Finding the Wrong Answer with Stokes' Theorem

[greg_rack]

Homework Statement

$\vec F=<x+y^2, y+z^2,z+x^2>$, C is the triangle with vertices (1,0,0), (0,1,0), (0,0,1). Compute $\int_{C}^{}\vec F\cdot d \vec r$

Relevant Equations

Stoke's theorem
Computable form of a surface integral

From Stokes' theorem: $\int_{C}^{}\vec F\cdot d\vec r=\iint_{S}^{}curl\vec F\cdot d\vec S=\iint_{D}^{}curl\vec F\cdot(\vec r_u \times \vec r_v)dA$
To get to the latter surface integral, I started by parametrizing the triangular surface in $uv$ coordinates as:
$$\vec r=<1-u-v,u,v>, 0\leq u\leq 1, 0\leq v\leq 1-u$$
I then computed the curl of the vector field, the partial derivatives in $u$ and $v$ of the above parametrization and their cross product:
$$curl\vec F=<-2z, -2x, 2(x-y)>, \vec r_u \times \vec r_v=<1,1,1>$$
Now we can dot the curl with the cross product of the partial derivatives and get to a computable form of the surface integral
$$curl\vec F(\vec r(u,v))\cdot (\vec r_u \times \vec r_v)=-2(u+v)\rightarrow -2\int_{0}^{1}\int_{0}^{1-u}(u+v)dv du$$
which leads to a wrong answer. What am I missing?

 

[ergospherical]

Check your curl. The $z$-component looks funny.

 

[Delta2]

Yes I agree with ergospherical, checked the curl with wolfram and the z component is wrong.

 

[greg_rack]

Thank you guys, I had indeed got the k hat component of the del cross F wrong!
It's cool now

 

[Orodruin]

Fun fact: The components of the field are just cyclic permutations $x\to y \to z \to x$ so the curl must have the same property. Therefore it is sufficient to compute one of the curl components and get the rest by cyclic permutation.