Stokes's Theorem for the Cone Consider the vector field $$F = \biggl \langle \sin x-\frac{y^3}3, \cos y+\frac{x^3}3, xyz \biggr \rangle.$$ Let $S$ be the surface given by the cone $$z^2 = x^2 + y^2 \text{ for } 0 \leq z \leq1.$$
I would like to verify Stokes's Theorem for the surface of the cone over the region $S.$ I have computed the curl $$\nabla \times F = \langle xz, -yz, x^2+y^2 \rangle,$$  but I get two different answers when I choose $z^2 = x^2 + y^2$ and $z = \sqrt{x^2 + y^2}.$

Recall that Stokes's Theorem says that $$\oint_{\partial S} \mathbf F \cdot d \mathbf r = \iint_S (\nabla \times \mathbf F) \, dS,$$ where $S$ is a parametrized surface; $\partial S$ is the boundary of $S$ parametrized by the curve $\mathbf r(t)$ on some closed interval $a \leq t \leq b;$ and $\nabla \times \mathbf F$ is the curl of the vector field $\mathbf F(x, y, z).$ Given that $$\mathbf F(x, y, z) = \biggl \langle \sin x - \frac{y^3}{3}, \cos y + \frac{x^3}{3}, xyz \biggr \rangle,$$ we find that $\nabla \times \mathbf F = \langle xz, -yz, x^2 + y^2 \rangle.$ Considering that $S$ is the cone $z^2 = x^2 + y^2$ for $0 \leq z \leq 1,$ we may parameterize $S$ by cylindrical coordinates $G(r, \theta) = \langle r \cos \theta, r \sin \theta, r \rangle$ over the region $U = [0, 1] \times [0, 2 \pi]$ in the $r \theta$-plane. We have  that $G_r(r, \theta) = \langle \cos \theta, \sin \theta, 1 \rangle$ and $G_\theta(r, \theta) = \langle -r \sin \theta, r \cos \theta, 0 \rangle$ so that the normal vector to $S$ is given by $$N(r, \theta) = G_r(r, \theta) \times G_\theta(r, \theta) = \langle -r \cos \theta, -r \sin \theta, r \rangle.$$ Ultimately, we find that the right-hand side of the original displayed equation is $$\iint_S (\nabla \times F) \, dS = \int_0^{2 \pi} \int_0^1 \langle r^2 \cos \theta, -r^2 \sin \theta, r^2 \rangle \cdot \langle -r \cos \theta, -r \sin \theta, r \rangle \, dr \, d \theta$$ $$= \int_0^{2 \pi} \int_0^1 2r^3 \sin^2 \theta \, dr \, d \theta = \frac{\pi}{2}. \phantom{We did it! Ya!}$$
Considering the picture of the cone $S,$ observe that $\partial S = \{(x, y, z) \,|\, x^2 + y^2 = 1 \text{ and } z = 1\}.$ Using the usual polar coordinates, we find that $\mathbf r(t) = \langle \cos t, \sin t, 1 \rangle$ parametrizes $\partial S$ for $0 \leq t \leq 2 \pi.$ We have therefore that $\mathbf r'(t) = \langle -\sin t, \cos t, 0 \rangle$ so that $$\oint_{\partial S} \mathbf F \cdot d \mathbf r = \int_0^{2 \pi} \biggl \langle \sin(\cos t) - \frac{\sin^3 t}{3}, \cos(\sin t) + \frac{\cos^3 t}{3}, \sin t \cos t \biggr \rangle \cdot \langle -\sin t, \cos t, 0 \rangle \, dt \phantom{!!}$$ $$= \int_0^{2 \pi} \biggl(-\sin t \sin(\cos t) + \frac{\sin^4 t}{3} + \cos t \cos(\sin t) + \frac{\cos^4 t}{3} \biggr) \, dt = \frac{\pi}{2}.$$ (One can compute this integral using $u$-substitution for the terms $-\sin t \sin(\cos t)$ and $\cos t \cos(\sin t)$ and the identity $\sin^4 t + \cos^4 t = \frac{\cos(4t) + 3}{4}.$) We have verified Stokes's Theorem.
