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17.8 The Divergence Theorem (part 1)

divergence: \( \text{div } \vec{F} = \vec{\nabla} \cdot \vec{F} \)

\[ \text{div } \vec{F} = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right\rangle \cdot \langle f, g, h \rangle = f_x + g_y + h_z \]

\( \iiint_E \text{div } \vec{F} \, dV \) accumulates the divergence inside an enclosed volume

Divergence Theorem

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

E: enclosed volume
S: the surface bounding the volume (closed surface)

for example, for a cube

Figure: A 3D cube with normal vectors S1 through S6 pointing outward from each face, labeled E for the volume inside.

S: the six sides of the cube
E: volume inside

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Why is the Theorem true?

divergence can be interpreted as a volume change of a small box in \( \vec{F} \)

Figure: A small red cube at the origin of a 3D coordinate system with green vectors representing a vector field.

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

Figure: An expanded red cube with green vectors pointing outward from its faces, illustrating positive divergence.

if box is rigid,

box expands if \( \text{div } \vec{F} > 0 \)

box shrinks if \( \text{div } \vec{F} < 0 \)

if box doesn't change volume but surface is porous, then

if \( \text{div } \vec{F} > 0 \rightarrow \) things flowing out
if \( \text{div } \vec{F} < 0 \rightarrow \) things flowing in

Figure: A 3D cube with green arrows passing through its faces, representing flow through the surface.

flow through sides relate directly to divergence

flux integral

\( \iint_S \vec{F} \cdot d\vec{S} \)

accumulated by

\( \iiint_E \text{div } \vec{F} \, dV \)

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the divergence theorem is usually only used if the surface is closed (no openings)

Example: \(\vec{F} = \langle x, y, z \rangle\)

\(S\): upper half of \(x^2 + y^2 + z^2 = 9\) and the circle enclosing the volume at \(z = 0\)

as usual, \(\vec{n}\) is positive pointing outward

Let's verify the theorem:

\[\iiint_E \text{div} \vec{F} \, dV = \iint_S \vec{F} \cdot d\vec{S}\]

\(S_1\): sphere part

\(S_2\): the circular base

outward normal means \(\vec{n}\) upward on the sphere part and downward on the circle

Figure: A 3D coordinate system showing a hemisphere S1 on the xy-plane with outward normal vectors n.

Surface integral first

\[\iint_S \vec{F} \cdot d\vec{S} = \iint_{S_1} \vec{F} \cdot d\vec{S}_1 + \iint_{S_2} \vec{F} \cdot d\vec{S}_2\]

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parametrize \(S_1\): Spherical

\(\rho = 3\), so let \(u = \phi\), \(v = \theta\)

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

\(x = \rho \sin \phi \cos \theta\)

\(0 \le u \le \pi/2\)
\(0 \le v \le 2\pi\)

\[\vec{r}_u = \langle 3 \cos u \cos v, \, 3 \cos u \sin v, \, -3 \sin u \rangle\]\[\vec{r}_v = \langle -3 \sin u \sin v, \, 3 \sin u \cos v, \, 0 \rangle\]\[\vec{r}_u \times \vec{r}_v = \langle 9 \sin^2 u \cos v, \, 9 \sin^2 u \sin v, \, 9 \cos u \sin u \rangle\]

direction correct? yes

parametrize \(S_2\): cylindrical / polar

\(u = r\), \(v = \theta\)

\[\vec{r}(u,v) = \langle u \cos v, \, u \sin v, \, 0 \rangle\]

\(0 \le u \le 3\)
\(0 \le v \le 2\pi\)

\[\vec{r}_u = \langle \cos v, \, \sin v, \, 0 \rangle\]\[\vec{r}_v = \langle -u \sin v, \, u \cos v, \, 0 \rangle\]\[\vec{r}_u \times \vec{r}_v = \langle 0, \, 0, \, u \rangle\]

direction correct? No, "out" on \(S_2\) is "down"

use \(\vec{r}_v \times \vec{r}_u = \langle 0, \, 0, \, -u \rangle\)

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now

\[ \iint_{S} \vec{F} \cdot (\vec{r}_u \times \vec{r}_v) \, dA \]

for each or \( \vec{r}_v \times \vec{r}_u \)

\[ \int_{0}^{2\pi} \int_{0}^{\pi/2} \langle 3\sin u \cos v, 3\sin u \sin v, 3\cos u \rangle \cdot \langle 9\sin^2 u \cos v, 9\sin^2 u \sin v, 9\cos u \sin v \rangle \, du \, dv \]

\( \vec{F} = \langle x, y, z \rangle \) using \( \vec{r}(u, v) \)

\[ + \int_{0}^{2\pi} \int_{0}^{3} \langle u \cos v, u \sin v, 0 \rangle \cdot \langle 0, 0, -u \rangle \, du \, dv \]

\[ = \dots = \boxed{54\pi} \]

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Divergence

\[ \iiint_{E} \text{div} \vec{F} \, dV = \iint_{S} \vec{F} \cdot d\vec{S} \]

\[ \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \rangle \cdot \langle x, y, z \rangle \]

now the equivalent triple integral: \( \text{div} \vec{F} = \vec{\nabla} \cdot \langle x, y, z \rangle = 3 \)

\[ \iiint_{E} 3 \, dV = 3 \iiint_{E} dV \]

Figure: 3D coordinate system showing a hemisphere of radius 3 centered at the origin in the upper half-space.

volume of enclosed space: half of a sphere radius 3

\[ = 3 \cdot \frac{1}{2} \cdot \frac{4}{3} \pi (3)^3 = 2\pi \cdot 27 = \boxed{54\pi} \]

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Example: Divergence Theorem

Given the vector field:

\[ \vec{F} = \langle -y^3 z^2, x^4, 4xy^2 \rangle \]

Let \( S \) be the surface of the space enclosed by:

\( x^2 + y^2 = 1 \) (cylinder)
\( z = 10 - y \) (plane)
\( z = 0 \) (xy-plane)

The normal is positive outward.

Figure: 3D sketch of a cylinder on the xy-plane intersected by a slanted plane z = 10 - y.

Figure: Diagram of the volume showing outward normal vectors n1, n2, and n3 for the three bounding surfaces.

Direct calculation would require:

3 surfaces
3 parametrizations
3 normals (3 cross products)

Since it is an enclosed space, we can use the Divergence Theorem to convert the surface integral into one triple integral:

\[ \iiint_E \text{div} \vec{F} \, dV \]

First, calculate the divergence of \( \vec{F} \):

\[ \begin{aligned} \text{div} \vec{F} &= \nabla \cdot \langle -y^3 z^2, x^4, 4xy^2 \rangle \\ &= \frac{\partial}{\partial x}(-y^3 z^2) + \frac{\partial}{\partial y}(x^4) + \frac{\partial}{\partial z}(4xy^2) \\ &= 0 + 0 + 0 \\ &= 0 \end{aligned} \]

Therefore, the integral is:

\[ \iiint_E \text{div} \vec{F} \, dV = \boxed{0} \]