26. Stokes' Theorem

Let \(S\) be a nice surface in \(\mathbb{R}^3\) with a nice properly oriented boundary, \(\partial S\), and let \(\vec{F}\) be a nice vector field on \(S\). Then \[ \iint_S \vec{\nabla}\times\vec{F}\cdot d\vec{S} =\oint_{\partial S} \vec{F}\cdot d\vec{s} \] Each piece of the boundary of the surface must be traversed counterclockwise as seen from the tip of the normal vector to the surface.

d. Applications

3. Path Independence

In either case, \(\partial S=C_1-C_2\), and so Stokes' Theorem says \[ \iint_S \vec{\nabla}\times\vec{F}\cdot d\vec{S} =\oint_{\partial S} \vec{F}\cdot d\vec{s} =\oint_{C_1} \vec{F}\cdot d\vec{s} -\oint_{C_2} \vec{F}\cdot d\vec{s} \] This may be rewritten as \[ \int_{C_1} \vec{F}\cdot d\vec{s} =\int_{C_2} \vec{F}\cdot d\vec{s} +\iint_S \vec{\nabla}\times\vec{F}\cdot d\vec{S} \qquad\qquad \text{(*)} \] In other words, to compute \(\displaystyle \int_{C_1} \vec{F}\cdot d\vec{s}\), you may alternatively compute \(\displaystyle \int_{C_2} \vec{F}\cdot d\vec{s}\) and \(\displaystyle \iint_S \vec{\nabla}\times\vec{F}\cdot d\vec{S}\), if that is easier.

Suppose that \(\vec F\) is a vector field satisfying \[ \int_{C_1} \vec{F}\cdot d\vec{s} =\int_{C_2} \vec{F}\cdot d\vec{s} \] whenever \(C_1\) can be continuously deformed into \(C_2\). Then we say \(\vec F\) has path independent line integrals.

In the special case that \(\vec\nabla\times\vec{F}=\vec 0\), then equation (*) says \(\vec F\) has path independent line integrals.

Let's compute a line integral in two different ways: Once directly and once using Stokes' Theorem and the relation \((*)\).

(1) directly along the curve \(C_1\), shown in blue, and
(2) using the relation \((*)\) where \(C_2\) is the straight line segment from \(A\) to \(B\), shown in red, and \(S\) is the region inside the semi-circle inside \(x^2+y^2=16\) for \(y\ge0\), shown in cyan.

The semicircle \(C_1\) is parameterized as \(\vec{r}(\theta)=(4\cos\theta, 4\sin\theta)\) for \(0 \le \theta \le \pi\). So the velocity is \[ \vec{v}(\theta) =\langle-4\sin\theta, 4\cos\theta\rangle. \] This is properly parametrized counterclockwise. Next we evaluate the vector field on the curve: \[ \vec{F} =\langle 2x+y,y-2\rangle =\langle 8\cos\theta+4\sin\theta, 4\sin\theta- 2\rangle \] So the integral is: (Be careful with the range of \(\theta\).) \[\begin{aligned} \int_{C_1} \vec{F}\cdot d\vec{s} &=\int_0^\pi \vec{F}\cdot\vec{v}\,d\theta \\ &=\int_0^\pi (-32\sin\theta\cos\theta- 16\sin^2\theta +16\sin\theta\cos\theta- 8\cos\theta)\,d\theta \\ &=\int_0^\pi (-16\sin\theta\cos\theta - 16\dfrac{1-\cos(2\theta)}{2}-8\cos\theta)\,d\theta \\ &=\left[- 8\sin^2\theta - 8\left(\theta- \dfrac{\sin(2\theta)}{2}\right) - 8\sin\theta\right]_0^\pi =- 8\pi \end{aligned}\]

The line \(C_2\) may be parameterized as: \(\vec{r}(t)=(t, 0)\) for \(-4\le t \le4\). The velocity is: \(\vec{v}=(1, 0)\). This points to the right, but we need it to point to the left. So we reverse it: \[ \vec{v}=(- 1, 0) \] The vector field on the line is: \[ \vec{F}(\vec{r}(t)) =\langle 2x+y,y-2\rangle =\langle 2t,- 2\rangle. \] So the line integral is: \[ \int_{C_2} \vec{F}\cdot d\vec{s} =\int_{-4}^4 \vec{F}\cdot\vec{v}\,dt =\int_{-4}^4 (-2t)\,dt =\left[\rule{0pt}{10pt}-t^2\right]_{-4}^4=0 \]

To use the formulas for Stokes' Theorem, we need to work in \(\mathbb{R}^3\). So the surface is the semi-circle \(x^2+y^2\le16\) for \(y\ge0\) with \(z=0\). It is best to work in cylindrical coordinates. So the surface may be parametrized as: \(\vec{R}=(r\cos\theta, r\sin\theta, 0)\) for which the normal is \(\vec{N}=(0,0,r)\). This points up along the \(z\)-axis. So the curve must be traversed counterclockwise which \(\partial S=C_1-C_2\) is.

The curl of the vector field \(\vec{F}=\langle 2x+y,y-2,0\rangle\) in rectangular coordinates is: \[ \vec{\nabla}\times\vec{F} =\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \partial_x & \partial_y & \partial_z \\ 2x+y & y-2 & 0 \\ \end{vmatrix} =\langle 0, 0,-1\rangle \] So the surface integral over \(S\) is: (Be careful with the range of \(\theta\).) \[\begin{aligned} \iint_S &\vec{F}\cdot d\vec{S} =\int_0^\pi\int_0^4 \vec{F}\cdot\vec{N}\,dr\,d\theta \\ &=\int_0^\pi\int_0^4 (-r)\,dr\,d\theta =-\pi\left[\dfrac{r^2}{2}\right]_0^4 =-8\pi \end{aligned}\] That was easy!

Finally we add the two integrals to get our answer: \[ \int_{C_1} \vec{F}\cdot d\vec{S} =\int_{C_2} \vec{F}\cdot d\vec{S} +\iint_S \vec{\nabla}\times\vec{F}\cdot d\vec{S} =0-8\pi=-8\pi \]

Notice the answers are the same. Clearly the integrals over the line and area were easier than the integral over the semi-circle.

Consider the integral \(\displaystyle I=\int_A^B \vec{F}\cdot d\vec{s}\) for \(\displaystyle \vec{F}=\langle e^x, e^y, e^z \rangle\), where \(A=(0,0,0)\) and \(B=(1,2,3)\).