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Vector Calculus: An Introduction

Daniel Daners

Section 13.2 The Theorem of Stokes

We have already discussed a version of Stokes’ theorem in the plane in Theorem 11.7. We can view a plane domain as a (flat) surface in space. Here we want to extend the theorem given there to “curved” surfaces lying in space. In Theorem 11.7 we have used the “curl” for plane vector fields instead of the “real” curl. If we view a plane domain as a surface in \(\mathbb R^3\) then we can consider the vector field \(\vect f=(f_1,f_2,0)\text{.}\) We saw in Remark 8.17 that in this case
\begin{equation*} \curl\vect f(\vect x)= \begin{bmatrix} 0 \\ 0\\ \dfrac{\partial}{\partial x_1}f_2(\vect x) -\dfrac{\partial}{\partial x_2}f_1(\vect x) \end{bmatrix}\text{.} \end{equation*}
If we assume that \(\vect n=(0,0,1)\) then the boundary of a plane domain is oriented counterclockwise by convention. Then the integrand on the left hand side of (11.3) becomes \((\curl\vect f)\cdot\vect n\text{.}\) For curved surfaces this turns out to be the right expression.

Proof.

Recall from Definition 7.15 that
\begin{equation*} \int_C\vect f\cdot\vect\tau\,ds =\int_C\vect f\cdot\,d\vect x\text{.} \end{equation*}
For simplicity we only give a proof for surfaces that can be completely described by one parametrisation. We therefore assume that
\begin{equation*} \vect g(\vect y) =\bigl(g_1(y_1,y_2),g_2(y_1,y_2),g_3(y_1,y_2)\bigr), \qquad (y_1,y_2)\in D \end{equation*}
is a regular parametrisation of \(S\text{.}\) Suppose that \((y_1(t),y_2(t))\text{,}\) \(t\in[a,b]\text{,}\) is a parametrisation of \(\partial D\) which is regular on every smooth part of \(\partial D\text{.}\) Then set
\begin{equation*} \vect\gamma(t):=\vect g(\vect y(t)),\qquad t\in[a,b]\text{.} \end{equation*}
Clearly the image of \(\vect\gamma\) contains \(\partial S\text{,}\) but it could be more. When deforming \(D\) into \(S\) we have to paste part of the edges together. Note however that \(\vect\gamma(t)\) runs twice along these `cuts’ but in opposite directions. Hence
\begin{equation} \int_{\partial S}\vect f\cdot\vect\tau\,ds =\int_a^b\vect f(\vect\gamma(t))\cdot\vect\gamma'(t)\,dt\text{.}\tag{13.2} \end{equation}
Examples are shown in Figure 13.3 and Figure 13.4. In the first example we cut a cylinder. The dotted parts of the boundary of \(D\) and \(S\) correspond to each other. Figure 13.4 shows a half sphere parametrised by spherical coordinates. We cut the half sphere along the dotted line and then deform it into a rectangle.
Figure 13.3. Cylindrical coordinates: the dotted lines correspond to each other.
Figure 13.4. Spherical coordinates: the dotted lines correspond to each other.
As with the proof of Green’s theorem and the divergence theorem we start with very simple vector fields, and then add up the results to get the formula for a general vector field. Hence assume that \(\vect f=(f_1,0,0)\text{.}\) Then by the previous formula
\begin{equation*} \int_{\partial S}\vect f\cdot\vect\tau\,ds =\int_a^b\vect f_1(\vect\gamma(t))\cdot\vect\gamma_1'(t)\,dt\text{.} \end{equation*}
Using the definition of \(\vect\gamma(t)\) and the chain rule (see Theorem 4.17) we get
\begin{equation*} \gamma_1'(t)=\grad f(\vect y(t))\cdot\vect y'(t)\text{.} \end{equation*}
Hence we get
\begin{align*} \int_{\partial S}\vect f\cdot\vect\tau\,ds \amp=\int_a^bf_1(\vect g(\vect y(t))) \grad g_1(\vect y(t))\cdot\vect y'(t)\,dt\\ \amp=\int_{\partial D}(f_1\circ\vect g)\grad g_1\cdot d\vect y\text{.} \end{align*}
Applying Green’s Theorem 10.3 to the vector field
\begin{equation*} (f_1\circ\vect g)\grad g_1 =\Bigl(f_1\circ\vect g\frac{\partial g_1}{\partial y_1}, f_1\circ\vect g\frac{\partial g_1}{\partial y_1}\Bigr) \end{equation*}
we see that
\begin{align*} \int_{\partial S}\vect f\cdot\vect\tau\,ds \amp=\int_{\partial D}(f_1\circ\vect g)\grad g_1\cdot d\vect y\\ \amp=\int_D\frac{\partial}{\partial y_1} \Bigl(f_1(\vect g(\vect y)) \frac{\partial g_1}{\partial y_2}(\vect y)\Bigr) -\frac{\partial}{\partial y_2} \Bigl(f_1(\vect g(\vect y) \frac{\partial g_1}{\partial y_1}(\vect y)\Bigr)\,d\vect y\text{.} \end{align*}
Now we compute the derivatives in the last integrand. Using the chain and product rules we get
\begin{align*} \frac{\partial}{\partial y_1} \amp\Bigl(f_1(\vect g(\vect y)) \frac{\partial g_1}{\partial y_2}(\vect y)\Bigr)\\ \amp=\grad f_1(\vect g(\vect y)) \cdot\frac{\partial\vect g}{\partial y_1}(\vect y) \frac{\partial g_1}{\partial y_2}(\vect y) +f_1(\vect g(\vect y))\frac{\partial^2\vect g_1}{\partial y_1\partial y_2}(\vect y) \end{align*}
and similarly
\begin{align*} \frac{\partial}{\partial y_2} \amp\Bigl(f_1(\vect g(\vect y)) \frac{\partial g_1}{\partial y_1}(\vect y)\Bigr)\\ \amp=\grad f_1(\vect g(\vect y)) \cdot\frac{\partial\vect g}{\partial y_2}(\vect y) \frac{\partial g_1}{\partial y_1}(\vect y) +f_1(\vect g(\vect y))\frac{\partial^2\vect g_1}{\partial y_2\partial y_1}(\vect y)\text{.} \end{align*}
Taking into account the symmetry of the second order partial derivatives (Proposition 4.38) we conclude that
\begin{align*} \frac{\partial}{\partial y_1} \amp\Bigl(f_1(\vect g(\vect y)) \frac{\partial g_1}{\partial y_2}(\vect y)\Bigr) -\frac{\partial}{\partial y_2} \Bigl(f_1(\vect g(\vect y) \frac{\partial g_1}{\partial y_1}(\vect y)\Bigr)\\ \amp=\grad f_1(\vect g(\vect y)) \cdot\Bigl(\frac{\partial\vect g}{\partial y_1}(\vect y) \frac{\partial g_1}{\partial y_2}(\vect y) -\frac{\partial\vect g}{\partial y_2}(\vect y) \frac{\partial g_1}{\partial y_1}(\vect y)\Bigr)\\ \amp=\frac{\partial f_1}{\partial x_1}(g(\vect y)) \Bigl(\frac{\partial g_1}{\partial y_1}(\vect y) \frac{\partial g_1}{\partial y_2}(\vect y) -\frac{\partial g_1}{\partial y_2}(\vect y) \frac{\partial g_1}{\partial y_1}(\vect y)\Bigr)\\ \amp\qquad+\frac{\partial f_1}{\partial x_2}(g(\vect y)) \Bigl(\frac{\partial g_2}{\partial y_1}(\vect y) \frac{\partial g_1}{\partial y_2}(\vect y) -\frac{\partial g_2}{\partial y_2}(\vect y) \frac{\partial g_1}{\partial y_1}(\vect y)\Bigr)\\ \amp\qquad+\frac{\partial f_1}{\partial x_3}(g(\vect y)) \Bigl(\frac{\partial g_3}{\partial y_1}(\vect y) \frac{\partial g_1}{\partial y_2}(\vect y) -\frac{\partial g_3}{\partial y_2}(\vect y) \frac{\partial g_1}{\partial y_1}(\vect y)\Bigr)\\ \amp=-\frac{\partial f_1}{\partial x_2}(g(\vect y)) \frac{\partial(g_1,g_2)}{\partial(y_1,y_2)}(\vect y) +\frac{\partial f_1}{\partial x_3}(g(\vect y)) \frac{\partial(g_3,g_1)}{\partial(y_1,y_2)}(\vect y) \end{align*}
Now observe that
\begin{equation*} \curl \begin{bmatrix} f_1 \\0\\0 \end{bmatrix} = \begin{bmatrix} 0 \\ \dfrac{\partial f_1}{\partial x_3} \\ -\dfrac{\partial f_1}{\partial x_2} \end{bmatrix}\text{.} \end{equation*}
Hence it follows from Proposition 9.31 that
\begin{align*} \int_{\partial S}(f_1,0,0)\cdot\vect\tau\,ds \amp=\int_D\frac{\partial f_1}{\partial x_3}(g(\vect y)) \frac{\partial(g_3,g_1)}{\partial(y_1,y_2)}(\vect y) -\frac{\partial f_1}{\partial x_2}(g(\vect y))\\ \frac{\partial(g_1,g_2)}{\partial(y_1,y_2)}(\vect y)\,d\vect y \amp=\int_S\curl(f_1,0,0)\cdot\vect n\,dS \end{align*}
Doing a cyclic permutation of the indices \(1\to 2\to 3\to 1\) we get
\begin{equation*} \int_{\partial S}(0,f_2,0)\cdot\vect\tau\,ds =\int_S\curl(0,f_2,0)\cdot\vect n\,dS \end{equation*}
and
\begin{equation*} \int_{\partial S}(0,0,f_3)\cdot\vect\tau\,ds =\int_S\curl(0,0,f_3)\cdot\vect n\,dS \end{equation*}
Adding the three identities we get (13.1), completing the proof of the theorem.
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