1. Line integral of vector fields: a summary 1.1. Definition (vector fields). Let D be a subset of R3.Avector field F on D is a function F : D → R3. We write F(x, y, z)=⟨M(x, y, z),N(x, y, z),P(x, y, z)⟩. So M,N,P are real valued functions on D. These are called the components of F. We say that F is continuous (resp. differentiable) if M,N,P are continuous (resp. differentiable). 1.2. Definition. Let A and B be two points of D.LetC be an “oriented” curve lying in D starting at A and ending at B.Hereoriented means that we have chosen a continuously varying nonzero tangent vector at each point of the curve. We shall indicate orientation on a curve by drawing an arrow along it. On a connected smooth curve, there are two orientations. Reversing the orientation means reversing the arrow. So the curve with reversed orientation starts at B and ends at A; this oriented curve will be denoted by −C. Let r :[a, b] → D be a “smooth” parametrization of the oriented curve C.Sor(a)=A and r(b)=B. We write r(t)=⟨g(t),h(t),k(t)⟩. Here smooth means that r′(t) exists and is continuous and nonzero for all t ∈ [a, b]. In other words, the parametrized curve has a continuously varying nonzero velocity vector at all time. Below, when talking about a curve C it is always assumed that C is oriented and piecewise smooth unless explicitly stated otherwise. A closed curve C is a curve that begin and end at the same point. In other words, r :[a, b] → R3 parametrizes a closed curve if and only if r(a)=r(b). 1.3. Definition. Given F : D → R3 and r :[a, b] → D as above, define the line integral of F on C,denoted C F · dr by ! b dr(t) F · dr = F(r(t)) · dt, "C "a dt or, more explicitly b ′ ′ ′ (1) F · dr = M(r(t))g (t)+N(r(t))h (t)+P (r(t))k (t)]dt. " " C a #
One can show that the line integral C F · dr exists if C is piecewise smooth and F is continuous on D. One also verifies! that the right hand side of (1) only depends on the oriented curve C and does not depend on the parametrization r :[a, b] → D. It will be useful for us to write the line integral in another way. For this, introduce variables (x, y, z) varying along the curve C giving the coordinates of a point on C. In other words ⟨x, y, z⟩ = r(t)=⟨g(t),h(t),k(t)⟩. Then formally, x = g(t) implies dx = g′(t)dt. Similarly, dy = h′(t)dt and dz = k′(t)dt. So equation (1), can be rephrased as
(2) F · dr = M(x, y, z)dx + N(x, y, z)dy + P (x, y, z)dz. "C "C 1 2
We shall say that the expression α = Mdx+ Ndy + Pdz is a differential 1-form on D and we further abbreviate equation (2) as
F · dr = α. "C "C All integrals encountered in this chapter can be realized as integrals of differential forms. When computing line or surface or volume integrals, we are integrating differential 1-forms or 2-forms or 3-forms respectively. In this course we shall not define differential forms precisely and we shall treat them as formalobjects.We shall contend ourselves with prescribing the rules for working with them. Before moving on we record a couple of basic properties of line integrals: Let C and C1 be smooth curves in D such that the end point of C is the starting point of C1.ThenC + C1 will denote the piecewise smooth curve obtained by following C and then following C1. From the definition, one easily verifies that
α = α + α and α = − α, "C+C1 "C "C1 "−C "C where, −C denotes the curve C with orientation reversed. 1.4 (Examples of vector fields). 1. Two important examples of vector fields arising in physics are velocity fields (in fluid dynamics) and force fields (say, in electromag- netism). If F is a force field, then the line integral C F · dr calculates the work done while moving along the curve C from A to B. ! 2. Let D be an open subset of R3 and let f : D → R be a function with continuous first partial derivatives. Then ∇f is a continuous vector field on D, called the gradient field of f. The gradient fields are going to play a key role in this chapter. 3. Let C be a curve with smooth parametrization r :[a, b] → R3.ThenT(t)= r′(t)/|r′(t)| is a vector field on C.ThevectorT(t) gives the unit tangent vector to C at r(t). 1.5 (Gradient fields:). Line integrals on gradient fields are easy to compute. Assume F : D → R3 be a vector field as in 1.1 and let C be a curve lying in D with a smooth parametrization r :[a, b] → D as in 1.2. Suppose there exists a differentiable function f : D → R such that F = ∇f.In this case, we say that f is a potential function for the field F.Define α :[a, b] → R by α(t)=f(r(t)) = f(g(t),h(t),k(t)). Using the chain rule, we obtain ′ ′ ′ ′ α (t)=(∂1f)(r(t))g (t)+(∂2f)(r(t))h (t)+(∂3f)(r(t))k (t).
From the definition of the line integral C ∇f · dr,wefindthat b ! ′ ′ ′ ∇f · dr = [(∂1f)(r(t))g (t)+(∂2f)(r(t))h (t)+(∂3f)(r(t))k (t)]dt "C "a b ′ = α (t)dt "a 3
b ′ The fundamental theorem of integral calculus implies a α (t)dt = α(b) − α(a)= f(r(b))−f(r(a)) = f(B)−f(A). Thus we have arrived! at the fundamental theorem of line integrals:
∇f · dr = f(B) − f(A), where C is any smooth oriented curve from A to B. "C From this fundamental theorem, we observe that if a vector field F has a potential function f, then the line integral of F on a curve from A to B only depends on the end points A and B and not on the curve itself. This leads to the next definition: 1.6. Definition. A continuous vector field F on D is called conservative if given any two points A and B in D and smooth curves C1 and C2 from A to B lying in
D,onehas C1 F · dr = C2 F · dr. In this situation, we say that the line integral of F is path independent! . ! 1.7. Lemma. AvectorfieldF is conservative on a connected region D if and only if C F · dr =0for all closed curve C lying in D. ! sketch of proof. Suppose F is conservative. Let C be a closed curve in D starting and ending at A.LetC1 denote the constant curve with parametrization r1(t)=A for t ∈ [0, 1]. Since F is conservative, we have C F · dr = C1 F · dr. The integral ′ ! ! C1 F · dr is clearly zero since r1(t) is identically zero. ! Conversely, suppose C F · dr = 0 for any closed curve C.LetC1 and C2 be two curves with same beginning! and end point. Then (C1 − C2) is a closed curve obtained by following C1 then following C2 in reverse. So 0 = C1−C2 F · dr = ! ! C1 F · dr − C2 F · dr.Hence C1 F · dr = C2 F · dr. ! ! ! ! The fundamental theorem of line integrals imply that if F is a gradient field of some differentiable function f, that is, if F = ∇f, then it is conservative. The converse is also true under appropriate condition and this is the nexttheorem. Before stating it, we need a definition: A subset D of Rn is called path connected if any two points of D can be joined by a continuous path lying in D. More precisely, given any two points A and B in D, there exists a continuous function r : [0, 1] → D such that r(0) = A and r(1) = B. 1.8. Theorem. AcontinuousvectorfieldF defined on an open path connected re- gion D is conservative if and only if F is a gradient field of some potential function. sketch of proof. Suppose F is conservative. We need to define a potential function f : D → R satisfying ∇f = F. For this, fix a point A ∈ D. Given a point P ∈ D, we choose any smooth curve C lying in D starting at A and ending at P and define
f(P )= F · dr. "C Since F is conservative, the line integral does not depend on the choice of the path C, so the function f is well defined. One verifies that ∇f = F (see book). The proves one implication. As already noted, the other implication follows from the fundamental theorem of line integrals. ! 1.9. Remark. All the results above can be obviously modified for vector fields in R2 and for curves in R2, rather than in R3. The proofs are similar and simpler. 4
The fundamental theorem of line integral makes computation of line integrals of conservative fields trivial once we know a potential function. So it is interesting to answer the question: How to determine if a vector field is conservative?. We shall investigate this question next. Before that, this is a good place to introduce the differential operator ∇ and three associated operators grad, curl and div, called gradient, curl and divergence operators, that play key role in our theory. 1.10. Definition. Define ∇ = ⟨∂1,∂2,∂3⟩. Let D be an open subset of R3.Letf : D → R be a scalar valued function and let F : D → R3 be a vector field on D.Wedefine
grad(f)=∇f = ⟨∂1f,∂2f,∂3f⟩
curl(F)=∇×F = ⟨∂2P − ∂3N,∂3M − ∂1P,∂1N − ∂2M⟩
div(F)=∇·F = ∂1M + ∂2N + ∂3P. We get back to the question:How to determine if a vector field is conservative?. An easy necessary condition is given by the next lemma. 1.11. Lemma. Let F = ⟨M,N,P⟩ be a conservative vector field on a path connected open set D in R3.AssumethatM, N, P have continuous first partial derivatives on D.ThentheyM,N,P must satisfy the following conditions
(3) ∂2P = ∂3N,∂3M = ∂1P,∂1N = ∂2M. In other words if a vector field F as above is conservative, then curl(F)=0. sketch of proof. From theorem 1.8 we know that F = ∇f for some potential func- tion f : D → R.SoM = ∂1f, N = ∂2f and P = ∂3f. The given conditions imply that f has continuous second partial derivatives on D. So the mixed second partial derivatives must agree, that is ∂1∂2f = ∂2∂1f and so on. It follows that
∂1N = ∂1∂2f = ∂2∂1f = ∂1M. The other two equalities follow similarly. ! 1.12. Remark. All the results above can be obviously modified for vector fields in R2 and for curves in R2, rather than in R3. The proofs are similar and simpler. In particular, let F(x, y)=⟨M(x, y),N(x, y)⟩ is a vector field on an open path connected subset D in the plane. If F is conservative, then the variant of lemma 1.11 implies that we must have
∂1N = ∂2M. However, the next important example shows that this necessary condition is not sufficient for general open path connected regions D in the plane. 1.13. Example. Let D be the subset of R2 obtained by omitting the origin (0, 0). Let F : D → R2 be the vector field F(x, y)=⟨M(x, y),N(x, y)⟩ where y x M(x, y)=− and N(x, y)= . x2 + y2 x2 + y2
Then one verifies easily that the necessary condition ∂1N = ∂2M is satisfied. Howeer we shall see that F is not conservative by showing that the integral of F over a closed curve is nonzero. Let Cd be the circle of radius d in the plane 5
2 with anticlockwise orientation. We parametrize Cd by r : [0, 2π] → R given by r(θ)=d⟨cos θ, sin θ⟩.Thenonecompute d sin θ d cos θ 1 F(r(θ)) = ⟨− d2 , d2 ⟩ = d ⟨− sin θ, cos θ⟩. and ′ r (t)=d⟨− sin θ, cos θ⟩ ′ 1 2 2 So F(r(θ)) · r (θ)=d. d ((− sin θ) +(cosθ) )=1and 2π 2π ′ F · dr = F(r(θ)) · r (θ)dθ = 1dθ =2π. "Cd "0 "0 Note that the region D in this example D has a “hole” at the origin and the vector field F cannot be extended to the origin since the functions M,N do not have limits as (x, y) → (0, 0). Rougly speaking, if the connected region D in R2 does not have any holes, then it turns out that the condition partial1N = ∂2M is actually sufficient to ensure that a continuous vector field ⟨M,N⟩ on D is conservative. This would follow from the Green’s theorem. Before stating the theorem we need to formalize the notion of a region with no holes to “simply connected regions” and define the canonical orientation on the boundary of a region. 1.14. Definition. The proofs of some of the assertions made in this section are beyond the scope of this course. But thankfully the statements are rather intuitive and their validity can be easily illustrated by simple examples. A simple closed curve C in the R2 is a closed curve that does not cross itself. In other words, C is a continuous simple closed curve in R2 if it has a parametrization r : [0, 1] → R2 where r is a continuous function such that r(0) = r(1) and r(t) ̸= r(s) for all 0 ≤ s Mdx+ Ndy = (∂1N − ∂2M)dxdy. "∂D "D rough sketch of proof. We shall only roughly indicate why the Green’s theorem is true. For an alternative argument see book. Step 1: The first step is to prove Green’s theorem for a rectangle. This is easy: Let R be the inside of a rectangle in the plane given by a ≤ x ≤ b and c ≤ y ≤ d. Let r1, r2, r3, r4 be the parametrized curves: where r1(t)=⟨t, c⟩; a ≤ t ≤ b r2(t)=⟨b, t⟩; c ≤ t ≤ d r3(t)=⟨t, d⟩; a ≤ t ≤ b r4(t)=⟨a, t⟩; c ≤ t ≤ d Let C1, C2, C3, C4 be the oriented curves with parametrization r1, r2, r3, r4 respectively. Then ∂R = C1 + C2 − C3 − C4. By fundamental theorem of integral calculus one has b ∂N(x, y) x=b dx = N(x, y) = N(b, y) − N(a, y). " x=a a ∂x $ For simplicity of notation,write αMdx + Ndy$ . Using Fubini’s theorem we obtain d b ∂N d d ∂1Ndxdy = dxdy = N(b, y)dy − N(a, y)dy = α − α. "D "c "a ∂x "c "c "C2 "C4 Similarly, one verifies that − ∂2Mdydx = α − α. "D "C1 "C3 The Green’s theorem for rectangle follows from this. Step 2: An uniform grid in the plane with width c is the set of of horizonal and vertical lines of the form x = mc and y = nc where m, n vary over all integers. A grid divides the plane into squares of side length c. Consider a connected region in the plane which is a union of finitely many of these squares R1, ··· ,Rk.Then one verifies easily that Γ=∂R = ∂R1 + ∂R2 + ···+ ∂Rn. Roughly speaking the reason for this is that if S is a segment of a grid line that is a boundary of some Rj but not a boundary of R, then there are exactly two squares 7 among R1, ··· ,Rn, (one on either side of S) whose boundary segment is S.So when we calculate the boundary of the Rj ’s each such S comes up twice and with opposite orientation. So the contribution from the segments in the line integral cancels out. The rest of the segments in the boundary of Rj ’s exactly give us the boundary of R. The Green’s theorem for R follows from the Green’s theorem for rectangles. Step 3: This is the main analytic step and this is where we shall skip the details. Suppose we are given a piecewise smooth simple closed curve with interior S. Consider the uniform grid of width 1/n and Let Sn be the union of the squares of this grid that intersect S (alternatively we could only consider the squares that are contained in S completely). The main point is that by taking n large (i.e. making the grid fine enough) we can “approximate the region S by the regions Sn arbitrarily well” and the boundary of Sn approximates the boundary of S.Now the Green’s theorem for S follows from the Green’s theorem for Sn by a limiting process. ! As an immediate corollary we obtain the following characterization of conserva- tive vector fields in simply connected regions: 1.16. Corollary. Let D be a connected and simply connected open subset of R2 whiose boundary is a piecewise smooth simple closed curve. Let F = ⟨M,N⟩ be a vector field on an open region containing D ∪ ∂D.AssumethefirstpartialsofM and N are continuous on an open set containing D ∪ ∂D.TheF is conservative if and only if ∂1N = ∂2M. sketch of proof. It suffices to show that C Mdx+Ndy = 0 for any piecewise smooth simple closed curve C lying in D (why?).! Let C be such a curve and let R be its inside. Since D is simply connected, it follows that R is entirely contained in D as well. So by Green’s theorem applied to R,wefindthat C Mdx + Ndy = ! R(∂1N − ∂2M)dxdy =0. ! ! 2. Calculus of differential forms We develop the language of differential forms a little bit because they help us pharse many results in this chapter in easy to memorize form and showthatmany of the results here are different incarnation of the same result called the Stoke’s theorem. We shall not define differential forms precisely and only prescribe a few rules for playing with them. Let D be an open subset of R3. A differential 0-form on D is simply a real valued function. A differential 1-form is a formal expression of the form Mdx+Ndy+Pdz where M,N,P are real valued functions on D. A differential 2-form is a formal expression of the form Mdy∧ dz + Ndz∧ dz + Pdx∧ dy. A differential 3-form on D is a formal expression of the form fdx∧ dy ∧ dz where f is a scalar valued function on D. For our purpose, it will be enough to think of dx, dy, dz as formal variable symbols. The symbol ∧ in dx ∧ dy defines a sort of product on differential forms, called the wedge product. Below we lay out some of the rules for working with the differential forms. The product of an m-form and an n-form is a (m + n)form. If ω,µ,ν are differential forms, then the wedge product satisfies the usual rules of 8 multiplication like ω ∧ (µ ∧ ν)=(ω ∧ µ) ∧ ν (ω + µ) ∧ ν = ω ∧ ν + µ ∧ ν The main non-obvious rule is that if ω is a m-form and µ is an n-form, then ω ∧ µ =(−1)mnµ ∧ ω. In particular, if ω and µ are 1-forms, then ω ∧ µ = −µ ∧ ω. For example, dx ∧ dy = −dy ∧ dx and so on. and if f is a 0-form and ν is any form then f ∧ ν = ν ∧ f. One writes f ∧ ν = fν. Finally, there is an operation d, called exterior derivative, that takes a k-form to a (k + 1) form. This operation is defined by the rules: If f is a 0-form, then df =(∂1f)dx +(∂2f)dy +(∂3f)dz. Next, if ω is an m-form and µ is an n-form, then d(ω ∧ µ)=(dω) ∧ µ +(−1)mω ∧ (dµ) Finally d(dω)=0. The above story of differential forms generalizes from R3 to Rn in obvious manner. We shall end this section with rephrasing two results we have seen in terms of differential forms. • If M = M(x, y)andN = N(x, y), and ω = Mdx + Ndy is an 1-form on the plane, then dω =(∂1N − ∂2M)dx ∧ dy. So the equation in the Green’s therorem reads ω = dω. "∂D ""D • Let f : D → R be a function on an open region D in R3 and assume f has continuous first partial derivatives. Let C be a piecewise smooth oriented curve in D.Thennotethat ∇f · dr = df . "C "C So the fundamental theorem of line integral reads f(B) − f(A)= df "C where A is the starting point of C and B is the endpoint. It is tempting to think of the left hand side as an “integral of f over the two point set ∂C = {A, B},whereB comes with positive orientation and A comes with negative orientation”. Then the fundamental theorem of line integral reads f = df . "∂C "C 9 3. Surface integrals and Stoke’s theorem 3.1. Definition. Let R be a open connected region in R2.LetS be a surface in R3 with parametrization r : R → R3. We write r(u, v)=⟨f(u, v),g(u, v),h(u, v)⟩. Assume that f,g,h have continuous first partial derivatives. Given a point (u0,v0) in R the vectors ru(u0,v0)andrv(u0,v0) are tangent along the curve at r(u0,v0). We shall say that r is a smooth parametrization of S if ru and rv are continuous and ru ×rv is nonzero at all points of R. A smooth surface is a surface that has a smooth parametrization. If r is a smooth parametrized surface, then (ru × rv)(u0,v0) is a nonzero normal vector to the surface S at the point r(u0,v0)andthenormalvector ru × rv varies continuously as we move along the surface. An orientation on the smooth surface S is a choice of continuously varying nonzero normal vector at every point of the surface. Let S be a surface in R3 with smooth parametrization r : R → R3 as above. Consider a little square T in the domain R where the first coordinate varies from u to u +∆u and the second coordinate varies from v to v +∆v.Thenonecanshow that a first order approximation of the image r(T ) is given by the parallelogram spanned by the two vectors ru∆u and rv∆v starting at r(u0,v0). The area of this parallelogram approximates the area of the little piece r(T ) of the surface and this area is equal to the norm of the vector ∆⃗σ =(ru × rv)∆u∆v. This leads us to define the differential 2-form d⃗σ =(ru × rv)du ∧ dv which is called the area element of the parametrized surface S. We also write dσ = |d⃗σ| = |ru × rv|du ∧ dv. Let q : S → R be a continuous function. Then we define the surface integral of a on S to be qdσ = q(r(u, v))|ru × rv|dudv. ""S ""R If q(r(u, v)) is the density of a thin sheet shaped like S,thentheabovesurface integral calculates the mass of the sheet. In particular S 1dσ is the surface area of S. !! Let F = ⟨M,N,P⟩ be a continuous vector field on an open set containing S. Then we define the integral of F across S to be F · d⃗σ = F(r(u, v)) · (ru(u, v) × rv(u, v))dudv. ""S ""R If F is the velocity field of a fluid flow, then the surface integral S F·d⃗σ calculates the flux of F across S, that is the volume of liquid flowing per unit!! time across the surface S. It is instructive to compare the above definitions of surface integrals with the definition of line integrals of scalar functions and vector fields. We also remark that the above definitions of surface integrals extend naturally to piecewise smooth parametrized surfaces. Introduce variables x, y, z that give the coordinates 10 on the surface S, that is, x = f(u, v), y = g(u, v), z = h(u, v). One can verify by direct computation the formal equality d⃗σ = ⟨dy ∧ dz, dz ∧ dx, dx ∧ dy⟩. It follows that F · d⃗σ = α ""S ""S where α is the differential 2 form α = Mdy ∧ dz + Ndz ∧ dx + Pdx∧ dy. Let r : R → R3 be a smooth parametrization of an orientable surface S.Assume that R is an closed region in R2 whose boundary ∂R is a finite union of piecewise smooth curves. Then r(∂R) is the called the boundary of the surface S and is denoted by ∂S. The boundary of such a parametrized S comes with a natural orientation defined as follows. Recall that r defines a natural orientation on S given by the normal vectors ru × rv. The orientation on ∂S is defined so that if we walk along ∂S with out head towards the normal ru × rv then the surface always stays to our left. Now we are ready to state the Stoke’s theorem for surfaces that relates the line integral of a vector field along the oriented boundary of.an oriented smooth surface with the surface integral of the curl of the vector field on the surface: 3.2. Theorem. Let S be a piecewise smooth oriented surface whose boundary ∂S is a finite union of piecewise smooth curves. Let F be a vector field on an open set containing S whose components have continuous first partial derivatives.Then F · dr = (∇×F) · d⃗σ. "∂S ""S It is easy to see that the Green’s theorem in the plane is a special case of Stoke’s theorem when the surface S is the planar region D. Write F = ⟨M,N,P⟩ as before. Then we have seen that the line integral on the left hand side of Stoke’s theorem can be expressed in the form ω "∂S where ω = Mdx+ Ndy + Pdz. One verifies that the surface integral in Stoke’s theorem is equal to S α where !! α =(∂2P − ∂3N)dy ∧ dz +(∂3M − ∂1P )dz ∧ dx +(∂1N − ∂2M)dx ∧ dy. By a direct calculation, one verifies that dω = α. So the equation in Stoke’s theorem reads ω = dω. "∂S ""S