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17.7 Stokes' Theorem (part 1)

Stokes' Theorem allows us to trade the surface integral of the curl of vector field for a line integral on the boundary of the surface.

Figure: A green hemispherical surface S with a red boundary curve C outlining the open base.

Surface \(S\) and curve \(C\) outlining the open part.

The surface must be open for Stokes' Theorem (has a hole).

Figure: A small green cup-like surface S with an open top boundary C.

cup w/o lid is open

Both the surface and the boundary curve are oriented; they obey the right-hand rule.

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Surface oriented by the normal vector

Align thumb of right hand w/ \(\vec{n}\). The other fingers naturally curl in the direction that is the positive orientation for the curve.

boundary curve \(C\)

Figure: A green bowl-shaped surface with a red normal vector n pointing downwards.

Another way to orient both:

Imagine walking along the boundary w/ head pointing toward positive normal, then the enclosed region is on your left.

Figure: A green hemisphere with a person walking on the boundary curve, head aligned with the normal vector n.

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Stokes' Theorem

\[ \iint_{S} \text{curl } \vec{F} \cdot d\vec{S} = \oint_{C} \vec{F} \cdot d\vec{r} \]

\( S \): Surface
\( C \): boundary curve

\[ d\vec{S} = \vec{n} dS \]\[ = (\vec{r}_u \times \vec{r}_v) dA \text{ or } (\vec{r}_v \times \vec{r}_u) dA \]

whichever is the positive normal (convention: upward or outward unless otherwise stated)

Why is \[ \iint_{S} \text{curl } \vec{F} \cdot d\vec{S} = \oint_{C} \vec{F} \cdot d\vec{r} \]

\( \iint_{S} \text{curl } \vec{F} \cdot d\vec{S} \)

accumulation of curl of vector field on surface

\( \oint_{C} \vec{F} \cdot d\vec{r} \)

circulation on boundary

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left side: \[ \iint_{S} \text{curl } \vec{F} \cdot d\vec{S} \]

Figure: A 2x3 grid of squares with internal arrows showing circulation that cancels at common edges.

In each grid, we accumulate curl.

Along common edge \( \rightarrow \) cancellation of flow.

Figure: A single large square boundary with arrows showing net circulation along the outer perimeter.

just the flow along the boundary \( \rightarrow \) line integral

\[ \oint_{C} \vec{F} \cdot d\vec{r} \]

Stokes' works just like how Green's works.

In fact, Green's is a special case of Stokes'.

Stokes' \( \rightarrow \) works for curved surface (e.g. hemisphere)
Green's \( \rightarrow \) flat surface only

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Example: Verifying Stokes' Theorem

Given the vector field:

\[ \vec{F} = \langle y, -x, 0 \rangle \]

Let \( S \) be the part of a sphere of radius 3 where \( z \geq 3/2 \). The normal is positive pointing outward.

Verify Stokes' Theorem:

\[ \iint_S \text{curl } \vec{F} \cdot d\vec{S} = \oint_C \vec{F} \cdot d\vec{r} \]

Left Side

Parametrize \( S \) first. Let's use spherical coordinates with \( \rho = 3 \). We will use parameters \( u = \phi \) and \( v = \theta \).

\[ \begin{aligned} x &= \rho \sin \phi \cos \theta = 3 \sin u \cos v \\ y &= \rho \sin \phi \sin \theta = 3 \sin u \sin v \\ z &= \rho \cos \phi = 3 \cos u \end{aligned} \]

Figure: 3D plot of a spherical cap with radius 3, cut at z=3/2, showing an outward normal vector.

To find the bounds for \( u \) (which is \( \phi \)):

\[ \begin{aligned} z &= \rho \cos \phi = 3 \cos \phi \\ \frac{3}{2} &= 3 \cos \phi \implies \cos \phi = \frac{1}{2} \implies \phi = \frac{\pi}{3} \end{aligned} \]

Thus, the parameter ranges are:

\[ 0 \leq u \leq \frac{\pi}{3}, \quad 0 \leq v \leq 2\pi \]

Figure: Geometric diagram showing the vertical cross-section of the sphere to determine the angle phi.

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Surface Parametrization and Normal Vector

\[ \vec{r}(u,v) = \langle 3 \sin u \cos v, 3 \sin u \sin v, 3 \cos u \rangle \]

Partial derivatives:

\[ \begin{aligned} \vec{r}_u &= \langle 3 \cos u \cos v, 3 \cos u \sin v, -3 \sin u \rangle \\ \vec{r}_v &= \langle -3 \sin u \sin v, 3 \sin u \cos v, 0 \rangle \end{aligned} \]

Cross product (normal vector):

\[ \vec{r}_u \times \vec{r}_v = \langle 9 \sin^2 u \cos v, 9 \sin^2 u \sin v, 9 \sin u \cos u \rangle \]

Is this "outward"? If so, \( z > 0 \).

Since \( 0 \leq u \leq \frac{\pi}{3} \implies \cos u, \sin u > 0 \), so \( z > 0 \). So, yes.

Evaluating the Surface Integral

Vector field and its curl:

\[ \vec{F} = \langle y, -x, 0 \rangle, \quad \text{curl } \vec{F} = \nabla \times \vec{F} = \dots = \langle 0, 0, -2 \rangle \]

The surface integral becomes:

\[ \begin{aligned} \iint_S \text{curl } \vec{F} \cdot d\vec{S} &= \iint_R \text{curl } \vec{F} \cdot (\vec{r}_u \times \vec{r}_v) \, dA \\ &= \int_0^{2\pi} \int_0^{\pi/3} \langle 0, 0, -2 \rangle \cdot \langle \dots, \dots, 9 \sin u \cos u \rangle \, du \, dv \\ &= \int_0^{2\pi} \int_0^{\pi/3} -18 \sin u \cos u \, du \, dv = \dots = -27\pi/2 \end{aligned} \]

\[ -27\pi/2 \]

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Stokes' Theorem Application

Stokes' says we can do \(\oint_C \vec{F} \cdot d\vec{r}\) instead.

Boundary is counterclockwise if viewed from above because \(\vec{n}\) is out/up.

\(C\): circle radius \(\frac{\sqrt{27}}{2}\)

Figure: 3D coordinate system showing a spherical cap with a boundary curve C and normal vector n pointing outwards.

\[\vec{r}(t) = \left\langle \frac{\sqrt{27}}{2} \cos t, \frac{\sqrt{27}}{2} \sin t, \frac{3}{2} \right\rangle\] \[0 \le t \le 2\pi\]

Surface and Boundary Equations

\(x^2 + y^2 + z^2 = 3^2\)

at \(z = 3/2\)

\(x^2 + y^2 = 9 - \frac{9}{4} = \frac{27}{4}\)

Vector Field and Differential

\[\vec{F} = \langle y, -x, 0 \rangle = \left\langle \frac{\sqrt{27}}{2} \sin t, -\frac{\sqrt{27}}{2} \cos t, 0 \right\rangle\] \[d\vec{r} = \vec{r}' dt = \left\langle -\frac{\sqrt{27}}{2} \sin t, \frac{\sqrt{27}}{2} \cos t, 0 \right\rangle\]

Line Integral Calculation

\[\oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi} \left( -\frac{27}{4} \sin^2 t - \frac{27}{4} \cos^2 t \right) dt = -\frac{27}{4} \int_0^{2\pi} dt\] \[= -\frac{27}{4} \cdot 2\pi = -\frac{27}{2} \pi\]

Result: \(-\frac{27}{2} \pi\)

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Stokes' Theorem

\[\iint_S \text{curl } \vec{F} \cdot d\vec{S} = \oint_C \vec{F} \cdot d\vec{r}\]

If different surfaces have the same boundary curve \(C\), then \(\iint_S \text{curl } \vec{F} \cdot d\vec{S}\) will be the same for all.

Figure: Diagram of a cone with a circular boundary curve C and a normal vector n.

Figure: Diagram of a hemisphere with the same circular boundary curve C and a normal vector n.

Figure: Diagram of a flat disk with the same circular boundary curve C and a normal vector n.

same boundary

all those surfaces have the same \(\iint_S \text{curl } \vec{F} \cdot d\vec{S}\)

Alternative Strategy

Another option: trade for a simpler surface w/ the same \(C\).

In previous example, we could have used a flat circle surface at \(z = 3/2\).