Stokes theorem

Let $S$ be an oriented smooth surface with piecewise smooth boundary, and let $F$ be a continuously differentiable vector field on a neighborhood of $S$. With the compatible boundary direction determined by the right-hand rule,For $G=(-y,x,0)$ on the horizontal unit disc with upward normal, the left side is $2$ times its area, giving $2\pi$. The counterclockwise boundary has tangential field of magnitude one and circumference $2\pi$, giving the same line integral. Reversing the surface normal reverses the boundary direction and both signs. The theorem replaces a circulation calculation with a surface integral of the curl, whichever is easier.

For sufficiently smooth scalar fields near an oriented surface, the product rule for the curl gives $\nabla \times (\varphi \nabla \psi )=\varphi \nabla \times \nabla \psi +\nabla \varphi \times \nabla \psi =\nabla \varphi \times \nabla \psi$. Stokes theorem then proves the displayed identity with consistent boundary orientation. A scalar multiple of a gradient need not itself be a gradient; the cross product of the two gradients detects its local circulation.

Stokes theorem applies to an oriented surface with several boundary components. Cutting a cylindrical side along one generator produces a parameter rectangle: the two copies of the cut are traversed in opposite directions and their line integrals cancel for a single-valued smooth vector field. The upper and lower rims retain opposite directions when viewed from the same side. This proves why cylindrical topology does not require a modification of Stokes theorem; it requires inclusion of the entire oriented boundary.

For a continuously differentiable vector field on an oriented surface with compatibly oriented boundary, this vector identity follows by dotting both sides with an arbitrary constant vector $a$ and applying Stokes theorem to $\mathbf{F}\times a$. The curl identity $\nabla \times (a\times \mathbf{F})=a\nabla \cdot \mathbf{F}-(a\cdot \nabla )\mathbf{F}$ gives the surface integrand. In components, $[(d\mathbf{S}\times \nabla )\times \mathbf{F}{]}_i=d{S}_j{\partial }_i{F}_j-d{S}_i{\partial }_j{F}_j$, with the derivatives acting on $\mathbf{F}$. For the position vector field $\mathbf{F}=\mathbf{r}$, it equals $-2d\mathbf{S}$; hence the line integral is minus twice the oriented vector area.

Let a closed oriented surface $S$ lose a patch $H$, leaving $S^{\prime }$. Assume a smooth vector field on the relevant surfaces and that its total curl flux through $S$ is zero. The Stokes theorem gives the title formula when the edge is oriented as the boundary of the remaining surface $S^{\prime }$. The boundary of the removed patch $H$, with its original normal, has the opposite direction and hence the opposite circulation.

For a cylindrical side of radius $R$, if $(\nabla \times \mathbf{v})\cdot \mathbf{n}=-\alpha R$ everywhere on the removed patch, the circulation around the boundary of the remaining surface is $\alpha R\mathrm{area}(H)$. This uses actual curved area, rather than its planar projection, and does not depend on the patch shape. For multiple boundary components, the formula concerns the sum with their induced orientations.

For an oriented smooth surface and compatible boundary orientation, apply Stokes theorem to $\varphi \mathbf{k}$ for an arbitrary constant vector $\mathbf{k}$. Since $\nabla \times (\varphi \mathbf{k})=\nabla \varphi \times \mathbf{k}$, its surface integral is $\mathbf{k}\cdot {\int }_{S}d\mathbf{S}\times \nabla \varphi$. The boundary integral is $\mathbf{k}\cdot \oint \varphi d\mathbf{x}$. Equality for every $\mathbf{k}$ proves the displayed vector identity. Reversing both orientations preserves the identity.

For an upward-oriented band of the circular elliptic paraboloid between radii $r_0$ and $R$, the vector field $B=(-y^3,x^3,z^3)$ has vertical curl $3(x^2+y^2)$. Its surface integral is $3\pi (R^4-r_0^4)/2$. Stokes theorem gives the same value from the counterclockwise outer circle minus the counterclockwise inner circle. Equivalently the induced inner boundary direction is clockwise. Forgetting the inner boundary or its reversed orientation changes the answer.