Stokes’ Theorem: Learn It 2

Applying Stokes’ Theorem

Stokes’ theorem translates between the flux integral of surface [latex]S[/latex] to a line integral around the boundary of [latex]S[/latex]. Therefore, the theorem allows us to compute surface integrals or line integrals that would ordinarily be quite difficult by translating the line integral into a surface integral or vice versa.

Calculate surface integral [latex]\displaystyle\iint_S\text{curl }{\bf{F}}\cdot{d}{\bf{S}}[/latex], where [latex]S[/latex] is the surface, oriented outward, in Figure 1 and [latex]{\bf{F}}=\langle{z},2xy,x+y\rangle[/latex].

A diagram of a complicated surface S in a three dimensional vector field. The surface is a cylindrical tube that twists about in the three-dimensional space arbitrarily. The upper end of the tube is an open circle leading to inside the tube. It is centered on the z-axis at a height of z=1 and has a radius of 1. The bottom end of the tube is closed with a hemispherical cap on the end. The vector arrows are best described by their components. The x component is positive everywhere and becomes larger as z increases. The y component is positive in the first and third octants and negative in the other two. The z component is zero when y=x and becomes more positive with more positive x and y values and more negative in the other direction.
Figure 1. A complicated surface in a vector field.

An amazing consequence of Stokes’ theorem is that if [latex]S'[/latex] is any other smooth surface with boundary [latex]C[/latex] and the same orientation as [latex]S[/latex], then [latex]\displaystyle\iint_S\text{curl }{\bf{F}}\cdot{d}{\bf{S}}=\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}=0[/latex] because Stokes’ theorem says the surface integral depends on the line integral around the boundary only.

Surface Independence

In the previous example, we calculated a surface integral simply by using information about the boundary of the surface. In general, let [latex]S_1[/latex] and [latex]S_2[/latex] be smooth surfaces with the same boundary [latex]C[/latex] and the same orientation. By Stokes’ theorem:

[latex]\displaystyle\iint_{S_1}\text{curl }{\bf{F}}\cdot{d}{\bf{S}}=\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\iint_{S_2}\text{curl }{\bf{F}}\cdot{d}{\bf{S}}[/latex].

Therefore, if [latex]\displaystyle\iint_{S_1}\text{curl }{\bf{F}}\cdot{d}{\bf{S}}[/latex] is difficult to calculate but [latex]\displaystyle\iint_{S_2}\text{curl }{\bf{F}}\cdot{d}{\bf{S}}[/latex] is easy to calculate, Stokes’ theorem allows us to calculate the easier surface integral. In the previous example, we could have calculated [latex]\displaystyle\iint_{S}\text{curl }{\bf{F}}\cdot{d}{\bf{S}}[/latex] by calculating [latex]\displaystyle\iint_{S^\prime}\text{curl }{\bf{F}}\cdot{d}{\bf{S}}[/latex], where [latex]S'[/latex] is the disk enclosed by boundary curve [latex]C[/latex] (a much simpler surface to work with).

Surface Independence and Path Independence

The equation [latex]{\displaystyle\iint_{S_1}\text{curl }{\bf{F}}\cdot{d}{\bf{S}}=\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\iint_{S_2}\text{curl }{\bf{F}}\cdot{d}{\bf{S}}}[/latex] shows that flux integrals of curl vector fields are surface independent in the same way that line integrals of gradient fields are path independent.

Recall that if [latex]\mathbf{F}[/latex] is a two-dimensional conservative vector field defined on a simply connected domain, [latex]f[/latex] is a potential function for [latex]\mathbf{F}[/latex], and [latex]C[/latex] is a curve in the domain of [latex]\mathbf{F}[/latex], then [latex]\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}[/latex] depends only on the endpoints of [latex]C[/latex]. Therefore if [latex]C'[/latex] is any other curve with the same starting point and endpoint as [latex]C[/latex], then [latex]\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\int_{C^\prime}{\bf{F}}\cdot{d}{\bf{r}}[/latex].

In other words, the value of the integral depends on the boundary of the path only; it does not depend on the path itself.

Analogously, suppose that [latex]S[/latex] and [latex]S'[/latex] are surfaces with the same boundary and same orientation, and suppose that [latex]\mathbf{G}[/latex] is a three-dimensional vector field that can be written as the curl of another vector field [latex]\mathbf{F}[/latex] (so that [latex]\mathbf{F}[/latex] is like a “potential field” of [latex]\mathbf{G}[/latex]). By Stokes’ theorem:

[latex]\displaystyle\int_S{\bf{G}}\cdot{d}{\bf{S}}=\displaystyle\iint_{S}\text{curl }{\bf{F}}\cdot{d}{\bf{S}}=\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\iint_{S^\prime}\text{curl }{\bf{F}}\cdot{d}{\bf{S}}=\displaystyle\iint_{S^\prime}\text{curl }{\bf{G}}\cdot{d}{\bf{S}}[/latex].

Therefore, the flux integral of [latex]{\bf{G}}[/latex] does not depend on the surface, only on the boundary of the surface. Flux integrals of vector fields that can be written as the curl of a vector field are surface independent in the same way that line integrals of vector fields that can be written as the gradient of a scalar function are path independent.

Calculate the line integral [latex]\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}[/latex], where [latex]{\bf{F}}=\langle{x}y,x^2+y^2+z^2,yz\rangle[/latex] and [latex]C[/latex] is the boundary of the parallelogram with vertices [latex](0, 0, 1)[/latex], [latex](0, 1, 0)[/latex], [latex](2, 0, -1)[/latex], and [latex](2, 1, -2)[/latex].