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17.8 The Divergence Theorem (part 2)

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

\( D \): space enclosed by \( S \)

assumed \( \vec{n} \) pointing outward

if \( \vec{n} \) is inward, then flip sign

\[ -\iiint_{D} \text{div} \vec{F} \, dV \]

this is useful if the enclosed space has another hollow space

A diagram showing a large blue cube with a smaller green cube nested inside it, representing a hollow volume.

\( D \): space/volume between cubes

remove a smaller cube from inside the big cube

\( S \): surface bounding \( D \) is the six outer faces and the six inner faces (that bound the small cube)

  • \( S_1 \): the outside surface (blue cube)
  • \( S_2 \): the inside surface (green)
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normal vector, as usual, is outward pointing

A diagram of a cube within a cube with red arrows showing normal vectors pointing out from the outer surface and into the inner cavity.

\( \rightarrow \) away from the enclosed volume

  • on outside, pointing away from volume \( \rightarrow \) out
  • on inside pointing away from volume \( \rightarrow \) in

between cubes

the flux integral:

\[ \iint_{S} \vec{F} \cdot d\vec{S} = \iint_{S} \vec{F} \cdot \vec{n} \, dS = \iint_{S_1} \vec{F} \cdot \vec{n} \, dS_1 - \iint_{S_2} \vec{F} \cdot \vec{n} \, dS_2 \]

due to inward normal on inside

the Divergence Theorem then gives

\[ \iiint_{D} \text{div} \vec{F} \, dV = \iint_{S} \vec{F} \cdot \vec{n} \, dS = \iint_{S_1} \vec{F} \cdot \vec{n} \, dS_1 - \iint_{S_2} \vec{F} \cdot \vec{n} \, dS_2 \]

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Divergence Theorem for Nested Regions

Let us define the regions for a nested cube configuration:

  • \(D_1\): space inside big cube
  • \(D_2\): space inside small cube

Apply Divergence Theorem again:

\[\iint_{S_1} \vec{F} \cdot \vec{n} dS_1 = \iiint_{D_1} \text{div } \vec{F} dV\]
\[\iint_{S_2} \vec{F} \cdot \vec{n} dS_2 = \iiint_{D_2} \text{div } \vec{F} dV\]
A diagram showing a small green cube nested inside a larger blue cube, illustrating regions D1 and D2.

Plug these into the equation on previous page, we get:

\[\iint_{S} \vec{F} \cdot d\vec{S} = \iiint_{D_1} \text{div } \vec{F} dV - \iiint_{D_2} \text{div } \vec{F} dV\]

Flux of hollow volume = volume integral of \(\text{div } \vec{F}\) of outer volume \(-\) volume integral of \(\text{div } \vec{F}\) of inner volume

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Example: Flux through a Spherical Shell

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

\(D\): between spheres radii 2 and 1, both centered at origin. Normal is positive outward.

\(D\): space between them

Choices:

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

\(\hookrightarrow\) set up bounds for the space between spheres: \(1 \le \rho \le 2\), etc.

or

\[\iint_{S} \vec{F} \cdot d\vec{S} = \iiint_{D_1} \text{div } \vec{F} dV - \iiint_{D_2} \text{div } \vec{F} dV\]

\(D_1\): big sphere

\(D_2\): small sphere

A 3D coordinate graph showing two concentric spheres centered at the origin with radii 1 and 2.

Let's try the first way first:

\[1 \le \rho \le 2\]

\[0 \le \phi \le \pi\]

\[0 \le \theta \le 2\pi\]

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\[ \text{div } \vec{F} = \vec{\nabla} \cdot \langle x, y, z \rangle = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3 \]
\[ \int_{0}^{2\pi} \int_{0}^{\pi} \int_{1}^{2} 3 \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]

Note: \( \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \) represents \( dV \).

\[ = 3 \int_{0}^{2\pi} \int_{0}^{\pi} \left. \frac{1}{3} \rho^3 \right|_{1}^{2} \sin \phi \, d\phi \, d\theta = \int_{0}^{2\pi} \int_{0}^{\pi} 7 \sin \phi \, d\phi \, d\theta \]
\[ = 14\pi \int_{0}^{\pi} \sin \phi \, d\phi = 14\pi \left. (-\cos \phi) \right|_{0}^{\pi} = \text{notebox: } 28\pi \]

Alternative Method

\[ \iiint_{D_1} \text{div } \vec{F} \, dV - \iiint_{D_2} \text{div } \vec{F} \, dV \quad \text{where } \text{div } \vec{F} = 3 \]
\[ = 3 \iiint_{D_1} dV - 3 \iiint_{D_2} dV = 3 \left( \frac{4}{3}\pi(2)^3 - \frac{4}{3}\pi(1)^3 \right) \]
  • \( \iiint_{D_1} dV \): volume of big sphere \( \rho = 2 \)
  • \( \iiint_{D_2} dV \): volume of small sphere \( \rho = 1 \)
Result: \( 28\pi \)
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Example

\( \vec{F} = \langle x^2, -y^2, z^2 \rangle \)

\( D \): space bounded by \( z = 6 - x - y \) and \( z = 3 - x - y \) in first octant

Bounds for \( D \)

\( 3 - x - y \le z \le 6 - x - y \)

\( 0 \le z \le 6 - x - y \) (each with their own \( x, y \) bounds)

Two integrals needed here, with bounds not simple to set up.

3D sketch of two triangular planes in the first octant, bounded by axes x, y, and z at intercepts 3 and 6.
2D graph in the xy-plane showing two linear boundaries with intercepts at (3,0), (0,3) and (6,0), (0,6).

Alternative Method

\[ \iiint_{D_1} \text{div } \vec{F} \, dV - \iiint_{D_2} \text{div } \vec{F} \, dV \]
  • \( D_1 \): volume enclosed by higher plane alone
  • \( D_2 \): volume enclosed by lower plane alone
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the second way seems a little easier

Region \(D_1\):

  • \(0 \le x \le 6\)
  • \(0 \le y \le 6 - x\)
  • \(0 \le z \le 6 - x - y\)
A 2D coordinate graph showing a line y = 6 - x intersecting the x and y axes at 6.

Region \(D_2\):

  • \(0 \le x \le 3\)
  • \(0 \le y \le 3 - x\)
  • \(0 \le z \le 3 - x - y\)

\(\text{div } \vec{F} = 2x - 2y + 2z\)

minus because \(D_2\) is inner volume

\[ \underbrace{\int_{0}^{6} \int_{0}^{6-x} \int_{0}^{6-x-y} (2x - 2y + 2z) \, dz \, dy \, dx}_{D_1} - \underbrace{\int_{0}^{3} \int_{0}^{3-x} \int_{0}^{3-x-y} (2x - 2y + 2z) \, dz \, dy \, dx}_{D_2} \]

\(= \dots =\)

\(\frac{405}{4}\)
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\[ \iiint_{D} \text{div } \vec{F} \, dV \text{ requires } \vec{F} \text{ to be defined throughout } D \]

if not, then Div. Theorem cannot be used directly

Example

\[ \vec{F} = \frac{\langle x, y, z \rangle}{\sqrt{x^2 + y^2 + z^2}} \]

\(S\): sphere radius 1, centered at origin

at origin, \(\vec{F}\) does not exist, so not \(\text{div } \vec{F}\), so can't use Div. Theorem directly

A 3D coordinate system showing a large sphere D and a smaller inner sphere with radius epsilon.

Work around:

make another smaller sphere enclosing origin, then take the limit as its radius \(\to 0\)

radius = \(\epsilon\), then \(\lim_{\epsilon \to 0}\)

in \(D\), \(\vec{F}\) exists everywhere

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\[ \text{div } \vec{F} = \frac{2}{\sqrt{x^2+y^2+z^2}} = \frac{2}{\rho} \text{ in spherical} \]

Region D:

\( \epsilon \leq \rho \leq 1 \)

\( 0 \leq \phi \leq \pi \)

\( 0 \leq \theta \leq 2\pi \)

\[ \iiint_D \text{div } \vec{F} \, dV = \int_0^{2\pi} \int_0^\pi \int_\epsilon^1 \frac{2}{\rho} \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]

Note: In the integral above, \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \).

\[ = \dots = 4\pi (1 - \epsilon^2) \]

now \( \lim_{\epsilon \to 0} 4\pi (1 - \epsilon^2) = \) \(4\pi\)