Stokes' theorem

Stokes' theorem can be thought of as a generalisation of Green's theorem in higher dimension. While Green's theorem relates a line integral to a double integral over some region $\mathcal{R}$, Stokes' theorem relates a line integral to a surface integral with the help of the curl.

Consider an open surface $\mathcal{S}$ enclosed by a simple, closed, positively-oriented curve $\mathcal{C}$, as shown in Fig. 2.14,

Figure 2.14: The curve $\mathcal{C}$ is a closed curve which forms the boundary of a surface $\mathcal{S}$.

Stokes' theorem

Let $\mathcal{S}$ be a an open surface whose edge is bounded by a simple, closed, positively-oriented curve $\mathcal{C}$. Then,

Note that the theorem can be useful in either direction. Sometimes the line integral is easier to evaluate than the surface integral, sometimes the other way around.

Figure 2.15: Sketch of the surface $\mathcal{S}$ given by $f(x, y, z)=0$ where $f(x, y, z)=$ $z+x^{2}+y^{2}-5$ above the plane $z=1$ for Example 2.17. The boundary curve $\mathcal{C}$ is where the surface intersects with the plane $z=1$, shown by the red curve. Example 2.17 Use Stokes' theorem to evaluate $\iint_{\mathcal{S}} \operatorname{curl} \boldsymbol{F} \cdot d \boldsymbol{S}$ where the vector field $\boldsymbol{F}$ is given by $\boldsymbol{F}=\left(z^{2},-3 x y, x^{3} y^{3}\right)$ and $\mathcal{S}$ is given by $f(x, y, z)=0$, where $f(x, y, z)=z+x^{2}+y^{2}-5$, and above the plane $z=1$.

Solution A sketch of the surface is shown in Fig. 2.15. The boundary curve $\mathcal{C}$ is where the curve $f=0$ and the plane $z=1$ intersect. With $f=0$, we have $z=5-x^{2}-y^{2}$ and intersection with $z=1$ gives:

This implies that the curve $\mathcal{C}$ is the circle centered at the origin with radius 2 that lies on the plane $z=1$ (as shown in Fig. 2.15). We can therefore use the following parametrisation: $x=2 \cos t, y=2 \sin t$ and $z=1$ to describe the position vector $\boldsymbol{r}(t)$ :

We use Stokes' theorem [Eq. (2.129)] from right to left to evaluate the surface integral. In combination with Eq. (2.56), we have:

To compute the integral we need to evaluate the vector field $\boldsymbol{F}$ at $\boldsymbol{r}(t)$ and compute the derivative of the parametrisation, $\boldsymbol{r}^{\prime}$. Using the components of the parametrisation as given above, yields:

And, differentiating Eq. (2.131) with respect to $t$ gives:

The dot product $\boldsymbol{F}(\boldsymbol{r}(t)) \cdot \boldsymbol{r}^{\prime}(t)$ is:

Finally, the integral is:

Applications of Stokes' theorem

We end this section by giving an application of Stokes' theorem in addition to the example give above of evaluating integrals. The theorem can also be used to prove other important theorems which find application in various physical settings. In particular we will use the theorem to show that an irrotational vector field is conservative.

Suppose we have a vector field $\boldsymbol{u}$. In Subsec. 2.2.4, we mentioned that an irrotational vector field has zero curl (i.e. the vorticity is zero). This is given by curl $\boldsymbol{u}=\nabla \times \boldsymbol{u}=0$. For any closed curve $\mathcal{C}$, with an irrotational vector field $\boldsymbol{u}$, by Stokes' theorem [Eq. (2.129)], we have:

for a surface $\mathcal{S}$ bounded by $\mathcal{C}$. Now, by Eq. (2.70), Eq. (2.134) implies that any irrotational vector field is conservative.