Divergence Theorem - Revision history

Lila at 21:48, 10 November 2021

2021-11-10T21:48:02Z

Lila at 21:47, 10 November 2021

2021-11-10T21:47:14Z

Lila: /* Example */

2021-11-10T21:46:42Z

Example

Lila: /* Example */

2021-11-10T21:45:45Z

Example

Lila: /* Example */

2021-11-10T21:45:02Z

Example

Lila at 21:43, 10 November 2021

2021-11-10T21:43:04Z

Lila at 21:41, 10 November 2021

2021-11-10T21:41:34Z

Lila at 21:40, 10 November 2021

2021-11-10T21:40:40Z

Lila: /* Corollaries */

2021-11-10T21:39:59Z

Corollaries

Lila: /* Corollaries */

2021-11-10T21:39:20Z

Corollaries

@@ Line 33: / Line 33: @@
 :<math>\Phi(V_\text{1}) + \Phi(V_\text{2}) =  \Phi_\text{1} + \Phi_\text{31} + \Phi_\text{2} + \Phi_\text{32}</math>
-where {{math|Φ<sub>1</sub>}} and {{math|Φ<sub>2</sub>}} are the flux out of surfaces {{math|''S''<sub>1</sub>}} and {{math|''S''<sub>2</sub>}},  {{math|Φ<sub>31</sub>}} is the flux through {{math|''S''<sub>3</sub>}} out of volume 1, and {{math|Φ<sub>32</sub>}} is the flux through {{math|''S''<sub>3</sub>}} out of volume 2.  The point is that surface {{math|''S''<sub>3</sub>}} is part of the surface of both volumes.  The "outward" direction of the [[normal vector]] <math>\mathbf{\hat n}</math> is opposite for each volume, so the flux out of one through {{math|''S''<sub>3</sub>}} is equal to the negative of the flux out of the other
+where {{math|Φ<sub>1</sub>}} and {{math|Φ<sub>2</sub>}} are the flux out of surfaces {{math|''S''<sub>1</sub>}} and {{math|''S''<sub>2</sub>}},  {{math|Φ<sub>31</sub>}} is the flux through {{math|''S''<sub>3</sub>}} out of volume 1, and {{math|Φ<sub>32</sub>}} is the flux through {{math|''S''<sub>3</sub>}} out of volume 2.  The point is that surface {{math|''S''<sub>3</sub>}} is part of the surface of both volumes.  The "outward" direction of the normal vector <math>\mathbf{\hat n}</math> is opposite for each volume, so the flux out of one through {{math|''S''<sub>3</sub>}} is equal to the negative of the flux out of the other
 :<math>\Phi_\text{31} = \iint_{S_3} \mathbf{F} \cdot \mathbf{\hat n} \; \mathrm{d}S = -\iint_{S_3} \mathbf{F} \cdot (-\mathbf{\hat n}) \; \mathrm{d}S = -\Phi_\text{32}</math>

@@ Line 138: / Line 138: @@
 : <math>\iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; \mathbf{F}\cdot\mathbf{n}\,\mathrm{d}S = 2\iiint_W\, dV = \frac{8\pi}{3},</math>
-because the unit ball {{mvar|W}} has volume <math>\frac{4''π''}{3}</math>.
+because the unit ball {{mvar|W}} has volume <math>\frac{4\pi}{3}</math>.
 ==Applications==

@@ Line 138: / Line 138: @@
 : <math>\iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; \mathbf{F}\cdot\mathbf{n}\,\mathrm{d}S = 2\iiint_W\, dV = \frac{8\pi}{3},</math>
-because the unit ball {{mvar|W}} has volume {{math|{{sfrac|4''π''|3}}}}.
+because the unit ball {{mvar|W}} has volume <math>\frac{4''π''}{3}</math>.
 ==Applications==

@@ Line 112: / Line 112: @@
 Suppose we wish to evaluate
-: <math>\scriptstyle S \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; \mathbf{F}\cdot\mathbf{n} \, \mathrm{d}S,</math>
+: <math>\iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; \mathbf{F}\cdot\mathbf{n} \, \mathrm{d}S,</math>
 where {{mvar|S}} is the unit sphere defined by
@@ Line 136: / Line 136: @@
 Therefore,
-: <math>\scriptstyle S \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; \mathbf{F}\cdot\mathbf{n}\,\mathrm{d}S = 2\iiint_W\, dV = \frac{8\pi}{3},</math>
+: <math>\iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; \mathbf{F}\cdot\mathbf{n}\,\mathrm{d}S = 2\iiint_W\, dV = \frac{8\pi}{3},</math>
 because the unit ball {{mvar|W}} has volume {{math|{{sfrac|4''π''|3}}}}.

@@ Line 112: / Line 112: @@
 Suppose we wish to evaluate
-:{{oiint
+: <math>\scriptstyle S \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; \mathbf{F}\cdot\mathbf{n} \, \mathrm{d}S,</math>
 where {{mvar|S}} is the unit sphere defined by
@@ Line 139: / Line 136: @@
 Therefore,
-:{{oiint
+: <math>\scriptstyle S \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; \mathbf{F}\cdot\mathbf{n}\,\mathrm{d}S = 2\iiint_W\, dV = \frac{8\pi}{3},</math>
 because the unit ball {{mvar|W}} has volume {{math|{{sfrac|4''π''|3}}}}.

@@ Line 21: / Line 21: @@
 Suppose {{mvar|V}} is a subset of <math>\mathbb{R}^n</math> (in the case of {{math|''n'' := 3, ''V''}} represents a volume in three-dimensional space) which is compact and has a piecewise smooth boundary {{mvar|S}} (also indicated with <math>\partial V = S</math>). If {{math|'''F'''}} is a continuously differentiable vector field defined on a neighborhood of {{mvar|V}}, then:
-<math>\iiint_V\left(\mathbf{\nabla}\cdot\mathbf{F}\right)\,\mathrm{d}V = \oiint\scriptstyle S (\mathbf{F}\cdot\mathbf{\hat{n}})\,\mathrm{d}S .</math>
+<math>\iiint_V\left(\mathbf{\nabla}\cdot\mathbf{F}\right)\,\mathrm{d}V = \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; (\mathbf{F}\cdot\mathbf{\hat{n}})\,\mathrm{d}S .</math>
 The left side is a volume integral over the volume {{mvar|V}}, the right side is the surface integral over the boundary of the volume {{mvar|V}}. The closed manifold <math>\partial V</math> is oriented by outward-pointing normals, and <math>\mathbf{\hat{n}}</math> is the outward pointing unit normal at each point on the boundary <math>\partial V</math>. (<math>\mathrm{d} \mathbf{S}</math> may be used as a shorthand for <math>\mathbf{n} \mathrm{d} S</math>.)  In terms of the intuitive description above, the left-hand side of the equation represents the total of the sources in the volume {{mvar|V}}, and the right-hand side represents the total flow across the boundary {{mvar|S}}.
@@ Line 72: / Line 72: @@
 * With <math>\mathbf{F}\rightarrow \mathbf{F}g</math> for a scalar function {{mvar|g}} and a vector field {{math|'''F'''}},
-::<math>\iiint_V\left[\mathbf{F}\cdot \left(\nabla g\right) + g \left(\nabla\cdot \mathbf{F}\right)\right] \mathrm{d}V = \oiint_S g\mathbf{F} \cdot \mathbf{n} \mathrm{d}S.</math>
+::<math>\iiint_V\left[\mathbf{F}\cdot \left(\nabla g\right) + g \left(\nabla\cdot \mathbf{F}\right)\right] \mathrm{d}V = \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; g\mathbf{F} \cdot \mathbf{n} \mathrm{d}S.</math>
@@ Line 79: / Line 79: @@
 * With <math>\mathbf{F}\rightarrow \mathbf{F}\times \mathbf{G}</math> for two vector fields {{math|'''F'''}} and {{math|'''G'''}}, where <math>\times</math> denotes a cross product,
-:: <math> \iiint_V \nabla \left( \mathbf{F} \times \mathbf{G}\right) \mathrm{d}V = \iiint_V \left[\mathbf{G}\cdot\left(\nabla\times\mathbf{F}\right) - \mathbf{F}\cdot \left( \nabla\times\mathbf{G}\right)\right]\, \mathrm{d}V = \oiint_S (\mathbf F\times\mathbf{G}) \cdot \mathbf{n} \mathrm{d}S.</math>
+:: <math> \iiint_V \nabla \left( \mathbf{F} \times \mathbf{G}\right) \mathrm{d}V = \iiint_V \left[\mathbf{G}\cdot\left(\nabla\times\mathbf{F}\right) - \mathbf{F}\cdot \left( \nabla\times\mathbf{G}\right)\right]\, \mathrm{d}V = \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; (\mathbf F\times\mathbf{G}) \cdot \mathbf{n} \mathrm{d}S.</math>
 * With <math>\mathbf{F}\rightarrow \mathbf{F}\cdot \mathbf{G}</math> for two vector fields {{math|'''F'''}} and {{math|'''G'''}}, where <math>\cdot </math> denotes a dot product,
-:: <math>\iiint_V \nabla \left( \mathbf{F} \cdot \mathbf{G}\right) \mathrm{d}V = \iiint_V \left[\mathbf{F}\cdot\left(\nabla \cdot \mathbf{G}\right) + \left( \nabla \cdot \mathbf{F}\right) \cdot \mathbf{G} \right]\, \mathrm{d}V = \oiint_S (\mathbf{F} \cdot \mathbf{G}) \cdot \mathbf{n} \mathrm{d}S.</math>
+:: <math>\iiint_V \nabla \left( \mathbf{F} \cdot \mathbf{G}\right) \mathrm{d}V = \iiint_V \left[\mathbf{F}\cdot\left(\nabla \cdot \mathbf{G}\right) + \left( \nabla \cdot \mathbf{F}\right) \cdot \mathbf{G} \right]\, \mathrm{d}V = \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; (\mathbf{F} \cdot \mathbf{G}) \cdot \mathbf{n} \mathrm{d}S.</math>
 * With <math>\mathbf{F}\rightarrow f\mathbf{c}</math> for a scalar function {{math|&thinsp;''f''&thinsp;}} and vector field '''c''':
-::\iiint_V \mathbf{c} \cdot \nabla f \, \mathrm{d}V = \oiint_S (\mathbf{c} f) \cdot \mathbf{n} \mathrm{d}S - \iiint_V f (\nabla \cdot \mathbf{c})\, \mathrm{d}V.</math>
+::\iiint_V \mathbf{c} \cdot \nabla f \, \mathrm{d}V = \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; (\mathbf{c} f) \cdot \mathbf{n} \mathrm{d}S - \iiint_V f (\nabla \cdot \mathbf{c})\, \mathrm{d}V.</math>
 :The last term on the right vanishes for constant <math>\mathbf{c}</math> or any divergence free (solenoidal) vector field,  e.g. Incompressible flows without sources or sinks such as phase change or chemical reactions etc. In particular, taking <math>\mathbf{c}</math> to be constant:
-:: <math>\iiint_V \nabla f \, \mathrm{d}V = \oiint_S f\mathbf{n} \mathrm{d}S.</math>
+:: <math>\iiint_V \nabla f \, \mathrm{d}V = \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; f\mathbf{n} \mathrm{d}S.</math>
 * With <math>\mathbf{F}\rightarrow \mathbf{c}\times\mathbf{F}</math> for vector field {{math|'''F'''}} and constant vector '''c''':
-:: <math>\iiint_V\mathbf{c} \cdot (\nabla\times\mathbf{F}) \, \mathrm{d}V = \oiint_S (\mathbf{F} \times \mathbf{c}) \cdot \mathbf{n} \mathrm{d}S.</math>
+:: <math>\iiint_V\mathbf{c} \cdot (\nabla\times\mathbf{F}) \, \mathrm{d}V = \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; (\mathbf{F} \times \mathbf{c}) \cdot \mathbf{n} \mathrm{d}S.</math>
 : By reordering the triple product on the right hand side and taking out the constant vector of the integral,
-:: <math> \iiint_V (\nabla \times\mathbf{F}) \, \mathrm{d}V \cdot \mathbf{c} = \oiint_S (\mathrm{d}\mathbf{S} \times \mathbf{F}) \cdot \mathbf{c}. </math>
+:: <math> \iiint_V (\nabla \times\mathbf{F}) \, \mathrm{d}V \cdot \mathbf{c} = \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; (\mathrm{d}\mathbf{S} \times \mathbf{F}) \cdot \mathbf{c}. </math>
 : Hence,
-:: <math>\iiint_V (\nabla \times\mathbf{F}) \, \mathrm{d}V = \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\;\mathbf{n} \times \mathbf{F} \mathrm{d}S. </math>
+:: <math>\iiint_V (\nabla \times\mathbf{F}) \, \mathrm{d}V = \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\; \mathbf{n} \times \mathbf{F} \mathrm{d}S. </math>
 == Example ==

@@ Line 104: / Line 104: @@
 : Hence,
-:: <math>\iiint_V (\nabla \times\mathbf{F}) \, \mathrm{d}V = \iint_S \!\!\!\!\!\!\!\!\! \large{\bigcirc} \;\;\;\mathbf{n} \times \mathbf{F} \mathrm{d}S. </math>
+:: <math>\iiint_V (\nabla \times\mathbf{F}) \, \mathrm{d}V = \iint_S \!\!\!\!\!\!\!\!\! \bigcirc \;\;\;\mathbf{n} \times \mathbf{F} \mathrm{d}S. </math>
 == Example ==