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17.7 Stokes' Theorem (part 2)

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

Let's take another look at the accumulation of curl.

curl: \( \vec{F} = \langle x, y, 0 \rangle \)

We see no rotation \( \rightarrow \) no curl.

If we place a paddle wheel w/ axis along z-axis here, no rotation is expected.

Vector field \vec{F} = \langle x, y, 0 \rangle showing vectors radiating outward from the origin with no rotation.

look at \( \vec{F} = \langle -y, x, 0 \rangle \)

This time, the same paddle wheel will spin (counterclockwise when viewed from above).

Vector field \vec{F} = \langle -y, x, 0 \rangle showing vectors circulating counterclockwise around the z-axis.
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If the axis of the paddle wheel is tilted:

Rotation is still expected but not as fast.

A paddle wheel tilted relative to the z-axis in a circulating vector field.

Now no rotation is expected.

A paddle wheel oriented such that its axis is perpendicular to the circulation of the field.

In \( \vec{F} = \langle x, y, 0 \rangle \)

\[ \text{curl } \vec{F} = \nabla \times \vec{F} = \langle 0, 0, 0 \rangle = \vec{0} \]

axis along any coordinate axis results in no rotation

In \( \vec{F} = \langle -y, x, 0 \rangle \)

\[ \text{curl } \vec{F} = \nabla \times \vec{F} = \langle 0, 0, 2 \rangle \]

This tells us that max. rotation is achieved if paddle wheel axis is along z-axis.

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\[ \vec{F} = \langle 5 - z^2, 0, 0 \rangle \]
\[ \text{curl } \vec{F} = \langle 0, -2z, 0 \rangle \]
  • max rotation achieved wheel axis is along y-axis
  • as long as part of the wheel's axis is along y-axis \(\rightarrow\) some spin
  • the component depends on \(z \rightarrow\) speed of rotation depends on \(z\)
  • high \(z \rightarrow\) fast spin

How is this related to Stokes' Theorem?

\[ \iint_S \text{curl } \vec{F} \cdot d\vec{S} = \iint_S \text{curl } \vec{F} \cdot \vec{n} \, dS \]

\(\vec{n}\) is the unit normal of surface \(S\)

\(\text{curl } \vec{F} \cdot \vec{n} \rightarrow\) projection of \(\text{curl } \vec{F}\) onto \(\vec{n}\)

So, \(\text{curl } \vec{F} \cdot \vec{n}\) is seeing how much alignment is between the paddle wheel's spinning axis and the unit normal.

Vector diagram showing curl F, unit normal n, and the projection curl F dot n.
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Example

\(S\): upper half of \(z^2 = a^2(1 - x^2 - y^2)\)

\(\vec{n}\) is positive upward

\(\vec{F} = \langle x - y, y + z, z - x \rangle \)

\(a\): some constant

Find \(a\) such that \(\iint_S \text{curl } \vec{F} \cdot d\vec{S}\) is maximized.

Shape of \(z^2 = a^2(1 - x^2 - y^2)\)?

Ellipsoid!

\[ z^2 = a^2(1 - x^2 - y^2) \]\[ \frac{z^2}{a^2} = 1 - x^2 - y^2 \]\[ x^2 + y^2 + \frac{z^2}{a^2} = 1 \]
3D coordinate graph showing an ellipsoid with different heights for high a and low a values.

\(a\) changes how \(\vec{n}\) is oriented

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Calculating Surface Integrals of Curl

Let's calculate:

\[ \iint_S \text{curl } \vec{F} \cdot d\vec{S} = \iint_S \text{curl } \vec{F} \cdot (\vec{r}_u \times \vec{r}_v) \, dA \]

or \( \vec{r}_v \times \vec{r}_u \), whichever points in the positive direction of \( \vec{n} \).

Parametrize Surface S

Given the surface equation:

\[ z^2 = a^2(1 - x^2 - y^2) \]

For the upper half (where \( a > 0 \)):

\[ z = a\sqrt{1 - x^2 - y^2} \]

Good coordinate system? Cylindrical

Using cylindrical coordinates:

  • \( x = r \cos \theta \)
  • \( y = r \sin \theta \)
  • \( z = z \)

Parameters: \( r, \theta, z \) (choose two). Let \( u = r \), \( v = \theta \).

A 2D coordinate graph showing a unit circle centered at the origin with intercepts at 1 on the x and y axes.
\[ x = u \cos v \]\[ y = u \sin v \]\[ z = a\sqrt{1 - u^2} \]

Note: \( z = a\sqrt{1 - (x^2 + y^2)} = a\sqrt{1 - r^2} = a\sqrt{1 - u^2} \)

Bounds of \( u, v \)?

  • \( 0 \le u \le 1 \)
  • \( 0 \le v \le 2\pi \)
\[ \vec{r}(u, v) = \langle u \cos v, u \sin v, a\sqrt{1 - u^2} \rangle \]
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Partial Derivatives and Cross Product

\[ \vec{r}_u = \left\langle \cos v, \sin v, \frac{-au}{\sqrt{1-u^2}} \right\rangle \]\[ \vec{r}_v = \langle -u \sin v, u \cos v, 0 \rangle \]\[ \vec{r}_u \times \vec{r}_v = \left\langle \frac{au^2 \cos v}{\sqrt{1-u^2}}, \frac{au^2 \sin v}{\sqrt{1-u^2}}, u \right\rangle \]

Positive? Yes.

Positive upward \( \rightarrow \) positive \( z \) component.

\( z: u \), where \( 0 \le u \le 1 \).

Vector Field and Curl

\[ \vec{F} = \langle x - y, y + z, z - x \rangle \]\[ \text{curl } \vec{F} = \langle 1, 1, 1 \rangle \]

Evaluating the Integral

\[ \iint_S \text{curl } \vec{F} \cdot d\vec{S} = \iint_S \text{curl } \vec{F} \cdot (\vec{r}_u \times \vec{r}_v) \, dA \]\[ = \int_0^{2\pi} \int_0^1 \langle 1, 1, 1 \rangle \cdot \left\langle \frac{au^2 \cos v}{\sqrt{1-u^2}}, \frac{au^2 \sin v}{\sqrt{1-u^2}}, u \right\rangle \, du \, dv \]
\[ \dots = \pi \]

Notice this does not depend on \( a \).

\( \iint_S \text{curl } \vec{F} \cdot d\vec{S} \) does not depend on \( a \).

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Stokes' Theorem Application

Stokes': \[ \iint_{S} \text{curl } \vec{F} \cdot d\vec{S} = \int_{C} \vec{F} \cdot d\vec{r} \]

\( C \) is the boundary curve at the open end of surface

3D plot with x, y, z axes showing a surface capped by a circle C of radius 1 in the xy-plane.
  • \( C \): always circle radius 1, regardless of \( a \)
  • Since \( C \) is independent of \( a \), by Stokes' Theorem we expect no dependency of \( \iint_{S} \text{curl } \vec{F} \cdot d\vec{S} \) on \( a \)

Let's work out \( \int_{C} \vec{F} \cdot d\vec{r} \)

\( C \): circle radius 1 at \( z = 0 \)

\( \vec{r}(t) = \langle \cos(t), \sin(t), 0 \rangle \) (xy-plane)

\( \vec{F} = \langle x-y, y+z, z-x \rangle = \langle \cos t - \sin t, \sin t, -\cos t \rangle \)

\( d\vec{r} = \vec{r}' dt = \langle -\sin t, \cos t, 0 \rangle dt \)

\[ \int_{C} \vec{F} \cdot d\vec{r} = \int_{0}^{2\pi} \langle \cos t - \sin t, \sin t, -\cos t \rangle \cdot \langle -\sin t, \cos t, 0 \rangle dt \]

\[ = \int_{0}^{2\pi} -\cos t \sin t + \sin^2 t + \sin t \cos t \, dt \]

\[ = \int_{0}^{2\pi} \sin^2 t \, dt = \pi \]