Stokes' Theorem Verification for Triangle with Given Vertices

[c0der]

Homework Statement

Verify Stokes' theorem for the following:

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

Around the triangle with vertices $(0,0,1),(1,0,1),(1,1,1)$

Homework Equations

$\int\int_S(curlF)\cdot ndA=\int_C F\cdot r' ds$

The Attempt at a Solution

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

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?

 

[Orodruin]

Homework Statement

Verify Stokes' theorem for the following:

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

Around the triangle with vertices $(0,0,1),(1,0,1),(1,1,1)$

Homework Equations

$\int\int_S(curlF)\cdot ndA=\int_C F\cdot r' ds$

The Attempt at a Solution

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

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.

 

[c0der]

Thank you, stupid me.