The vector field
[tex]\vec F(x,y,z)=\langle x^2-y^2,z^2-x^2,y^2-z^2\rangle[/tex]
has curl
[tex]\nabla\times\vec F(x,y,z)=\langle2y-2z,0,2y-2x\rangle[/tex]
Stokes' theorem says the line integral of [tex]\vec F[/tex] along [tex]C[/tex] is equal to the integral of the curl of [tex]\vec F[/tex] over a surface [tex]S[/tex] with [tex]C[/tex] as its boundary. Parameterize [tex]S[/tex] by
[tex]\vec s(x,y)=\langle x,y,0\rangle[/tex]
with [tex]-16\le x\le16[/tex] and [tex]-16\le y\le16[/tex]. Take the normal vector to [tex]S[/tex] to be
[tex]\dfrac{\partial\vec s}{\partial x}\times\dfrac{\partial\vec s}{\partial y}=\langle0,0,1\rangle[/tex]
Then the line integral reduces to
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S[/tex]
[tex]=\displaystyle\int_{-16}^{16}\int_{-16}^{16}\langle2y-2z,0,2y-2x\rangle\cdot\langle0,0,1\rangle\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle2\int_{-16}^{16}\int_{-16}^{16}(y-x)\rangle\,\mathrm dx\,\mathrm dy=\boxed0[/tex]
