Verifying Stokes' theorem (orientation?)

In summary, the conversation discusses verifying Stokes' theorem for a given surface and determining the correct orientation for the parameterization in order to use the theorem correctly.
  • #1
Feodalherren
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Homework Statement


F= <y,z,x>
S is the hemisphere x^2 + y^2 + z^2 = 1, y ≥ 0, oriented in the direction of the positive y-axis.
Verify Stokes' theorem.

Homework Equations


The Attempt at a Solution


So I completed the surface integral part. I'm trying to do the line integral part of Stokes' theorem and end up with the same answer.
Where I get confused is there parametrization part.

I said that r(t) = <cos t, 0, sin t>, 0≤t≤2∏.
Apparently that's the wrong orientation. But when I "grab" the y-axis with my thumb in the positive y-direction and curl my fingers they go from the z axis to the x-axis counter clockwise. Isn't that the CORRECT orientation?
I guess what I'm asking is how do I determine the orientation when I'm using Stokes' theorem. I assume I want the same counter clockwise orientation that I do for Green's theorem.
 
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  • #2
Your parameterization goes in the opposite direction. As t goes from 0 to pi/2, r(t) goes from <1, 0, 0> to <0, 0, 1> — in other words, from the x-axis to the z-axis.
 
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Related to Verifying Stokes' theorem (orientation?)

What is Stokes' theorem and why is it important?

Stokes' theorem is a fundamental theorem in vector calculus that relates the integral of a vector field over a surface to the line integral of the vector field around the boundary of the surface. It is important because it provides a powerful tool for calculating flux through surfaces and for solving problems in fluid mechanics, electromagnetism, and other fields.

How do you verify Stokes' theorem?

To verify Stokes' theorem, you must first ensure that the surface and the curve bounding it are oriented in the same direction. Then, you can use the formula for Stokes' theorem to calculate the line integral and surface integral separately, and check if they are equal. If they are equal, then Stokes' theorem is verified.

What is the role of orientation in verifying Stokes' theorem?

The orientation of the surface and the curve bounding it is crucial in verifying Stokes' theorem. This is because the formula for the theorem involves taking the dot product of the vector field and the surface's normal vector, which depends on the orientation. If the orientations are different, the resulting integral will be negative, and the theorem will not hold.

Can Stokes' theorem be applied to any surface and curve?

No, Stokes' theorem can only be applied to surfaces that are smooth and bounded by a simple closed curve. Additionally, the curve must be piecewise smooth and oriented in the same direction as the surface. If these conditions are not met, the theorem cannot be applied.

Are there any limitations to Stokes' theorem?

Stokes' theorem is limited to vector fields that are continuously differentiable, and it cannot be applied to surfaces with holes or self-intersections. Additionally, it only applies to surfaces in three-dimensional space, and cannot be extended to higher dimensions. It also assumes that the surface and curve are oriented in a consistent manner, which may not always be the case in real-world applications.

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