Your description of the surface is incomplete. But it looks like you're considering some subset of the paraboloid z = 1 - x ² - y ², so I'll go ahead and assume it's the part of said paraboloid above the x,y-plane, so that the boundary is a circle centered at the origin with radius 1.
By Stokes' theorem, the line integral of F along this boundary (∂S) is equal to the surface integral of curl(F ) over the surface itself (S). We have
F(x, y, z) = z ² i + 2x j + y ² k
which has curl
curl(F ) = (∂(y ²)/∂y - ∂(2x)/∂z) i - (∂(y ²)/∂x - ∂(z ²)/∂z) j + (∂(2x)/∂x - ∂(z ²)/∂y) k
curl(F ) = 2y i + 2z j + 2k
Parameterize S by the vector function,
r(u, v) = u cos(v) i + u sin(v) j + (1 - u ²) k
with 0 ≤ u ≤ 1 and 0 ≤ v ≤ 2π.
Take the upward-pointing normal vector to S to be
n = ∂r/∂v × ∂r/∂u
n = (-u sin(v) i + u cos(v) j ) × (cos(v) i + sin(v) j - 2u k)
n = 2u ² cos(v) i + 2u ² sin(v) j + u k
Then the integral of curl(F ) over S - and hence the line integral of F over ∂S - is
[tex]\displaystyle \iint_S \mathrm{curl}(\mathbf F(x,y,z))\cdot\mathbf S \\\\ = \iint_S \mathrm{curl}(\mathbf F(\mathbf r(u,v)))\cdot\mathbf n\,\mathrm du\,\mathrm dv \\\\ = \int_0^{2\pi}\int_0^1 \left(2u\sin(v)\,\mathbf i + 2(1-u^2)\,\mathbf j + 2\,\mathbf k\right)\cdot\left(2u^2\cos(v)\,\mathbf i+2u^2\sin(v)\,\mathbf j+u\,\mathbf k\right)\,\mathrm du\,\mathrm dv \\\\ = \int_0^{2\pi}\int_0^1 \left(4u^3\sin(v)\cos(v)+4(1-u^2)u^2\sin(v)+2u\right)\,\mathrm du\,\mathrm dv \\\\ = \int_0^{2\pi}\left(\sin(v)\cos(v)+\frac8{15}\sin(v)+1\right)\,\mathrm dv \\\\ = \boxed{2\pi}[/tex]
Just to confirm this result, we can compute the line integral directly, since it's not so difficult to deal with. Parameterize ∂S by the vector function
r(t) = cos(t ) i + sin(t ) j
with 0 ≤ t ≤ 2π. (Note that there is a k component, but its coefficient is 0.) Then
dr/dt = -sin(t ) i + cos(t ) j
and the line integral is again
[tex]\displaystyle \int_{\partial S}\mathbf F(x,y,z)\cdot\mathrm d\mathbf r \\\\ = \int_{\partial S} \mathbf F(\mathbf r(t))\cdot\frac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt \\\\= \int_0^{2\pi} (\cos(t)\,\mathbf i+\sin(t)\,\mathbf j)\cdot(-\sin(t)\,\mathbf i+\cos(t)\,\mathbf j)\,\mathrm dt \\\\ = \int_0^{2\pi}2\cos^2(t)\,\mathrm dt \\\\ = \boxed{2\pi}[/tex]
