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An oriented surface is a surface equipped with a chosen orientation.

Take the normal vector outward from the cylinder-and-cone solid. The surface consists of the cylindrical side and conical roof, with no bottom disk; this convention fixes the otherwise unspecified orientation of a surface. The boundary is the radius-two circle at $z=-2$. Reversing every normal vector reverses the final sign.

For $\mathbf{F}=(y{z}^2,0,0)$, its curl is $(0,2yz,-z^2)$. For the cylindrical parametrized surface, use ${\mathbf{r}}_c=(2\cos \theta ,2\sin \theta ,z)$, with $-2\le z\le 2$, and use ${\mathbf{r}}_{c,\theta }\times {\mathbf{r}}_{c,z}=(2\cos \theta ,2\sin \theta ,0)$ as the outward area vector. Its flux isFor the cone, let $q=4-z$, $2\le z\le 4$ and ${\mathbf{r}}_k=(q\cos \theta ,q\sin \theta ,z)$. Its outward area vector is ${\mathbf{r}}_{k,\theta }\times {\mathbf{r}}_{k,z}=(q\cos \theta ,q\sin \theta ,q)$. ThusThe total surface integral is therefore $-16\pi$ with the outward convention, or $+16\pi$ for the opposite convention.

For Stokes theorem, the induced bottom-circle orientation is counterclockwise as viewed from above, opposite to the orientation of an outward bottom cap. Use $\mathbf{r}=(2\cos \theta ,2\sin \theta ,-2)$ with increasing $\theta$. Thenagreeing with the two parametrized fluxes. The seam at $z=2$ is internal and cancels between the two pieces; the cone tip does not supply another boundary curve.

WriteThe regularity of the profile makes $q>0$. The coordinate tangent vectors areso the first fundamental form of a surface of revolution has coefficientsChoose the unit normalTaking scalar products with the second derivatives of $\varphi$ gives the second fundamental form coefficientsThus the coordinate directions are principal directions, with principal curvaturesThe curvatures of a parametrized surface of revolution are thereforeReversing the orientation of a surface reverses the sign of $H$ but leaves $K$ unchanged.

On the part with $1\le z\le 2$, writeThis is the upper portion of a hyperboloid of one sheet. Choosing the outward orientation, whose normal points radially away from the $z$-axis, the vector area element isIts radial component is positive and its vertical component is negative. Reversing the normal reverses all the fluxes below.

Forthe curl isOn the surface, $z^2={\rho }^2-1$, and thereforeDirect integration gives

To verify this with the Stokes theorem, note that on a circle of fixed $\rho ,z$,The induced boundary orientation runs in the negative $\varphi$ direction on the top circle and the positive $\varphi$ direction on the bottom circle. Hence

For $S^{\prime }$, Stokes' theorem again reduces the flux to its two boundary circles. The outward orientation gives positive $\varphi$ direction at $z=-1$, where ${\rho }^2=2$, and negative $\varphi$ direction at $z=\sqrt{2}$, where ${\rho }^2=3$. Thus

Choose the orientation of a surface represented by a smooth Gauss map $N:S\to {\mathbb{S}}^2$. The shape operator at $p$ isIf its principal curvatures are $k_1,k_2$, then by definition

Let $E(x)=Ax+b$ be a proper Euclidean motion of Euclidean three-space, so $A\in SO(3)$. Orient $\tilde{S}=E(S)$ by transporting the normal:The chain rule gives the orthogonal conjugacyThe determinant and trace are invariant under similarity, so the Euclidean invariance of the shape operator yields

If the surface orientation is reversed, then $N$ and $S_p$ are multiplied by $-1$. Therefore the orientation reversal of surface curvature says

For an oriented surface with Gauss map $N$, the shape operator at $p$ is the self-adjoint linear map