Physics
Stokes Theorem
Stokes' theorem is a fundamental theorem in vector calculus that relates the surface integral of the curl of a vector field to the line integral of the vector field around the boundary curve. It states that the circulation of a vector field around a closed curve is equal to the flux of the curl of the vector field through any surface bounded by the curve.
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12 Key excerpts on "Stokes Theorem"
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All the Math You Missed
(But Need to Know for Graduate School)
- Thomas A. Garrity(Author)
- 2021(Publication Date)
- Cambridge University Press(Publisher)
On the left-hand side we have an integral of the vector field F over the boundary. On the right-hand side we have an integral of the function div(F) (which involves derivatives of the vector field) over the interior. In Stokes’ Theorem, the interval becomes a surface, so that the boundary is a curve, and the function again becomes a vector field. The role of the derivative though will now be played by the curl of the vector field. 94 Classical Stokes’ Theorems Theorem 5.2.3 (Stokes’ Theorem) Let M be a surface in R 3 with compact boundary curve ∂M. Let n(x,y,z) be the unit normal vector field to M and let T(x,y,z) denote the induced unit tangent vector to the curve ∂M. If F(x,y,z) is any vector field, then ∂M F · T ds = M curl(F) · n dS . As with the Divergence Theorem, a sketch of the proof will be given later in this chapter. Again, on the left-hand side we have an integral involving a vector field F on the boundary while on the right-hand side we have an integral on the interior involving the curl of F (which is in terms of the various derivatives of F). Although both the Divergence Theorem and Stokes’ Theorem were proven independently, their similarity is more than a mere analogy; both are special cases, as is the Fundamental Theorem of Calculus, of one very general theorem, which is the goal of the next chapter . The proofs of each are also quite similar. There are in fact two basic methods for proving these types of theorems. The first is to reduce to the Fundamental Theorem of Calculus, f (b) − f (a) = b a df dx dx . This method will be illustrated in our sketch of the Divergence Theorem. The second method involves two steps. Step one is to show that given two regions R 1 and R 2 that share a common boundary, we have ∂R 1 function + ∂R 2 function = ∂(R 1 ∪R 2 ) function. Step two is to show that the theorem is true on infinitesimally small regions. - eBook - ePub
Two and Three Dimensional Calculus
with Applications in Science and Engineering
- Phil Dyke(Author)
- 2018(Publication Date)
- Wiley(Publisher)
To resolve this, a common-sense approach is used. The theorems are stated and proved in a way to show, it is hoped, what is going on. Once this understanding is established then there follows arguments that show how the theorems are extended. Finally, the derived relationships are applied to practical problems. Let us start with the relationship between surface and path integrals, called Green's theorem in the plane in Chapter 10 which is now generalised. 11.2 Stokes' Theorem The starting point for Stokes' theorem is to restate Green's theorem in the plane: The first task is to write this in vectorial terms. Defining, then gives The integrand on the right-hand side can be written and is the element of the flat rectangular region in the plane, so Green's theorem in the plane can be written: 11.1 Changing this slightly to with where now is a curved surface, as if made of thin flexible plastic then hardened, and and are curvilinear co-ordinates with the first two unit vectors embedded in. Equation 11.1 now reads exactly the same but is now more general and is termed Stokes' theorem. This is an example of a co-ordinate-free equation derived in a particular way that because it now is free of the co-ordinates, immediately generalises. It does however warrant a proof. Here it is as a formal theorem. Theorem (Stokes' Theorem) For an orientable surface with a bounding curve and vector field continuous and differentiable on, it is true that Proof There are many ways to approach the proof of Stokes' theorem. Pure mathematicians couch the theorem in general terms and the proof then needs knowledge of such esoteric topological concepts as compact manifolds and de Rham cohomologies. This is unsuitable here, as is the slick, few-line proof using tensors. Some are quite content to believe the truth of Stokes' theorem based on the preliminary vectorisation of Green's theorem in the plane, Equation 11.1 - eBook - PDF
Mathematical Methods and Physical Insights
An Integrated Approach
- Alec J. Schramm(Author)
- 2022(Publication Date)
- Cambridge University Press(Publisher)
Unlike the divergence theorem, which identifies the scalar-valued divergence as a volume density, Stokes’ theorem reveals the vector-valued curl as an oriented surface density. Its flux, through a surface S bounded by C, is the circulation around C. Recall that a surface-independent field has zero divergence. But the vector field whose flux is being evaluated on the right-hand side of Stokes’ theorem, (15.32), is a curl — and ∇ · ( ∇ × B) ≡ 0. Thus any field of the form F = ∇ × B is surface independent: Stokes’ theorem holds for any surface bounded by a given curve C. The standard expression of Stokes’ theorem is given in (15.32). However, there are other, vector- valued versions which are also useful — and which you’re asked to derive in Problem 15.12: C d = S ˆ n × ∇ da ≡ S d a × ∇ (15.33a) C d × E = S ( ˆ n × ∇) × E da ≡ S (d a × ∇) × E. (15.33b) 206 15 THE THEOREMS OF GAUSS AND STOKES Example 15.8 B = x 2 yˆ ı − xy 2 ˆ j y x P 2 P 1 C 3 C 2 C 1 q We’ll demonstrate Stokes’ theorem with the field of Examples 14.13 and 14.15. C 2 − C 1 : We found before that the work along C 1 is −1/3 while along C 2 it’s 0. Thus along the closed path C 2 − C 1 we have = +1/3. To verify Stoke’s theorem, we must integrate the curl of B over the enclosed triangular surface. Going from P 1 to P 2 along C 2 and back along C 1 gives the circulation a clockwise orientation, so d a = − ˆ kdx dy. Then S ( ∇ × B) · d a = S −(x 2 + y 2 ) ˆ k · (− ˆ k) dxdy = + 1 0 dx x 0 dy(x 2 + y 2 ) = 4 3 1 0 x 3 = 1 3 . C 1 − C 3 : In cylindrical coordinates, x = 1 − r cos φ, y = r sin φ, and a quick calculation shows the curl to be −(r 2 − 2r cos φ + 1) ˆ k. The circulation is counter-clockwise, so d a = + ˆ k r dr dφ. Then integrating over the quarter-circle gives S ( ∇ × B) · d a = − 1 0 rdr π/2 0 (r 2 − 2r cos φ + 1) dφ = 2 3 − 3π 8 , which agrees with the sum of the line integrals computed earlier. - eBook - PDF
- James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
Stokes’ Theorem Let S be an oriented piecewise-smooth surface that is bounded by a simple, closed, piecewise-smooth boundary curve C with positive orientation. Let F be a vector field whose components have continuous partial derivatives on an open region in R 3 that contains S. Then y C F dr - y S y curl F dS Since y C F dr - y C F T ds and y S y curl F dS - y S y curl F n dS Stokes’ Theorem says that the line integral around the boundary curve of S of the tangen- tial component of F is equal to the surface integral over S of the normal component of the curl of F. The positively oriented boundary curve of the oriented surface S is often written as -S, so Stokes’ Theorem can be expressed as 1 y S y curl F dS - y -S F dr There is an analogy among Stokes’ Theorem, Green’s Theorem, and the Fundamental Theorem of Calculus. As before, there is an integral involving derivatives on the left side of Equation 1 (recall that curl F is a sort of derivative of F) and the right side involves the values of F only on the boundary of S. In fact, in the special case where the surface S is flat and lies in the xy-plane with upward orientation, the unit normal is k, the surface integral becomes a double integral, and Stokes’ Theorem becomes y C F dr - y S y curl F dS - y S y scurl Fd k dA This is precisely the vector form of Green’s Theorem given in Equation 16.5.12. Thus we see that Green’s Theorem is really a special case of Stokes’ Theorem. Although Stokes’ Theorem is too difficult for us to prove in its full generality, we can give a proof when S is a graph and F, S, and C are well behaved. - eBook - PDF
Dynamics of the Atmosphere
A Course in Theoretical Meteorology
- Wilford Zdunkowski, Andreas Bott(Authors)
- 2003(Publication Date)
- Cambridge University Press(Publisher)
M6.3 Integral theorems M6.3.1 Stokes’ integral theorem The Stokes integral theorem relating a surface and a line integral is based on the following general expression: S d S × ∇ ( · · · ) = d r ( · · · ) (M6 . 20) where the term ( · · · ) indicates that this formula may be applied to tensors of arbitrary rank, that is scalars, vectors, dyadics etc. The derivation of this important equation will be omitted here since it is given in many mathematical textbooks on vector analysis. Figure M6.6 depicts the parameters occurring in (M6.20). If the surface S is closed then the border line vanishes and the right hand side of (M6.20) is zero. Of great importance are the first scalar and the vector of (M6.20). The first scalar results in the Stokes integral theorem S d S × ∇ · A = S ∇ × A · d S = d r · A = A T dr (M6 . 21) where A T is the tangential component of A along the border line . (M6.21) is a fundamental formula not only in fluid dynamics but also in many other branches of science. Using Grassmann’s rule (M1.44), we find for the vector of the dyadic S ( d S × ∇ ) × A = S [ ∇ A − ( ∇ · A ) E ] · d S = d r × A (M6 . 22) where we have used the property of the conjugate dyadics. M6.3 Integral theorems 91 In the following we present some important applications of the Stokes integral theorem. In the first example we apply (M6.21) to a very small surface element S = S e 3 so that ∇ × A may be viewed as a constant vector and the integral on the left-hand side of (M6.21) may be replaced by ∇ × A · e 3 S . This yields the coordinate-free definition of the rotation , ∇ × A · e 3 = 1 S d r · A (M6 . 23) In the next application we consider the horizontal flow in the Cartesian coordinate system. In this case we write the differentials d S = k dx dy and d r = i dx + j dy . - eBook - ePub
- Brian R. Martin, Graham Shaw(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
the circulation of V around the loop C. Thus the theorem states that the surface integral of is equal to the circulation of V around the bounding curve C. This is closely related to the interpretation of curl. To see this, we apply (12.68) to a loop C that encloses a small surface element, which shrinks to a point when d s → 0. In this limit, the variation of V and can be neglected on d s, so that the left-hand side of (12.68) becomes, implying (12.69) In other words, at a point r is the circulation per unit area around the boundary of an infinitesimal surface d s containing the point r. For example, let us again consider a vector field V = ρ v, where ρ is the density and v is the velocity field of a fluid. Then for a uniform flow pattern, such as that shown in Figure 12.12 a, and V is said to be irrotational. On the other hand, at the centre of a vortex, like that shown in Figure 12.12 b, clearly. It is also non-zero in a non-uniform parallel motion, as shown in Figure 12.12 c, since the velocities on either side of a point are different. Essentially, when there is rotational motion in addition to, or opposed to, translational motion. A practical viewpoint is to consider what would happen if one inserted a small ‘paddle wheel’, which is free to rotate about its axis. In the flow pattern of Figure 12.12 a, where, it would not rotate: the motion is irrotational. In Figures 12.12 b and 12.12 c, where, it would rotate. Figure 12.12 Flow of a fluid. The coloured lines are the flow lines; the arrows show the direction of the vector field V. Their lengths show the relative magnitudes of V. In the rest of this section we will first derive Stokes' theorem, and then consider some applications. 12.4.1 Proof of Stokes' theorem We start by considering a closed curve C surrounding a plane surface S parallel to the x–y plane, so that z is constant. Then and But by Green's theorem in the plane (11.20), we have so that (12.70) where, a unit vector in the z- direction - eBook - ePub
- Richard L. Eisenman(Author)
- 2013(Publication Date)
- Dover Publications(Publisher)
[2]. We will show how Stokes’ theorem transforms this Faraday law into one of Maxwell’s fundamental equations. If we combine[1]and[2], Faraday’s law becomes . This equates a line integral and a surface integral, and we may appeal to the Stokes Theorem to put the two on common ground. Apply the Stokes Theorem to get . (Note that it is not convenient to apply the Stokes Theorem to the right side of[3]because[5].) Substituting[4]in[3]gives . Since this relation must hold regardless of the surface selected, the integrand itself in[6]must be zero; i.e.,[7]. This statement is one of[8]equations.EXERCISE [38]:The Divergence theorem in fluid flow. Here we intend to show how conservation of mass and the Divergence theorem lead to the basic law of continuity in fluid flow. The vocabulary required is the velocity , whose dimensions are[1], and the density ρ, whose dimensions are[2]. Then the quantity has dimensions[3]and measures the net loss of mass per unit of time through the surface σ. On the other hand, the volume integral of the time rate of change of density, in symbols[4], measures the net gain in mass per unit of time from the volume bounded by the surface. According to conservation of mass, the gain should be[5]the loss so that . Here a surface integral and a volume integral are equated, and we may use a transformation theorem, namely[7], to put them on common ground. It would be difficult to transform the volume integral to a surface integral because[8]. Thus we apply the Divergence theorem to . Now, combining, . This integral must be zero no matter what volume is selected, so the integrand itself must be zero, whence we have the celebrated law of continuity,[11] - eBook - PDF
Calculus
Multivariable
- William G. McCallum, Deborah Hughes-Hallett, Andrew M. Gleason, David O. Lomen, David Lovelock, Jeff Tecosky-Feldman, Thomas W. Tucker, Daniel E. Flath, Joseph Thrash, Karen R. Rhea, Andrew Pasquale, Sheldon P. Gordon, Douglas Quinney, Patti Frazer Lock(Authors)
- 2014(Publication Date)
- Wiley(Publisher)
In this section we will examine some consequences of these theorems. 20.3 THE THREE FUNDAMENTAL THEOREMS 1085 Fundamental Theorem of Calculus for Line Integrals C gradf · d r = f (Q) − f (P ). Stokes’ Theorem S curl F · d A = C F · d r . Divergence Theorem W div F dV = S F · d A . Notice that, in each case, the region of integration on the right is the boundary of the region on the left (except that for the first theorem we simply evaluate f at the boundary points); the integrand on the left is a sort of derivative of the integrand on the right; see Figure 20.20. C P Q The boundary of the curve C consists of the points P and Q S C Boundary of surface S is curve C Boundary of region W is surface S W S Figure 20.20: Regions and their boundaries for the three fundamental theorems The Gradient and the Curl Suppose that F is a smooth gradient field, so F = grad f for some function f . Using the Funda- mental Theorem for Line Integrals, we saw in Chapter 18 that C F · d r = 0 for any closed curve C. Thus, for any unit vector n circ n F = lim Area→0 C F · d r Area of C = lim Area→0 0 Area = 0, where the limit is taken over circles C in a plane perpendicular to n , and oriented by the right-hand rule. Thus, the circulation density of F is zero in every direction, so curl F = 0 , that is, curl grad f = 0 . (This formula can also be verified using the coordinate definition of curl. See Problem 27 on page 1076.) Is the converse true? Is any vector field whose curl is zero a gradient field? Suppose that curl F = 0 and let us consider the line integral C F · d r for a closed curve C contained in the domain of F . If C is the boundary curve of an oriented surface S that lies wholly in the domain of curl F , then Stokes’ Theorem asserts that C F · d r = S curl F · d A = S 0 · d A = 0. - eBook - ePub
Vector Analysis and Cartesian Tensors
Third Edition
- Donald Edward Bourne(Author)
- 2018(Publication Date)
- Chapman and Hall/CRC(Publisher)
Integral theorems 6 6.1 INTRODUCTIONIn Chapter 4 , scalar and vector fields were defined and the properties of gradient, divergence, and curl were discussed; and in Chapter 5 we were concerned with various integrals of scalar and vector fields, and the techniques whereby such integrals are evaluated. The ground has thus been prepared for the two central theorems in vector analysis: (i) the divergence theorem (also called Gauss’s theorem), which relates the integral of a vector field F over a closed surface S to the volume integral of div F over the region bounded by S; and (ii) Stokes’s theorem which relates the integral of a vector field F around a closed curve C to the integral of curl F over any open surface S bounded by C . In this chapter we shall prove these theorems and some related results.6.2 THE DIVERGENCE THEOREM (GAUSS’S THEOREM) Definitions1. A simple closed surface S is said to be convex if any straight line meeting it does so in at most two points.2. A simple closed surface S will be called a semi-convex surface if axes Oxyz can be so chosen that any line drawn parallel to one of the axes and meeting S either (i) does so in just one or just two points, or (ii) has a portion of finite length in common with S.The divergence theoremA closed region V is bounded by a simple closed surface S. If the vector field F is defined and continuously differentiable throughout V, then
Proof=∬ SF. d S∭ Vdiv F d V .(6.1) The result will first be proved, in part (a), for a region V bounded by a convex surface S. The theorem will then be extended to more general regions in parts (b), (c) and (d).(a) Take rectangular axes Oxyz and denote by R the projection of S on to the xy-plane. Divide S into upper and lower parts, SU and SL , as viewed from R (Fig. 6.1 ). Let a line drawn parallel to Oz from a point in R cut SL at P and SU at Q, and denote the z-coordinates of P, Q by zP , zQ (zP < zQ ). All points (x, y, z) in V are covered if x and y range over all values in R and the corresponding values of z range from zP to zQ - P C Kendall, D.E. Bourne(Authors)
- 2017(Publication Date)
- Routledge(Publisher)
Ω is a continuously differentiable scalar field, prove the following corollary of Stokes’s theorem:[Hint . Consider Ω a , where a is a constant vector field, and proceed as in Example 2, section 6.2 .]6.15 Prove the following corollary of Stokes’s theorem, and state the conditions to be satisfied by F , S and :[Hint . Put F = F 1 e 1 + F 2 e 2 + F 3 e 3 , where e 1 , e 2 , e 3 are the basic unit vectors of a rectangular cartesian coordinate system, and use the result of Exercise 6.14.]6.16 Two non-intersecting simple closed curves lie in the xy-plane, with enclosing , and are described in the anticlockwise sense. Show that, for the vector field F = (zy 2 , 2xyz , x ),6.17 Consider the remarks made in Section 6.2 about the extension of Gauss’s theorem to functions whose partial derivatives are not continuous. Verify that Stokes’s theorem (6.25 ) extends in a similar way, provided that any discontinuities in the derivatives of F are finite and are confined to a finite number of simple curves on the surface S .6.5 Limit Definitions of Div F and Curl F
Div FLet P be a point in a region V bounded by a simple closed surface S , throughout which the vector field F and its divergence are defined. If V 0 denotes the volume of V , the mean value theorem for integrals* shows that there exists a point P’ in V such thatThus, by the divergence theorem,If the surface S is now allowed to collapse towards P’ , the linear dimensions of V become arbitrarily small, and P’ must approach P (Fig. 6.15(a) ). It follows that at PThis relationship is very often used to define div F .Curl FLet F and curl F be defined throughout a region V containing a given point P . Draw a plane surface S through P and contained in V , with its normal in the direction of a given unit vector n and bounded by a correspondingly oriented simple closed curve (Fig. 6.15(b) ). Let A be the area of the surface S . Then, using Stokes’s theorem followed by the mean value theorem, there exists a point P’ on S such thatFig. 6.15Allowing to shrink towards P , it follows that at P- eBook - PDF
- Dominique Barthes-Biesel(Author)
- 2012(Publication Date)
- CRC Press(Publisher)
The same technique can also be used for problems involving interactions between flowing particles. For more details, see the book by Kim and Karrila [32]. 8.4 Integral Form of the Stokes Equations n n D ∂ D FIGURE 8.11 Definition of the fluid domain. When the flow geometry is complex, it is difficult to obtain an analytical solution and we thus have to find a numerical solution to the problem. If we wish to know the pressure and velocity at each point of the domain D , the classical numerical techniques developed for the Navier–Stokes equations can be used (finite differences, finite elements, volume of fluid, etc). However, it may happen that we are interested in the flow in some parts of D only or that we only want to determine the velocity of the domain boundary and the forces acting on it. If such is the case, it is not necessary to compute the flow everywhere in D . The boundary integral method is then very efficient as it General Solutions of the Stokes Equations 143 is based on the determination of the force and velocity distribution on the domain boundary. A further advantage of the method is that it allows us to include moving and/or deformable boundaries (liquid interfaces, elastic walls, etc) without having to mesh again the whole flow domain. 8.4.1 Velocity Field in D We consider a fluid domain D bounded by a piecewise continuous boundary ∂ D . The unit normal vector to ∂ D pointing inside D is denoted n (Figure 8.11). We seek the solution ( u , σ ) of the Stokes equations in D : ∇· u = 0 and ∇· σ = 0 (8.66) with mixed boundary conditions u = U ( x ) for x ∈ ∂ D 1 (8.67) σ · n = f B ( x ) for x ∈ ∂ D 2 (8.68) where ∂ D = ∂ D 1 ∪ ∂ D 2 and where U ( x ) and f B ( x ) are given. We use the integral form (2.51) of the general reciprocal theorem. Flow (1) is the real flow ( u , σ ) that we seek to determine. Flow (2) is the stokeslet flow created by a point force F located at x (Equations 8.4 to 8.7). - eBook - ePub
Tensor and Vector Analysis
With Applications to Differential Geometry
- C. E. Springer(Author)
- 2013(Publication Date)
- Dover Publications(Publisher)
(32) without symbols as follows. The integral, over any plane region, of the normal component of the curl of a vector field is equal to the circulation of the vector field (taken in the appropriate direction) around the boundary of the region. In the generalization to Stokes’ theorem, the plane region becomes a region on a curved surface.Equation (27) can also be interpreted as a special case of the divergence theorem of Gauss which will be proved in Section 11–8 for three-space. To do this, let the vector field F be given by U 2 i − U 1 j + U 3 k with U 3 ≡ 0. The divergence of F is given bywhich is the integrand of the right-hand member in (27) . Consider the component of F along the normal to the curve Γ. The unit vector along the normal is (dy /ds )i − (dx /ds )j. Hence, F·N ds = U 1 dx + U 2 dy , which appears in the line integral of (27) . Therefore, Green’s theorem in the plane can be written in the formIf the region is generalized to a region D of three-space and the bounding curve Γ is generalized to the bounding surface S of the region D , one might hazard the conjecture thatwhere dσ in (33) becomes the differential of volume dV in (34) and ds in (33) becomes the differential of area dσ on S in (34) . Equation (34) expresses the Gauss divergence theorem in space which will appear in Section 11–8 .EXERCISES
1) Verify Green’s theorem in the plane for the finite region bounded by 2y = x and 4y = x 2 , where the line integral is . Ans . The common value of the two integrals is .2) Verify Green’s theorem in the plane for taken around the circle x 2 + y 2 = a 2 . Ans . The common value is 3πa 2 .3) Evaluate around the ellipse b 2 x 2 + a 2 y 2 = a 2 b 2 . (Hints: Use Green’s theorem, and save further time by using πab for the area of the ellipse.) Ans . 4πab .4) Show that around any simple closed plane curve gives the area enclosed by the curve. Do the same for . Then show that
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