Stokes Theorem

1 Stokes Theorem

Let \(F (x,y, z) \) be a space vector field, \(p \) a point in the domain of \(F\), and \(\rm {v}\) a unit vector based at \(p\). Let \(C\subset \mathbb {R}^2 \) be a very small circle centered at \(p\), around \(\rm {v}\), which travels counterclockwise when seen from the tip of \(\rm {v}\). Then the integral

\[ \int _C F \cdot d \gamma \]

measures how much rotation is generated by \(F\) around the axis \(\rm {v}\). It turns out we have another quantity that measures precisely that:

\[ \text {Curl}(F)(p) \cdot \rm {v} . \]

Overall, if we have an oriented surface \( \Sigma \subset \mathbb {R}^3\) with boundary \(\partial \Sigma \) oriented counterclockwise, the two integrals

\begin{gather*} \int _{\partial \Sigma } F \cdot d \gamma \\ \iint _{\Sigma } ( \text {Curl}(F) \cdot N )\, dS \end{gather*}

measure how much rotation the vector field \(F\) generates over \(\Sigma \) (with axis perpendicular to the surface). Stokes Theorem states that they agree.

Let \(C \subset \mathbb {R}^3\) be the circle \(x^2 + y^2 = 25\) in the \(xy\)-plane, oriented counterclockwise, when seen from above. Compute
\[ \int _C \langle x^2 - 2z - 5 , 3x - e^y , z^3 - xy + 1 \rangle \cdot d \gamma \]

Set \(F = \langle x^2 - 2z - 5 , 3x - e^y , z^3 - xy + 1 \rangle \) and compute

\[ \text {Curl} (F) = \langle - x , -2 + y , 3 \rangle \]

If we define \(\Sigma \) to be disk \(x^2 + y^2 \leq 25 \) in the \(xy\)-plane, oriented upward, we have \(\partial \Sigma = C\), and by Stokes Theorem,

\begin{align*} \int _C \langle & x^2 - 2z - 5 , 3x - e^y , z^3 - xy + 1 \rangle \cdot d \gamma \\ & = \iint _{\Sigma } \langle - x , -2 + y , 3 \rangle \cdot dS \\ & = \iint _{\Sigma } \langle - x , -2 + y , 3 \rangle \cdot \langle 0,0,1 \rangle \, dS \\ & = \iint _{\Sigma } 3 \, dS \\ & = 75 \pi \end{align*}
Let \(C\subset \mathbb {R}^3\) be the curve that travels along straight lines first from \((1,0,0)\) to \((0,0,1)\), then from \((0,0,1)\) to \((0,1,0)\), and then from \((0,1,0)\) to \((1,0,0)\). Compute
\[ \int _C \langle 3x + y, e^y - 2x , 5 z - 4x - 1 \rangle \cdot d \gamma \]

Set \(F = \langle 3x + y, e^y - 2x , 5 z - 4x - 1 \rangle \) and compute

\[ \text {Curl} (F) = \langle 0, 4 , -3 \rangle \]

If we define \(\Sigma \) to be the triangle with vertices \((1,0,0)\), \((0,1,0)\), \((0,0,1)\), oriented upward, by Stokes Theorem we have

\[ \int _C \langle 3x + y, e^y - 2x , 5 z - 4x - 1 \rangle \cdot d \gamma = - \iint _{\Sigma } \langle 0,4,-3\rangle \cdot dS \]

where the minus sign appears because when we look at \(C\) from the tip of \(N = \langle 1,1,1\rangle \), it goes clockwise. Note that \(\langle 0,4,-3 \rangle \cdot N = 4-3 = 1\), so

\[ \iint _{\Sigma } \langle 0,4,-3\rangle \cdot dS = \iint _{\Sigma } 1 \, dS = \text {Area}(\Sigma ) \]

The triangle is equilateral with side \(\sqrt {2}\), so its area is \(\frac {1}{2}\), and

\[ \int _C \langle 3x + y, e^y - 2x , 5 z - 4x - 1 \rangle \cdot d \gamma = - \frac {1}{2} \]
Let \(C\subset \mathbb {R}^3\) be the curve that goes from \((1,0)\) to \((-1,0)\) along the arc \(x^2 + y^2 = 1\), \(y \geq 0\), in the \(xy\)-plane, followed by the curve that goes back to \((1,0)\) along the arc \(x^2 + z^2 = 1\), \(z\geq 0\), in the \(xz\)-plane. Compute
\[ \int _C \langle 3x + 2 + z^2 , 4 \cos y - z^2 , 2y - 5 \rangle \cdot d \gamma \]

Set \(F = \langle 3x + 2 + z^2 , 4 \cos y , 2y - 5 \rangle \). We compute

\[ \text {Curl} (F) = \langle 2 , 2z , 0 \rangle \]

If we define \(\Sigma \) to be the portion of the sphere \(x^2 + y^2 + z^2 = 1\) with \(y \geq 0\), \(z \geq 0\), oriented upward, we get \(\partial \Sigma = C\). Then by Stokes Theorem,

\[ \int _C \langle 3x + 2 + z^2 , 4 \cos y , 2y - 5 \rangle \cdot d \gamma = \iint _{\Sigma } \langle 2 , 2z , 0 \rangle \cdot dS \]

Parametrizing \(\Sigma \), we use \(\varphi : [0, \pi ] \times [0, \pi / 2] \to \Sigma \) with

\[ \varphi (u,v) = ( \cos u \sin v , \sin u \sin v , \cos v ) , \]

then

\begin{align*} \frac {\partial \varphi }{\partial u } & = \langle -\sin u \sin v , \cos u \sin v , 0 \rangle \\ \frac {\partial \varphi }{\partial v } & = \langle \cos u \cos v , \sin u \cos v , - \sin v \rangle \\ \frac {\partial \varphi }{\partial u } \times \frac {\partial \varphi }{\partial v } & = \langle - \cos u \sin ^2 v , - \sin u \sin ^2 v , - \sin v \cos v \rangle \end{align*}

Note that \( \frac {\partial \varphi }{\partial u } \times \frac {\partial \varphi }{\partial v } \) points inwards, so \(\varphi \) is NOT compatible with the orientation. The integrand becomes

\begin{align*} \langle 2 , 2 \cos v ,0 \rangle & \cdot \langle - \cos u \sin ^2 v , - \sin u \sin ^2 v , - \sin v \cos v \rangle \\ & = -2 \cos u \sin ^2 v - 2 \sin u \cos v \sin ^2 v \end{align*}

Then (recalling that \(\varphi \) was not compatible with the orientation), we conclude

\begin{align*} \iint _{\Sigma } \langle 2 , 2z , 0 \rangle \cdot dS & = - \int _{0 }^{\pi } \int _0 ^{\pi /2} [-2 \cos u \sin ^2 v - 2 \sin u \cos v \sin ^2 v ] \, d v du \\ & = 0 + 4 \int _0 ^{ \pi / 2 } \cos v \sin ^2 v \, dv \\ & = 4 \left [ \frac {\sin ^3 v}{3 } \right ] _{v = 0} ^{ \pi /2} \\ & = \frac {4}{3} \end{align*}