Misplaced Pages

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79-400: More precisely, the divergence theorem states that the surface integral of a vector field over a closed surface, which is called the "flux" through the surface, is equal to the volume integral of the divergence over the region enclosed by the surface. Intuitively, it states that "the sum of all sources of the field in a region (with sinks regarded as negative sources) gives the net flux out of

158-395: A neighborhood of V , then: The left side is a volume integral over the volume V , and the right side is the surface integral over the boundary of the volume V . The closed, measurable set ∂ V {\displaystyle \partial V} is oriented by outward-pointing normals , and n ^ {\displaystyle \mathbf {\hat {n}} }

237-400: A neighbourhood (or neighborhood ) is one of the basic concepts in a topological space . It is closely related to the concepts of open set and interior . Intuitively speaking, a neighbourhood of a point is a set of points containing that point where one can move some amount in any direction away from that point without leaving the set. If X {\displaystyle X} is

316-404: A neighbourhood of S {\displaystyle S} is a set V {\displaystyle V} that includes an open set U {\displaystyle U} containing S {\displaystyle S} , S ⊆ U ⊆ V ⊆ X . {\displaystyle S\subseteq U\subseteq V\subseteq X.} It follows that

395-402: A surface integral is a generalization of multiple integrals to integration over surfaces . It can be thought of as the double integral analogue of the line integral . Given a surface, one may integrate over this surface a scalar field (that is, a function of position which returns a scalar as a value), or a vector field (that is, a function which returns a vector as value). If

474-661: A topological space and p {\displaystyle p} is a point in X , {\displaystyle X,} then a neighbourhood of p {\displaystyle p} is a subset V {\displaystyle V} of X {\displaystyle X} that includes an open set U {\displaystyle U} containing p {\displaystyle p} , p ∈ U ⊆ V ⊆ X . {\displaystyle p\in U\subseteq V\subseteq X.} This

553-466: A differential form (in terms of a divergence) and an integral form (in terms of a flux). Any inverse-square law can instead be written in a Gauss's law -type form (with a differential and integral form, as described above). Two examples are Gauss's law (in electrostatics), which follows from the inverse-square Coulomb's law , and Gauss's law for gravity , which follows from the inverse-square Newton's law of universal gravitation . The derivation of

632-462: A host of physical laws can be written in both a differential form (where one quantity is the divergence of another) and an integral form (where the flux of one quantity through a closed surface is equal to another quantity). Three examples are Gauss's law (in electrostatics ), Gauss's law for magnetism , and Gauss's law for gravity . Continuity equations offer more examples of laws with both differential and integral forms, related to each other by

711-441: A point p {\displaystyle p} (sometimes called a punctured neighbourhood ) is a neighbourhood of p , {\displaystyle p,} without { p } . {\displaystyle \{p\}.} For instance, the interval ( − 1 , 1 ) = { y : − 1 < y < 1 } {\displaystyle (-1,1)=\{y:-1<y<1\}}

790-446: A region R is not flat, then it is called a surface as shown in the illustration. Surface integrals have applications in physics , particularly with the theories of classical electromagnetism . Assume that f is a scalar, vector, or tensor field defined on a surface S . To find an explicit formula for the surface integral of f over S , we need to parameterize S by defining a system of curvilinear coordinates on S , like

869-584: A set V {\displaystyle V} is a neighbourhood of a point p {\displaystyle p} if there exists an open ball with center p {\displaystyle p} and radius r > 0 , {\displaystyle r>0,} such that B r ( p ) = B ( p ; r ) = { x ∈ X : d ( x , p ) < r } {\displaystyle B_{r}(p)=B(p;r)=\{x\in X:d(x,p)<r\}}

SECTION 10

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948-411: A set V {\displaystyle V} is a neighbourhood of S {\displaystyle S} if and only if it is a neighbourhood of all the points in S . {\displaystyle S.} Furthermore, V {\displaystyle V} is a neighbourhood of S {\displaystyle S} if and only if S {\displaystyle S}

1027-1005: A smaller value near the poles of a sphere, where the lines of longitude converge more dramatically, and latitudinal coordinates are more compactly spaced). The surface integral can also be expressed in the equivalent form where g is the determinant of the first fundamental form of the surface mapping r ( s , t ) . For example, if we want to find the surface area of the graph of some scalar function, say z = f ( x , y ) , we have where r = ( x , y , z ) = ( x , y , f ( x , y )) . So that ∂ r ∂ x = ( 1 , 0 , f x ( x , y ) ) {\displaystyle {\partial \mathbf {r} \over \partial x}=(1,0,f_{x}(x,y))} , and ∂ r ∂ y = ( 0 , 1 , f y ( x , y ) ) {\displaystyle {\partial \mathbf {r} \over \partial y}=(0,1,f_{y}(x,y))} . So, which

1106-559: A surface is split into pieces, on each piece a parametrization and corresponding surface normal is chosen, and the pieces are put back together, we will find that the normal vectors coming from different pieces cannot be reconciled. This means that at some junction between two pieces we will have normal vectors pointing in opposite directions. Such a surface is called non-orientable , and on this kind of surface, one cannot talk about integrating vector fields. Neighbourhood (mathematics) In topology and related areas of mathematics ,

1185-545: A tangential and a normal component, then only the normal component contributes to the flux. Based on this reasoning, to find the flux, we need to take the dot product of v with the unit surface normal n to S at each point, which will give us a scalar field, and integrate the obtained field as above. In other words, we have to integrate v with respect to the vector surface element d s = n d s {\displaystyle \mathrm {d} \mathbf {s} ={\mathbf {n} }\mathrm {d} s} , which

1264-748: A variant of the straightening theorem for vector fields , we may choose O {\displaystyle O} so that ∂ ∂ x n {\displaystyle {\frac {\partial }{\partial x_{n}}}} is the inward unit normal − N {\displaystyle -N} at ∂ Ω {\displaystyle \partial \Omega } . In this case g ( x ′ , 0 ) d x ′ = g ∂ Ω ( x ′ ) d x ′ = d S {\displaystyle {\sqrt {g(x',0)}}\,dx'={\sqrt {g_{\partial \Omega }(x')}}\,dx'=dS}

1343-872: A vector equation of C {\displaystyle C} is At a point P {\displaystyle P} on C {\displaystyle C} : Therefore, Because M = R e ( F ) = 2 y {\displaystyle M={\mathfrak {Re}}(\mathbf {F} )=2y} , we can evaluate ∂ M ∂ x = 0 {\displaystyle {\frac {\partial M}{\partial x}}=0} , and because N = I m ( F ) = 5 x {\displaystyle N={\mathfrak {Im}}(\mathbf {F} )=5x} , ∂ N ∂ y = 0 {\displaystyle {\frac {\partial N}{\partial y}}=0} . Thus Surface integral In mathematics , particularly multivariable calculus ,

1422-979: Is C 1 {\displaystyle C^{1}} on an open neighborhood O {\displaystyle O} of Ω ¯ {\displaystyle {\overline {\Omega }}} , that is, u ∈ C 1 ( O ) {\displaystyle u\in C^{1}(O)} , then for each i ∈ { 1 , … , n } {\displaystyle i\in \{1,\dots ,n\}} , ∫ Ω u x i d V = ∫ ∂ Ω u ν i d S , {\displaystyle \int _{\Omega }u_{x_{i}}\,dV=\int _{\partial \Omega }u\nu _{i}\,dS,} where ν : ∂ Ω → R n {\displaystyle \nu :\partial \Omega \to \mathbb {R} ^{n}}

1501-594: Is a C 1 {\displaystyle C^{1}} vector field on Ω ¯ {\displaystyle {\overline {\Omega }}} . Then ( grad ⁡ u , X ) = − ( u , div ⁡ X ) + ∫ ∂ Ω u ⟨ X , N ⟩ d S , {\displaystyle (\operatorname {grad} u,X)=-(u,\operatorname {div} X)+\int _{\partial \Omega }u\langle X,N\rangle \,dS,} where N {\displaystyle N}

1580-493: Is a neighbourhood for the set N {\displaystyle \mathbb {N} } of natural numbers , but is not a uniform neighbourhood of this set. The above definition is useful if the notion of open set is already defined. There is an alternative way to define a topology, by first defining the neighbourhood system , and then open sets as those sets containing a neighbourhood of each of their points. A neighbourhood system on X {\displaystyle X}

1659-440: Is a neighbourhood of p = 0 {\displaystyle p=0} in the real line , so the set ( − 1 , 0 ) ∪ ( 0 , 1 ) = ( − 1 , 1 ) ∖ { 0 } {\displaystyle (-1,0)\cup (0,1)=(-1,1)\setminus \{0\}} is a deleted neighbourhood of 0. {\displaystyle 0.} A deleted neighbourhood of

SECTION 20

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1738-494: Is a subset of the interior of V . {\displaystyle V.} A neighbourhood of S {\displaystyle S} that is also an open subset of X {\displaystyle X} is called an open neighbourhood of S . {\displaystyle S.} The neighbourhood of a point is just a special case of this definition. In a metric space M = ( X , d ) , {\displaystyle M=(X,d),}

1817-685: Is a uniform neighbourhood, and that a set is a uniform neighbourhood if and only if it contains an r {\displaystyle r} -neighbourhood for some value of r . {\displaystyle r.} Given the set of real numbers R {\displaystyle \mathbb {R} } with the usual Euclidean metric and a subset V {\displaystyle V} defined as V := ⋃ n ∈ N B ( n ; 1 / n ) , {\displaystyle V:=\bigcup _{n\in \mathbb {N} }B\left(n\,;\,1/n\right),} then V {\displaystyle V}

1896-595: Is contained in V . {\displaystyle V.} V {\displaystyle V} is called a uniform neighbourhood of a set S {\displaystyle S} if there exists a positive number r {\displaystyle r} such that for all elements p {\displaystyle p} of S , {\displaystyle S,} B r ( p ) = { x ∈ X : d ( x , p ) < r } {\displaystyle B_{r}(p)=\{x\in X:d(x,p)<r\}}

1975-628: Is contained in V . {\displaystyle V.} Under the same condition, for r > 0 , {\displaystyle r>0,} the r {\displaystyle r} -neighbourhood S r {\displaystyle S_{r}} of a set S {\displaystyle S} is the set of all points in X {\displaystyle X} that are at distance less than r {\displaystyle r} from S {\displaystyle S} (or equivalently, S r {\displaystyle S_{r}}

2054-424: Is coordinate free, it shows that the divergence does not depend on the coordinates used. We are going to prove the following: Theorem  —  Let Ω ⊂ R n {\displaystyle \Omega \subset \mathbb {R} ^{n}} be open and bounded with C 1 {\displaystyle C^{1}} boundary. If u {\displaystyle u}

2133-414: Is equal to the sum of the flux through its two faces, so the sum of the flux out of the two parts is where Φ 1 and Φ 2 are the flux out of surfaces S 1 and S 2 , Φ 31 is the flux through S 3 out of volume 1, and Φ 32 is the flux through S 3 out of volume 2. The point is that surface S 3 is part of the surface of both volumes. The "outward" direction of

2212-411: Is equal to the volume rate of fluid crossing this surface, i.e., the surface integral of the velocity over the surface. Since liquids are incompressible, the amount of liquid inside a closed volume is constant; if there are no sources or sinks inside the volume then the flux of liquid out of S is zero. If the liquid is moving, it may flow into the volume at some points on the surface S and out of

2291-478: Is equivalent to the point p ∈ X {\displaystyle p\in X} belonging to the topological interior of V {\displaystyle V} in X . {\displaystyle X.} The neighbourhood V {\displaystyle V} need not be an open subset of X . {\displaystyle X.} When V {\displaystyle V}

2370-421: Is given by the first fundamental form of the surface. Consider a vector field v on a surface S , that is, for each r = ( x , y , z ) in S , v ( r ) is a vector. The integral of v on S was defined in the previous section. Suppose now that it is desired to integrate only the normal component of the vector field over the surface, the result being a scalar, usually called the flux passing through

2449-1702: Is identified with an open set in R + n = { x ∈ R n : x n ≥ 0 } {\displaystyle \mathbb {R} _{+}^{n}=\{x\in \mathbb {R} ^{n}:x_{n}\geq 0\}} . We zero extend u {\displaystyle u} and X {\displaystyle X} to R + n {\displaystyle \mathbb {R} _{+}^{n}} and perform integration by parts to obtain ( grad ⁡ u , X ) = ∫ O ⟨ grad ⁡ u , X ⟩ g d x = ∫ R + n ∂ j u X j g d x = ( u , − div ⁡ X ) − ∫ R n − 1 u ( x ′ , 0 ) X n ( x ′ , 0 ) g ( x ′ , 0 ) d x ′ , {\displaystyle {\begin{aligned}(\operatorname {grad} u,X)&=\int _{O}\langle \operatorname {grad} u,X\rangle {\sqrt {g}}\,dx\\&=\int _{\mathbb {R} _{+}^{n}}\partial _{j}uX^{j}{\sqrt {g}}\,dx\\&=(u,-\operatorname {div} X)-\int _{\mathbb {R} ^{n-1}}u(x',0)X^{n}(x',0){\sqrt {g(x',0)}}\,dx',\end{aligned}}} where d x ′ = d x 1 … d x n − 1 {\displaystyle dx'=dx_{1}\dots dx_{n-1}} . By

Divergence theorem - Misplaced Pages Continue

2528-472: Is open (resp. closed, compact, etc.) in X , {\displaystyle X,} it is called an open neighbourhood (resp. closed neighbourhood, compact neighbourhood, etc.). Some authors require neighbourhoods to be open, so it is important to note their conventions. A set that is a neighbourhood of each of its points is open since it can be expressed as the union of open sets containing each of its points. A closed rectangle, as illustrated in

2607-413: Is that sometimes surfaces do not have parametrizations which cover the whole surface. The obvious solution is then to split that surface into several pieces, calculate the surface integral on each piece, and then add them all up. This is indeed how things work, but when integrating vector fields, one needs to again be careful how to choose the normal-pointing vector for each piece of the surface, so that when

2686-412: Is the assignment of a filter N ( x ) {\displaystyle N(x)} of subsets of X {\displaystyle X} to each x {\displaystyle x} in X , {\displaystyle X,} such that One can show that both definitions are compatible, that is, the topology obtained from the neighbourhood system defined using open sets

2765-919: Is the original one, and vice versa when starting out from a neighbourhood system. In a uniform space S = ( X , Φ ) , {\displaystyle S=(X,\Phi ),} V {\displaystyle V} is called a uniform neighbourhood of P {\displaystyle P} if there exists an entourage U ∈ Φ {\displaystyle U\in \Phi } such that V {\displaystyle V} contains all points of X {\displaystyle X} that are U {\displaystyle U} -close to some point of P ; {\displaystyle P;} that is, U [ x ] ⊆ V {\displaystyle U[x]\subseteq V} for all x ∈ P . {\displaystyle x\in P.} A deleted neighbourhood of

2844-413: Is the outward pointing unit normal at almost each point on the boundary ∂ V {\displaystyle \partial V} . ( d S {\displaystyle \mathrm {d} \mathbf {S} } may be used as a shorthand for n d S {\displaystyle \mathbf {n} \mathrm {d} S} .) In terms of the intuitive description above, the left-hand side of

2923-1819: Is the outward pointing unit normal vector to ∂ Ω {\displaystyle \partial \Omega } . Equivalently, ∫ Ω ∇ u d V = ∫ ∂ Ω u ν d S . {\displaystyle \int _{\Omega }\nabla u\,dV=\int _{\partial \Omega }u\nu \,dS.} Proof of Theorem. x ′ = ( x 1 , … , x n − 1 ) , {\displaystyle x'=(x_{1},\dots ,x_{n-1}),} it holds that U = { x ∈ R n : | x ′ | < r  and  | x n − g ( x ′ ) | < h } {\displaystyle U=\{x\in \mathbb {R} ^{n}:|x'|<r{\text{ and }}|x_{n}-g(x')|<h\}} and for x ∈ U {\displaystyle x\in U} , x n = g ( x ′ ) ⟹ x ∈ ∂ Ω , − h < x n − g ( x ′ ) < 0 ⟹ x ∈ Ω , 0 < x n − g ( x ′ ) < h ⟹ x ∉ Ω . {\displaystyle {\begin{aligned}x_{n}=g(x')&\implies x\in \partial \Omega ,\\-h<x_{n}-g(x')<0&\implies x\in \Omega ,\\0<x_{n}-g(x')<h&\implies x\notin \Omega .\\\end{aligned}}} We are going to prove

3002-575: Is the outward-pointing unit normal vector to ∂ Ω {\displaystyle \partial \Omega } . Proof of Theorem. We use the Einstein summation convention. By using a partition of unity, we may assume that u {\displaystyle u} and X {\displaystyle X} have compact support in a coordinate patch O ⊂ Ω ¯ {\displaystyle O\subset {\overline {\Omega }}} . First consider

3081-515: Is the same as the surface integral of the vector field which has as components f x {\displaystyle f_{x}} , f y {\displaystyle f_{y}} and f z {\displaystyle f_{z}} . Various useful results for surface integrals can be derived using differential geometry and vector calculus , such as the divergence theorem , magnetic flux , and its generalization, Stokes' theorem . Let us notice that we defined

3160-411: Is the standard formula for the area of a surface described this way. One can recognize the vector in the second-last line above as the normal vector to the surface. Because of the presence of the cross product, the above formulas only work for surfaces embedded in three-dimensional space. This can be seen as integrating a Riemannian volume form on the parameterized surface, where the metric tensor

3239-451: Is the union of all the open balls of radius r {\displaystyle r} that are centered at a point in S {\displaystyle S} ): S r = ⋃ p ∈ S B r ( p ) . {\displaystyle S_{r}=\bigcup \limits _{p\in {}S}B_{r}(p).} It directly follows that an r {\displaystyle r} -neighbourhood

Divergence theorem - Misplaced Pages Continue

3318-495: Is the unit circle, C {\displaystyle C} , that can be represented parametrically by: such that 0 ≤ s ≤ 2 π {\displaystyle 0\leq s\leq 2\pi } where s {\displaystyle s} units is the length arc from the point s = 0 {\displaystyle s=0} to the point P {\displaystyle P} on C {\displaystyle C} . Then

3397-411: Is the vector normal to S at the given point, whose magnitude is d s = ‖ d s ‖ . {\displaystyle \mathrm {d} s=\|\mathrm {d} {\mathbf {s} }\|.} We find the formula The cross product on the right-hand side of this expression is a (not necessarily unital) surface normal determined by the parametrisation. This formula defines

3476-515: Is the volume element on ∂ Ω {\displaystyle \partial \Omega } and the above formula reads ( grad ⁡ u , X ) = ( u , − div ⁡ X ) + ∫ ∂ Ω u ⟨ X , N ⟩ d S . {\displaystyle (\operatorname {grad} u,X)=(u,-\operatorname {div} X)+\int _{\partial \Omega }u\langle X,N\rangle \,dS.} This completes

3555-475: The determinant of the Jacobian of the transition function from ( s , t ) {\displaystyle (s,t)} to ( x , y ) {\displaystyle (x,y)} . The transformation of the other forms are similar. Then, the surface integral of f on S is given by where is the surface element normal to S . Let us note that the surface integral of this 2-form

3634-412: The latitude and longitude on a sphere . Let such a parameterization be r ( s , t ) , where ( s , t ) varies in some region T in the plane . Then, the surface integral is given by where the expression between bars on the right-hand side is the magnitude of the cross product of the partial derivatives of r ( s , t ) , and is known as the surface element (which would, for example, yield

3713-431: The normal vector n ^ {\displaystyle \mathbf {\hat {n}} } is opposite for each volume, so the flux out of one through S 3 is equal to the negative of the flux out of the other so these two fluxes cancel in the sum. Therefore: Since the union of surfaces S 1 and S 2 is S This principle applies to a volume divided into any number of parts, as shown in

3792-519: The Application of Mathematical Analysis to the Theories of Electricity and Magnetism , Siméon Denis Poisson in 1824 in a paper on elasticity, and Frédéric Sarrus in 1828 in his work on floating bodies. To verify the planar variant of the divergence theorem for a region R {\displaystyle R} : and the vector field: The boundary of R {\displaystyle R}

3871-463: The Gauss's law-type equation from the inverse-square formulation or vice versa is exactly the same in both cases; see either of those articles for details. Joseph-Louis Lagrange introduced the notion of surface integrals in 1760 and again in more general terms in 1811, in the second edition of his Mécanique Analytique . Lagrange employed surface integrals in his work on fluid mechanics. He discovered

3950-446: The answer to this question is simple; the value of the surface integral will be the same no matter what parametrization one uses. For integrals of vector fields, things are more complicated because the surface normal is involved. It can be proven that given two parametrizations of the same surface, whose surface normals point in the same direction, one obtains the same value for the surface integral with both parametrizations. If, however,

4029-1653: The case where the patch is disjoint from ∂ Ω {\displaystyle \partial \Omega } . Then O {\displaystyle O} is identified with an open subset of R n {\displaystyle \mathbb {R} ^{n}} and integration by parts produces no boundary terms: ( grad ⁡ u , X ) = ∫ O ⟨ grad ⁡ u , X ⟩ g d x = ∫ O ∂ j u X j g d x = − ∫ O u ∂ j ( g X j ) d x = − ∫ O u 1 g ∂ j ( g X j ) g d x = ( u , − 1 g ∂ j ( g X j ) ) = ( u , − div ⁡ X ) . {\displaystyle {\begin{aligned}(\operatorname {grad} u,X)&=\int _{O}\langle \operatorname {grad} u,X\rangle {\sqrt {g}}\,dx\\&=\int _{O}\partial _{j}uX^{j}{\sqrt {g}}\,dx\\&=-\int _{O}u\partial _{j}({\sqrt {g}}X^{j})\,dx\\&=-\int _{O}u{\frac {1}{\sqrt {g}}}\partial _{j}({\sqrt {g}}X^{j}){\sqrt {g}}\,dx\\&=(u,-{\frac {1}{\sqrt {g}}}\partial _{j}({\sqrt {g}}X^{j}))\\&=(u,-\operatorname {div} X).\end{aligned}}} In

SECTION 50

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4108-407: The diagram. Since the integral over each internal partition (green surfaces) appears with opposite signs in the flux of the two adjacent volumes they cancel out, and the only contribution to the flux is the integral over the external surfaces (grey) . Since the external surfaces of all the component volumes equal the original surface. The flux Φ out of each volume is the surface integral of

4187-561: The differential forms transform as So d x d y {\displaystyle \mathrm {d} x\mathrm {d} y} transforms to ∂ ( x , y ) ∂ ( s , t ) d s d t {\displaystyle {\frac {\partial (x,y)}{\partial (s,t)}}\mathrm {d} s\mathrm {d} t} , where ∂ ( x , y ) ∂ ( s , t ) {\displaystyle {\frac {\partial (x,y)}{\partial (s,t)}}} denotes

4266-538: The divergence theorem in 1762. Carl Friedrich Gauss was also using surface integrals while working on the gravitational attraction of an elliptical spheroid in 1813, when he proved special cases of the divergence theorem. He proved additional special cases in 1833 and 1839. But it was Mikhail Ostrogradsky , who gave the first proof of the general theorem, in 1826, as part of his investigation of heat flow. Special cases were proven by George Green in 1828 in An Essay on

4345-399: The divergence theorem says that the integral is equal to: where W is the unit ball : Since the function y is positive in one hemisphere of W and negative in the other, in an equal and opposite way, its total integral over W is zero. The same is true for z : Therefore, because the unit ball W has volume ⁠ 4 π / 3 ⁠ . As a result of the divergence theorem,

4424-507: The divergence theorem. In fluid dynamics , electromagnetism , quantum mechanics , relativity theory , and a number of other fields, there are continuity equations that describe the conservation of mass, momentum, energy, probability, or other quantities. Generically, these equations state that the divergence of the flow of the conserved quantity is equal to the distribution of sources or sinks of that quantity. The divergence theorem states that any such continuity equation can be written in

4503-447: The equation represents the total of the sources in the volume V , and the right-hand side represents the total flow across the boundary S . The divergence theorem follows from the fact that if a volume V is partitioned into separate parts, the flux out of the original volume is equal to the algebraic sum of the flux out of each component volume. This is true despite the fact that the new subvolumes have surfaces that were not part of

4582-404: The example of the velocity field of a fluid , such as a gas or liquid. A moving liquid has a velocity—a speed and a direction—at each point, which can be represented by a vector , so that the velocity of the liquid at any moment forms a vector field. Consider an imaginary closed surface S inside a body of liquid, enclosing a volume of liquid. The flux of liquid out of the volume at any time

4661-416: The figure, is not a neighbourhood of all its points; points on the edges or corners of the rectangle are not contained in any open set that is contained within the rectangle. The collection of all neighbourhoods of a point is called the neighbourhood system at the point. If S {\displaystyle S} is a subset of a topological space X {\displaystyle X} , then

4740-446: The following: Theorem  —  Let Ω ¯ {\displaystyle {\overline {\Omega }}} be a C 2 {\displaystyle C^{2}} compact manifold with boundary with C 1 {\displaystyle C^{1}} metric tensor g {\displaystyle g} . Let Ω {\displaystyle \Omega } denote

4819-698: The immersed surface, where d S {\displaystyle \mathrm {d} S} is the induced volume form on the surface, obtained by interior multiplication of the Riemannian metric of the ambient space with the outward normal of the surface. Let be a differential 2-form defined on a surface S , and let be an orientation preserving parametrization of S with ( s , t ) {\displaystyle (s,t)} in D . Changing coordinates from ( x , y ) {\displaystyle (x,y)} to ( s , t ) {\displaystyle (s,t)} ,

SECTION 60

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4898-505: The infinitesimal dV , the part in parentheses becomes the divergence, and the sum becomes a volume integral over V ∬ S ( V ) F ⋅ n ^ d S = ∭ V div ⁡ F d V {\displaystyle \;\iint _{S(V)}\mathbf {F} \cdot \mathbf {\hat {n}} \;\mathrm {d} S=\iiint _{V}\operatorname {div} \mathbf {F} \;\mathrm {d} V\;} Since this derivation

4977-476: The integral on the left (note the dot and the vector notation for the surface element). We may also interpret this as a special case of integrating 2-forms, where we identify the vector field with a 1-form, and then integrate its Hodge dual over the surface. This is equivalent to integrating ⟨ v , n ⟩ d S {\displaystyle \left\langle \mathbf {v} ,\mathbf {n} \right\rangle \mathrm {d} S} over

5056-562: The last equality we used the Voss-Weyl coordinate formula for the divergence, although the preceding identity could be used to define − div {\displaystyle -\operatorname {div} } as the formal adjoint of grad {\displaystyle \operatorname {grad} } . Now suppose O {\displaystyle O} intersects ∂ Ω {\displaystyle \partial \Omega } . Then O {\displaystyle O}

5135-1063: The manifold interior of Ω ¯ {\displaystyle {\overline {\Omega }}} and let ∂ Ω {\displaystyle \partial \Omega } denote the manifold boundary of Ω ¯ {\displaystyle {\overline {\Omega }}} . Let ( ⋅ , ⋅ ) {\displaystyle (\cdot ,\cdot )} denote L 2 ( Ω ¯ ) {\displaystyle L^{2}({\overline {\Omega }})} inner products of functions and ⟨ ⋅ , ⋅ ⟩ {\displaystyle \langle \cdot ,\cdot \rangle } denote inner products of vectors. Suppose u ∈ C 1 ( Ω ¯ , R ) {\displaystyle u\in C^{1}({\overline {\Omega }},\mathbb {R} )} and X {\displaystyle X}

5214-486: The normals for these parametrizations point in opposite directions, the value of the surface integral obtained using one parametrization is the negative of the one obtained via the other parametrization. It follows that given a surface, we do not need to stick to any unique parametrization, but, when integrating vector fields, we do need to decide in advance in which direction the normal will point and then choose any parametrization consistent with that direction. Another issue

5293-438: The original volume's surface, because these surfaces are just partitions between two of the subvolumes and the flux through them just passes from one volume to the other and so cancels out when the flux out of the subvolumes is summed. See the diagram. A closed, bounded volume V is divided into two volumes V 1 and V 2 by a surface S 3 (green) . The flux Φ( V i ) out of each component region V i

5372-429: The part in parentheses below, does not in general vanish but approaches the divergence div F as the volume approaches zero. As long as the vector field F ( x ) has continuous derivatives, the sum above holds even in the limit when the volume is divided into infinitely small increments As | V i | {\displaystyle |V_{\text{i}}|} approaches zero volume, it becomes

5451-456: The pieces are put back together, the results are consistent. For the cylinder, this means that if we decide that for the side region the normal will point out of the body, then for the top and bottom circular parts, the normal must point out of the body too. Last, there are surfaces which do not admit a surface normal at each point with consistent results (for example, the Möbius strip ). If such

5530-401: The proof. By replacing F in the divergence theorem with specific forms, other useful identities can be derived (cf. vector identities ). Suppose we wish to evaluate where S is the unit sphere defined by and F is the vector field The direct computation of this integral is quite difficult, but we can simplify the derivation of the result using the divergence theorem, because

5609-418: The rate of liquid removed by the sink. If there are multiple sources and sinks of liquid inside S , the flux through the surface can be calculated by adding up the volume rate of liquid added by the sources and subtracting the rate of liquid drained off by the sinks. The volume rate of flow of liquid through a source or sink (with the flow through a sink given a negative sign) is equal to the divergence of

5688-463: The region". The divergence theorem is an important result for the mathematics of physics and engineering , particularly in electrostatics and fluid dynamics . In these fields, it is usually applied in three dimensions. However, it generalizes to any number of dimensions. In one dimension, it is equivalent to the fundamental theorem of calculus . In two dimensions, it is equivalent to Green's theorem . Vector fields are often illustrated using

5767-410: The surface S . The flux outward through S equals the volume rate of flow of fluid into S from the pipe. Similarly if there is a sink or drain inside S , such as a pipe which drains the liquid off, the external pressure of the liquid will cause a velocity throughout the liquid directed inward toward the location of the drain. The volume rate of flow of liquid inward through the surface S equals

5846-552: The surface integral by using a parametrization of the surface S . We know that a given surface might have several parametrizations. For example, if we move the locations of the North Pole and the South Pole on a sphere, the latitude and longitude change for all the points on the sphere. A natural question is then whether the definition of the surface integral depends on the chosen parametrization. For integrals of scalar fields,

5925-491: The surface. For example, imagine that we have a fluid flowing through S , such that v ( r ) determines the velocity of the fluid at r . The flux is defined as the quantity of fluid flowing through S per unit time. This illustration implies that if the vector field is tangent to S at each point, then the flux is zero because the fluid just flows in parallel to S , and neither in nor out. This also implies that if v does not just flow along S , that is, if v has both

6004-869: The vector field F ( x ) over the surface The goal is to divide the original volume into infinitely many infinitesimal volumes. As the volume is divided into smaller and smaller parts, the surface integral on the right, the flux out of each subvolume, approaches zero because the surface area S ( V i ) approaches zero. However, from the definition of divergence , the ratio of flux to volume, Φ ( V i ) | V i | = 1 | V i | ∬ S ( V i ) F ⋅ n ^ d S {\displaystyle {\frac {\Phi (V_{\text{i}})}{|V_{\text{i}}|}}={\frac {1}{|V_{\text{i}}|}}\iint _{S(V_{\text{i}})}\mathbf {F} \cdot \mathbf {\hat {n}} \;\mathrm {d} S} ,

6083-410: The velocity field at the pipe mouth, so adding up (integrating) the divergence of the liquid throughout the volume enclosed by S equals the volume rate of flux through S . This is the divergence theorem. The divergence theorem is employed in any conservation law which states that the total volume of all sinks and sources, that is the volume integral of the divergence, is equal to the net flow across

6162-413: The volume at other points, but the amounts flowing in and out at any moment are equal, so the net flux of liquid out of the volume is zero. However if a source of liquid is inside the closed surface, such as a pipe through which liquid is introduced, the additional liquid will exert pressure on the surrounding liquid, causing an outward flow in all directions. This will cause a net outward flow through

6241-444: The volume's boundary. Suppose V is a subset of R n {\displaystyle \mathbb {R} ^{n}} (in the case of n = 3, V represents a volume in three-dimensional space ) which is compact and has a piecewise smooth boundary S (also indicated with ∂ V = S {\displaystyle \partial V=S} ). If F is a continuously differentiable vector field defined on

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