Verify Stokes' theorem

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Homework Statement


Verify Stokes' theorem for the following:

[itex]F=[y^2, x^2, -x+z][/itex]

Around the triangle with vertices [itex](0,0,1),(1,0,1),(1,1,1)[/itex]

Homework Equations


[itex]\int\int_S(curlF)\cdot ndA=\int_C F\cdot r' ds[/itex]

The Attempt at a Solution


[/B]
For the LHS:
[itex]curlF\cdot n=2x-2y[/itex]
[itex]\int\int_S(curlF)\cdot ndA=\int_0^1 \int_0^{1-x}2x-2ydydx [/itex]

This gives zero. Also integrating over the three curves of the triangle gives zero. However, the book's answer is 1/3. Any idea what the mistake is?
 

Answers and Replies

  • #2
Orodruin
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Homework Statement


Verify Stokes' theorem for the following:

[itex]F=[y^2, x^2, -x+z][/itex]

Around the triangle with vertices [itex](0,0,1),(1,0,1),(1,1,1)[/itex]

Homework Equations


[itex]\int\int_S(curlF)\cdot ndA=\int_C F\cdot r' ds[/itex]

The Attempt at a Solution


[/B]
For the LHS:
[itex]curlF\cdot n=2x-2y[/itex]
[itex]\int\int_S(curlF)\cdot ndA=\int_0^1 \int_0^{1-x}2x-2ydydx [/itex]

This gives zero. Also integrating over the three curves of the triangle gives zero. However, the book's answer is 1/3. Any idea what the mistake is?
First of all, the area of integration is wrong. The triangle is bounded by the line x = y, the x axis, and the line y = 1.

Second, in this area, the integrand is 2x - 2y = 2(x-y). For the entire area, this is positive (except for at the boundary x = y, where it is zero). The result must therefore be positive. Try drawing the points (in the x-y-plane, the z-coordinate is constant) on a piece of paper. In the region you have integrated over, the integrand is antisymmetric with respect to x=y and the the area is symmetric with respect to this and should therefore give a zero result.

In short, you have verified Stokes' theorem, but for a different curve.
 
  • #3
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Thank you, stupid me.
 

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