The Theorem of Stokes

Recall from Definition 7.15 that

\int_{C} f \cdot τ d s = \int_{C} f \cdot d x .

For simplicity we only give a proof for surfaces that can be completely described by one parametrisation. We therefore assume that

g (y) = (g_{1} (y_{1}, y_{2}), g_{2} (y_{1}, y_{2}), g_{3} (y_{1}, y_{2})), (y_{1}, y_{2}) \in D

is a regular parametrisation of

S .

Suppose that

(y_{1} (t), y_{2} (t)),

t \in [a, b],

is a parametrisation of

\partial D

which is regular on every smooth part of

\partial D .

Then set

γ (t) := g (y (t)), t \in [a, b] .

Clearly the image of

γ

contains

\partial S,

but it could be more. When deforming

D

into

S

we have to paste part of the edges together. Note however that

γ (t)

runs twice along these `cuts’ but in opposite directions. Hence

\begin{matrix} (13.2) & \int_{\partial S} f \cdot τ d s = \int_{a}^{b} f (γ (t)) \cdot γ^{'} (t) d t . \end{matrix}

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

f = (f_{1}, 0, 0) .

Then by the previous formula

\int_{\partial S} f \cdot τ d s = \int_{a}^{b} f_{1} (γ (t)) \cdot γ_{1}^{'} (t) d t .

Using the definition of

γ (t)

and the chain rule (see Theorem 4.17) we get

γ_{1}^{'} (t) = grad f (y (t)) \cdot y^{'} (t) .

Hence we get

\begin{aligned} \int_{\partial S} f \cdot τ d s & = \int_{a}^{b} f_{1} (g (y (t))) grad g_{1} (y (t)) \cdot y^{'} (t) d t \\ = \int_{\partial D} (f_{1} \circ g) grad g_{1} \cdot d y . \end{aligned}

Applying Green’s Theorem 10.3 to the vector field

(f_{1} \circ g) grad g_{1} = (f_{1} \circ g \frac{\partial g_{1}}{\partial y_{1}}, f_{1} \circ g \frac{\partial g_{1}}{\partial y_{1}})

we see that

\begin{aligned} \int_{\partial S} f \cdot τ d s & = \int_{\partial D} (f_{1} \circ g) grad g_{1} \cdot d y \\ = \int_{D} \frac{\partial}{\partial y_{1}} (f_{1} (g (y)) \frac{\partial g_{1}}{\partial y_{2}} (y)) - \frac{\partial}{\partial y_{2}} (f_{1} (g (y) \frac{\partial g_{1}}{\partial y_{1}} (y)) d y . \end{aligned}

Now we compute the derivatives in the last integrand. Using the chain and product rules we get

\begin{aligned} \frac{\partial}{\partial y_{1}} & (f_{1} (g (y)) \frac{\partial g_{1}}{\partial y_{2}} (y)) \\ = grad f_{1} (g (y)) \cdot \frac{\partial g}{\partial y_{1}} (y) \frac{\partial g_{1}}{\partial y_{2}} (y) + f_{1} (g (y)) \frac{\partial^{2} g_{1}}{\partial y_{1} \partial y_{2}} (y) \end{aligned}

and similarly

\begin{aligned} \frac{\partial}{\partial y_{2}} & (f_{1} (g (y)) \frac{\partial g_{1}}{\partial y_{1}} (y)) \\ = grad f_{1} (g (y)) \cdot \frac{\partial g}{\partial y_{2}} (y) \frac{\partial g_{1}}{\partial y_{1}} (y) + f_{1} (g (y)) \frac{\partial^{2} g_{1}}{\partial y_{2} \partial y_{1}} (y) . \end{aligned}

Taking into account the symmetry of the second order partial derivatives (Proposition 4.38) we conclude that

\begin{aligned} \frac{\partial}{\partial y_{1}} & (f_{1} (g (y)) \frac{\partial g_{1}}{\partial y_{2}} (y)) - \frac{\partial}{\partial y_{2}} (f_{1} (g (y) \frac{\partial g_{1}}{\partial y_{1}} (y)) \\ = grad f_{1} (g (y)) \cdot (\frac{\partial g}{\partial y_{1}} (y) \frac{\partial g_{1}}{\partial y_{2}} (y) - \frac{\partial g}{\partial y_{2}} (y) \frac{\partial g_{1}}{\partial y_{1}} (y)) \\ = \frac{\partial f_{1}}{\partial x_{1}} (g (y)) (\frac{\partial g_{1}}{\partial y_{1}} (y) \frac{\partial g_{1}}{\partial y_{2}} (y) - \frac{\partial g_{1}}{\partial y_{2}} (y) \frac{\partial g_{1}}{\partial y_{1}} (y)) \\ + \frac{\partial f_{1}}{\partial x_{2}} (g (y)) (\frac{\partial g_{2}}{\partial y_{1}} (y) \frac{\partial g_{1}}{\partial y_{2}} (y) - \frac{\partial g_{2}}{\partial y_{2}} (y) \frac{\partial g_{1}}{\partial y_{1}} (y)) \\ + \frac{\partial f_{1}}{\partial x_{3}} (g (y)) (\frac{\partial g_{3}}{\partial y_{1}} (y) \frac{\partial g_{1}}{\partial y_{2}} (y) - \frac{\partial g_{3}}{\partial y_{2}} (y) \frac{\partial g_{1}}{\partial y_{1}} (y)) \\ = - \frac{\partial f_{1}}{\partial x_{2}} (g (y)) \frac{\partial (g_{1}, g_{2})}{\partial (y_{1}, y_{2})} (y) + \frac{\partial f_{1}}{\partial x_{3}} (g (y)) \frac{\partial (g_{3}, g_{1})}{\partial (y_{1}, y_{2})} (y) \end{aligned}

Now observe that

curl [\begin{matrix} f_{1} \\ 0 \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ \frac{\partial f_{1}}{\partial x_{3}} \\ - \frac{\partial f_{1}}{\partial x_{2}} \end{matrix}] .

Hence it follows from Proposition 9.31 that

\begin{aligned} \int_{\partial S} (f_{1}, 0, 0) \cdot τ d s & = \int_{D} \frac{\partial f_{1}}{\partial x_{3}} (g (y)) \frac{\partial (g_{3}, g_{1})}{\partial (y_{1}, y_{2})} (y) - \frac{\partial f_{1}}{\partial x_{2}} (g (y)) \\ \frac{\partial (g_{1}, g_{2})}{\partial (y_{1}, y_{2})} (y) d y & = \int_{S} curl (f_{1}, 0, 0) \cdot n d S \end{aligned}

Doing a cyclic permutation of the indices

1 \to 2 \to 3 \to 1

we get

\int_{\partial S} (0, f_{2}, 0) \cdot τ d s = \int_{S} curl (0, f_{2}, 0) \cdot n d S

and

\int_{\partial S} (0, 0, f_{3}) \cdot τ d s = \int_{S} curl (0, 0, f_{3}) \cdot n d S

Adding the three identities we get (13.1), completing the proof of the theorem.

Vector Calculus: An Introduction

Section 13.2 The Theorem of Stokes

Theorem 13.2. Stokes’ Theorem.

Proof.