{2\pi} [(1+cost)^2 + e^{1+cost}](cost)dt$$

$$C_3 = \int_{0}^{2\pi} (1-cost-sint)^2e^{1-cost-sint}(sint-cost)dt$$

BUT How can I solve ##C_1##?

To sum up; these are the points I want to understand:

1) How to know the projection on the xy plane of the surface enclosed by C.
2) What does ##\int_{a} (\vec \nabla \times \vec F) \cdot d \vec a = \oint_{C} \vec F \cdot d \vec l = 0## mean? (##\vec F## is not a conservative field, so that is not the reason...).
3) How to solve ##C_1## kind of integrals (any trigonometric trick?). I tried to solve it using https://www.integral-calculator.com/ but founds no primitive...

Thanks.
 

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Orodruin

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How can I know the projection on the xy plane of the surface enclosed by C?
How do you usually find a projection of any point onto the xy plane?

3) You do not always need a primitive to solve an integral. Some terms you can throw away based on symmetry arguments.

2) It does not "mean" anything other than that the integrals are zero. That can happen even if the field is not conservative. The point with a conservative field is that it holds regardless of the curve C.

The unit normal vector to the circle is:


^n=(x−1)^i+(y+1)^jn^=(x−1)i^+(y+1)j^​

\hat n = (x-1) \hat i + (y+1) \hat j

Setting up the integral:


d→a=^kdxdy​
This is not correct. It is unclear to me what you are trying to do with ##\hat n##, but it certainly is not a unit normal to anything. Also, ##d\vec a## is not in the ##\hat k## direction based on what the plane is. Or are you taking some integration surface that is not in the plane ##x+y+z = 3##? You should specify what surface you are using.
 
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Let me reply point per point.

How do you usually find a projection of any point onto the xy plane?
With vectors I see how could I find the projection; let's say we have:

$$\vec A = A_x \hat i + A_y \hat j + A_z \hat k$$

And we want the projection of this vector on the z axis (##A_z##). Then we just have to take the dot product with a unit vector in the z direction:

$$A_z = A \cdot \hat k$$

I am wondering how to do the same but with surfaces.
 

LCKurtz

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@JD_PM : Before starting a new thread how about finishing these?:


and

 

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