Chapter 5 Line Integrals

Chapter 5

A basic problem in higher dimensions is the following. Suppose we are given a function (scalar field) g on Rn and a bounded curve C in Rn, can one define a suitable integral of g over C? We can try the following at first. Since C is bounded, we can enclose it in a closed rectangular box R and then define a functiong ˜ on R to be g on C and 0 outside. Clearly, R R R g˜ = C g, but for n > 1, this will give only zero because C will have content zero. So we have to do something different. Before giving the “right” definition, let us review the basic properties of a curve C in Rn, parametrized by an interval in R. Recall that a (parametrized) curve C in Rn is the image of a continuous map α :[a, b] → Rn, for some a < b in R. Such a curve is said to be C1 if and only if α0(t) exists and is continuous everywhere on the interval. We like this condition because it implies that the curve C has a well defined unit tangent vector T at every point, and moreover, that this T moves continuously as we move along the curve. In practice we will be able to allow a finite number of points where the curve develops corners. This leads to the definition of a 1 1 piecewise C curve C to be a curve which is a finite union of C curves C1,C2,...,Cr.

The arc length of a C1 curve C parametrized by α :[a, b] → Rn is deﬁned at any point t ∈ [a, b] to be Z t s(t) = kα0(t)k dt. a s(b) is called the arc length of the whole curve. We will now describe the heuristic reason for this deﬁnition.

Let Q be a partition of [a, b], i.e., the datum a = t0 < t1 < ··· < tm = b. Let Li denote

1 n the line joining α(ti−1) to α(ti) in R , ∀ i ≥ 1. Put

m X `m = length(Li), i=1 Pm which equals i=1 kα(ti) − α(ti−1)k.

Clearly, as m is very large, `m provides a good approximation for s(b), and a proper deﬁnition must surely express s(b) as the limit of `m as m tends to inﬁnity.

For each i, we may apply the mean value theorem to α(t) restricted to [ti−1, ti] and ﬁnd a point zi in that subinterval such that

0 α(ti) − α(ti−1) = (ti − ti−1)α (zi).

This yields the equality m X 0 `m = (ti − ti−1)kα (zi)k. i=1 Since by assumption α0 is continuous, so is kα0k. Hence kα0k is integrable and we may let m go to inﬁnity and get Z b 0 lim `m = kα (t)k dt. m→∞ a

Examples: (1) C = circle in R2 with radius r and center 0. It is parametrized by α(t) = (r cos t, r sin t), 0 ≤ t ≤ 2π. Since sin t and cos t are continuously diﬀerentiable, C is C1. Moreover, α0(t) = (−r sin t, r cos t), and so kα0(t)k = rp(− sin t)2 + (cos t)2 = r. Hence the R 2π arc length of C is 0 r dt = 2π, as expected. (It is always good to do such a simple example at ﬁrst, because it tells us that the theory is seemingly on the right track.) (2) C = the ellipse with minor and major semi-axes a and b, respectively. This curve is parametrized by α(t) = (a cos t, b sin t), 0 ≤ t ≤ 2π, and the arc length of C is

Z 2π p Z 2π p a2 sin2 t + b2 cos2 t dt = b 1 − k2 sin2 t dt. 0 0

q a2 This is a so called elliptic integral. As a function of k = 1 − b2 it cannot be expressed in terms of elementary functions.

3 (3) (Helix) Let C in R be parametrized by α(t) = (cos t, sin t, t), 0 ≤ t ≤ 4π. Again, α√is (certainly) C1, and α0(t) = (− sin t, cos t, 1). So kα0(t)k = p(− sin t)2 + (cos t)2 + 1 = 2, R 4π √ √ and the arc length of C is 0 2 dt = 4π 2.

2 2 3 1 (4) Let C be defined by α(t) = (2t, t , log t) ∈ R for t ∈ [ 4 , 2000]. Suppose we need to find the arc length ` between the points (2, 1, 0) and (4, 4, log 2). Note first that (2, 1, 0) = α(1) and (4, 4, log 2) = α(2). So we have:

Z 2 ` = kα0(t)k dt. 1 q 0 1 0 2 1 1 Also, α (t) = (2, 2t, t ), so kα (t)k = 4 + 4t + t2 = (2t + t ). (Note that since t > 0, we do not need to take the absolute value while taking the square root.) Hence

Z 2 Z 2 1 2 2 ` = 2t dt + dt = t2 + log t = 3 + log 2. 1 1 t 1 1

(5) This example deals with a piecewise C1 curve. Note that the deﬁnition of arc length goes over in the obvious way for such curves.

Take C to be the unit (upper) semicircle C1 centered at 0, together with the ﬂat diameter C2, then we can parametrize C1 in the usual way by setting α1(t) = (cos t, sin t), for t in [0, π]. Let us parametrize C2 by a function α2 on [π, 2π] as follows. Write α2(t) = (dt + e, 0) with the requirement α2(π) = (−1, 0) and α2(2π) = (1, 0). Then we need dπ + e = −1 and 2 2 2dπ + e = 1, which yields d = π and e = −3. So we have: α2(t) = ( π t − 3, 0) for t ∈ [π, 2π]. 1 1 Clearly, α1 and α2 are both C , so C is piecewise C . Suppose we want to ﬁnd the arc length `(C) of the whole curve C. It is clear from geometry that `(C1) = π and `(C2) = 2, so `(C) = π + 2. We can also derive this from 0 0 2 2 Calculus. Indeed, kα1(t)k = 1 and kα2(t)k = k π k = π . So Z π Z 2π 2 `(C) = 1 dt + dt = π + 2. 0 π π

Now we are ready to give a way to integrate functions (scalar fields) over curves in n-space. Let g : D → R be a scalar field, with C ⊆ D ⊆ Rn, where C is a C1 curve parametrized by α :[a, b] → Rn. Then we define the line integral of g over C with respect to arc length to be Z Z b g ds = g(α(t))kα0(t)k dt. C a Note that by the definition of s(t), s0(t) is none other than kα0(t)k; so kα0(t)k dt may be thought of as ds.

3 There is yet another type of line integral. Suppose f : D → Rn is a vector ﬁeld with D ⊆ Rn. (Note that the source space and the target space of f have the same dimension.) Then we may deﬁne the line integral of f over C to be

Z Z b f · dα = f(α(t)) · α0(t) dt. C a These two integrals are related as follows. Let T denote the unit tangent vector to C at α(t). α0(t) 0 0 0 0 Then T = kα0(t)k (assuming α (t) 6= 0). Since kα (t)k = s (t), we can identify T with α (s) dα (= ds ). Put g(α(t)) = f(α(t)) · T , which deﬁnes a scalar ﬁeld. Then we have Z Z Z f · dα = g(α(t))kα0(t)k dt = g ds. C C C

The following lemma is immediate from the deﬁnition. R R R Lemma 1. (i) C f1 · dα + C f2 · dα = C (f1 + f2) · dα R R R (ii) f · dα1 + f · dα2 = f · dα, where C is the union of C1 and C2, parametrized C1 C2 C by α(t) deﬁned by putting together α1 and α2.

It is important to note that each parametrized curve C comes with an orientation (or direction). As we traverse the interval [a, b] in the positive direction, α(t) traverses the curve C in a certain direction (which we cannot see by just looking at the image set C = α([a, b])). This was not important when we calculated the arc length, but it is quite important to be aware of when line integrals of vector ﬁelds are involved. Suppose C is parametrized in two diﬀerent ways, say by α :[a, b] → Rn and β :[c, d] → Rn. We say that α is equivalent to β, and write α ∼ β, if there exists a change of parameter C1 function u :[a, b] → [c, d] such that

(i) u is bijective, i.e. one-to-one and onto,

(ii) u0(t) > 0 for all t ∈ [a, b], and

(iii) β(u(t)) = α(t), for any t ∈ [a, b].

It is not hard to see using the chain rule that Z Z f · dβ = f · dα, C C

4 so our integral is independent of the parametrization. The condition (ii) assures that the orientation of the parametrized curve is not changed. If u is a reparametrization with u0(t) < 0 for all t then we have Z Z f · dβ = − f · dα. C C For example, this is the case when α(t) = β(−t + a + b), a ≤ t ≤ b. One can view the arc length function s(t) as a particular reparametrization mapping [a, b] onto [0, length(C)]. This gives for any parametrized curve C a canonical parametrization where the parameter of a point x = α(t) ∈ C is just the arc length along C from α(a) to x = α(t).

5.1 An application

Line integrals arise all over the place. We will content ourselves with describing the example of a thin wire W in the shape of a C1 curve C parametrized by a 1-1 path α :[a, b] → Rn. Suppose the mass density of W is given by a function g(x), x = (x1, . . . , xn). Then the total mass of W is given by Z M = g ds. C

Its center of mass x¯ = (¯x1, x¯2,..., x¯n) is given by 1 Z x¯j = xjg ds, 1 ≤ j ≤ n. M C When g is a constant field, W is said to be uniform andx ¯ is called the centroid. Suppose L is any fixed line in Rn. Denote by δ(x) the distance from x to L. Then the moment of inertia about L is defined to be Z 2 IL = δ g ds. C

Example. Consider the wire W in the shape of C = C1 ∪ C2, with C1 being parametrized by α1(t) = (cos t, sin t), 0 ≤ t ≤ π, and C2 by α2(t) = (t, 0), −1 ≤ t ≤ 1. Suppose W has p 2 2 mass density given by g(x, y) = x + y . Then g(α1(t)) = 1 and g(α2(t)) = |t|. Also, 0 0 kα1(t)k = 1 and kα2(t)k = 1. Thus the total mass is given by Z π Z 0 Z 1 M = 1 dt − t dt + t dt = π + 1. 0 −1 0

5 The coordinates of the center of mass are given by

1 Z π Z 0 Z 1 x¯ = cos t dt − t2 dt + t2 dt = 0, π + 1 0 −1 0 and 1 Z π 2 y¯ = sin t dt + 0 = . π + 1 0 π + 1 The distance from (x, y) to the x-axis (resp. y-axis) is simply |y| (resp. |x|). Hence the moment of inertia about the x-axis is Z π Z π 2 1 − cos 2t π Ix = sin t dt + 0 = dt = . 0 0 2 2 The moment of inertia about the y-axis is

Z π Z 0 Z 1 2 3 3 π 1 Iy = cos t dt − t dt + t dt = + . 0 −1 0 2 2

5.2 Gradient ﬁelds

In one-variable calculus one has the following basic relationship between the integral and the derivative. If g(t) is a continuously differentiable function on some open interval containing a and b, then Z b g0(t)dt = g(b) − g(a). a This is sometimes called the second fundamental theorem of calculus. Here we will be concerned with a generalization of this fact to line integrals. Given a differentiable scalar field g on an open set D in Rn, we may of course consider its gradient field ∇g. It turns out, if this gradient is a continuous function of x ∈ D, the line integrals R C ∇g · dα can be evaluated simply. Theorem 1. (Second fundamental theorem of calculus for line integrals) Let g be a differentiable scalar field with continuous gradient ∇g on an open set D in Rn. Then, for any two points P,Q ∈ D joined by a piecewise C1 path C completely lying in D and parametrized by α :[a, b] → D with α(a) = P and α(b) = Q, we have Z ∇g · dα = g(Q) − g(P ). C

6 Remark. The stunning thing about this result is that the integral depends only on the end points P and Q, and not on the path C connecting them. This is pretty revolutionary, if you think about it!

Corollary 1. Let g, D be as in the Theorem. Then, for any point P in D, and any piecewise C1 path C connecting P to itself, i.e., with α(a) = P = α(b), we have Z ∇g · dα = 0. C

Such a path C whose beginning and end points are the same is called a closed path, and H the line integral of a vector field f over a closed path C is usually denoted C f · dα. Proof of Theorem. We will prove it for C a C1 curve, and leave the piecewise C1 extension, which is routine, to the reader. By definition, Z Z b ∇g · dα = ∇g(α(t)) · α0(t) dt. C a Put h(t) = g(α(t)), for all t ∈ [a, b]. Then by the chain rule, which we can apply since both α and g are differentiable, we have h0(t) = ∇g(α(t)) · α0(t). So Z Z b ∇g · dα = h0(t) dt. C a Note that h0(t) is continuous on (a, b) by the continuity hypothesis on ∇g. For each u ∈ R u 0 [a, b], set G(u) = a h (t) dt. Then by one variable calculus, G is continuous on [a, b] and differentiable on (a, b) with G0(u) = h0(u) there. So G − h is constant on (a, b) and hence also on [a, b] by continuity. To find this constant we set u = a and use G(a) = 0 to conclude G(u) − h(u) = −h(a). Consequently we have Z ∇g · dα = G(b) = h(b) − h(a) = g(α(b)) − g(α(a)) C as needed.

5.3 Criterion for path independence

Let D be an open set in Rn. The natural question raised by the result of the previous section is to what extent we can characterize vector ﬁelds on f on D whose line integrals are path

7 independent, i.e., depend only on the end points. In this section we give the (extremely satisfying) answer, sometimes called the first fundamental theorem of calculus for line integrals. Recall that in one-variable calculus, given any continuous function f(t) and defining Z t F (t) = f(s)ds, a the first fundamental theorem says that F 0(t) = f(t). Before stating and proving the result, we need to introduce an important concept called connectivity. We need to discuss when two points x, y in our open set D can be joined by a path α :[a, b] → D with α(a) = x and α(b) = y. Even for n = 1 this is only always possible when D is an open interval but not in general. Given an open set D ⊆ Rn we say x, y ∈ D lie in the same path component of D if there is a (piecewise C1) path α :[a, b] → D with α(a) = x and α(b) = y.

Lemma 2. Let D ⊆ Rn be an open set. ”Lying in the same path component” is an equivalence relation on D. The equivalence classes, which are called the path components of D, are themselves open sets.

Proof. We need to check the axioms for an equivalence relation. Clearly x is joined to x by the constant path α : [0, 1] → D with value x (this path is C1 with derivative α0(t) ≡ 0). Reversing the orientation of a path (as in the example at the end of section 5.1) gives us symmetry. Transitivity follows by joining two paths (recall our paths only need to be 1 piecewise C ). In any open ball Ba(r) we have the straight line from the center to any other point in Ba(r). So if y lies in the path component of x, any open ball around y which lies in D then also lies in the path component of x. So this path component is open.

In general D might have several (even inﬁnitely many) path components. We say D is (path) connected if it has only one path component. For any two points x, y ∈ D there is then a path in D joining them. Any open set can be written in a unique way as the disjoint union of connected open sets. n n Examples. Any open ball Ba(r), any open box (a, b), R , R \{0} for n > 1 are connected. The set R2 \ x-axis is not connected. It has two path components.

Theorem 2. (First fundamental theorem of calculus for line integrals) Let D be an open set in Rn, and let f : D → Rn be a continuous vector field. Suppose the line integral of f over any piecewise C1 path C in D depends only on the end points of C. Then there exists a differentiable scalar field ϕ on D, called a potential function of f, such that f = ∇ϕ on D. Moreover, the potential function ϕ can be explicitly defined as follows. In each path

8 component E of D ﬁx any point P ∈ E and set Z x ϕ(x) = f · dα, P where the integral is taken over any piecewise C1 path C in D connecting P to x. Then ϕ is well deﬁned and represents a potential function for f.

Putting the ﬁrst and second fundamental theorems for line integrals together, we easily obtain the following useful

Corollary 2. Let D be an open set in Rn, and let f : D → Rn be a continuous vector ﬁeld. Then the following properties are equivalent:

(i) f = ∇ϕ, for some potential function ϕ

(ii) The line integrals of f over piecewise C1 curves in D are path independent.

(iii) The line integrals of f over closed, piecewise C1 curves in D are zero.

Definition. A vector field f satisfying any of the (equivalent) properties of Corollary is said to be conservative. In the textbook the theorem and its corollary are stated only for connected open sets. Clearly the two versions are equivalent by treating each path component separately. Proof of Theorem. Let ϕ be defined as in Theorem. That it is well defined is a consequence of the path independence hypothesis for line integrals of f in D. Write f = (f1, f2, . . . , fn), where fj is, for each j ≤ n, the j-th component (scalar) field of f. We have to show: (∀j) ∂ϕ exists and equals fj. (∗) ∂xj

For each x in D, ∃ r > 0 such that the vector x + hej lies in D whenever |h| ∈ (0, r). We may write Z x+hej ϕ(x + hej) − ϕ(x) = f · dα, x where the integral is taken over the line segment C connecting x to x + hej, parametrized 0 by α(t) = x + thej, for t ∈ [0, 1]. Since α (t) = hej, we get

Z 1 ϕ(x + hej) − ϕ(x) = f(x + thej) · ej dt, h 0

9 which equals

Z 1 1 Z h fj(x + thej) dt = fj(x + uej) du 0 h 0 1 = (g(h) − g(0)), h R t where g(t) = 0 fj(x + uej) du, ∀ t ∈ (−r, r). Letting h go to zero, we then get the limit 0 ∂ϕ 0 0 g (0). Thus (x) exists and equals g (0). But by construction, g (0) = fj(x), and so we are ∂xj done.

5.4 A necessary condition to be conservative

Not all vector ﬁelds f are conservative. It is also quite hard (except in special cases) to check that f is conservative. Luckily, we at least have a test which can be used in many cases to rule out some f from being conservative.

1 n ∂fi Theorem 3. Let f = (f1, . . . , fn) be a C vector ﬁeld on an open set D in , i.e. exist R ∂xj and are continuous for all 1 ≤ i, j ≤ n. Suppose f is a conservative ﬁeld. Then we must have (everywhere on D) ∂f ∂f i = j , for all i, j. ∂xj ∂xi

∂ϕ Proof. Suppose f = ∇ϕ, for a potential function ϕ on D, so that fj = , for all j ≤ n. ∂xj Since f is diﬀerentiable, all the partial derivatives of each fj exist, and we get (∀ i, j) ∂f ∂ ∂ϕ ∂2ϕ j = = , ∂xi ∂xi ∂xj ∂xi∂xj and ∂f ∂ ∂ϕ ∂2ϕ i = = . ∂xj ∂xj ∂xi ∂xj∂xi 1 Since f is C , each partial derivative of fi is continuous for every i. Then the mixed partial 2 2 derivatives ∂ ϕ and ∂ ϕ are also continuous, and must be equal by an earlier theorem. ∂xi∂xj ∂xj ∂xi

Examples. (1) Let f : R3 → R3 be the vector ﬁeld given by f(x, y, z) = (y, x, 0). Denote by C the oriented curve parametrized by α(t) = (tet, cos t, sin t), for t ∈ [0, π]. Compute the R line integral I = C f · dα.

10 By deﬁnition this line integral I is Z π Z π f(α(t)) · α0(t) dt = (cos t, tet, 0) · ((1 + t)et, − sin t, cos t) dt 0 0 Z π = [(1 + t) cos t et − t sin t et] dt. 0

This integral can be calculated, but with some trouble. It would have been better had we ∂ϕ first checked to see if f could be conservative, i.e., if there is a function ϕ such that y = ∂x , ∂ϕ ∂ϕ x = ∂y and 0 = ∂z . The last equation says that ϕ is independent of z, and the first two can be satisfied by taking ϕ(x, y, z) = xy, for all (x, y, z) ∈ R3. Thus f = ∇ϕ everywhere on R3, and so by the second fundamental theorem for line integrals, I = ϕ(α(π)) − ϕ(α(0)) = ϕ(πeπ, −1, 0) − ϕ(0, 1, 0) = −πeπ. (2) (Physics example) Consider the force field f(x, y, z) = (x, y, z) (=x¯i + y¯j + zk¯). Find the work done in moving a particle along the parabola C = {(x, y, z) ∈ R3 | y = x2, z = 0} from x = −1 to x = 2. We can parametrize C by the C1 function α(t) = (t, t2, 0). We are interested in those t lying in the interval [−1, 2]. The work done is given by

Z Z 2 W = f · dα = f(α(t)) · α0(t) dt C −1 Z 2 Z 2 = (t, t2, 0) · (1, 2t, 0) dt = (t + 2t3) dt −1 −1 t2 t4 2 = + = 9. 2 2 −1 Alternatively, as in example (1), we could have looked for a potential function ϕ satisfying ∂ϕ ∂ϕ ∂ϕ x2+y2+z2 ∂x = x, ∂y = y and ∂z = z. An immediate solution is given by ϕ(x, y, z) = 2 . So, by the second fundamental theorem for line integrals,

W = ϕ(α(2)) − ϕ(α(−1)) = ϕ(2, 4, 0) − ϕ(−1, 1, 0) 22 + 42 (−1)2 + 12 = − = 9. 2 2

(3) Let f : R3 → R3 be the vector ﬁeld sending (x, y, z) to (x2, xy, 1). Determine its line integral over the curve C parametrized by α(t) = (1, t, et), for t ∈ [0, 1]. First let us see if f could be conservative. Since f is diﬀerentiable with continuous partial derivatives, we can apply the necessary criterion proved in the previous section. Then, for f

11 ∂fi ∂fj 2 to be conservative, we would need = , for all i, j. Here f1(x, y, z) = x , f2(x, y, z) = xy ∂xj ∂xi ∂f1 and f3(x, y, z) = 1. (As usual, we are writing (x, y, z) for (x1, x2, x3).) But ∂y = 0, while ∂f2 ∂x = y. So the criterion fails, and f cannot be conservative. We could have also seen this ∂ϕ 2 directly by trying to find a potential function ϕ(x, y, z). We would have needed ∂x = x , ∂ϕ ∂ϕ x3 ∂y = xy and ∂z = 1. The first equation gives (by integrating): ϕ(x, y, z) = 3 + ψ(y, z), ∂ψ ∂ϕ with ψ independent of x. Then the second equation forces ∂y , which is ∂y , to equal xy, implying that ψ is not independent of x. So there is no potential function and f is not conservative. In any case, we can certainly compute the line integral from first principles:

Z Z 1 f · dα = f(1, t, et) · (0, 1, et) dt C 0 Z 1 Z 1 = (1, t, 1) · (0, 1, et) dt = (t + et) dt 0 0 t2 1 1 1 = + e2 = + e − 1 = e − . 2 0 2 2

The question arises whether the condition of Theorem 3 is also sufficient to ensure that f is conservative. It turns out that it is under further assumptions on D. Any path component of D should be not only connected but what is called ”simply connected”, a notion that will be discussed in the next chapter for n = 2. Intuitively it means that such a region has ”no holes” in the sense that any closed curve in D can be contracted inside D to a point. An example of a connected but not simply connected set is D = R2 \{0}. On this D we have the following Examples. Let x y f(x, y) = , . x2 + y2 x2 + y2 Then ∂f 2xy ∂f 1 = − = 2 ∂y (x2 + y2)2 ∂x and the criterion of Theorem 3 is satisfied. Indeed, the vector field f is conservative and has the potential function p φ(x, y) = log( x2 + y2). Now let −y x g(x, y) = , . x2 + y2 x2 + y2

12 Then ∂g y2 − x2 ∂g 1 = = 2 ∂y (x2 + y2)2 ∂x and the criterion of Theorem 3 is again satisﬁed. However, if C is the closed path parametrized by α(t) = (cos(t), sin(t)), 0 ≤ t ≤ 2π we have

Z Z 2π Z 2π g · dα = (− sin(t), cos(t)) · (− sin(t), cos(t))dt = 1 dt = 2π. C 0 0 So the vector ﬁeld g is not conservative. One might think that the function

φ(x, y) = arctan(y/x)

is a potential function since we have ∂φ 1 y −y = − = ∂x 1 + (y/x)2 x2 x2 + y2

and ∂φ 1 1 x = = . ∂y 1 + (y/x)2 x x2 + y2 The problem is that φ isn’t deﬁned at any point on the y-axis, i.e. where x = 0. Noting that π 2 arctan(z) approaches 2 as z tends to ∞, we can extend φ to the region R \{(0, y)|y ≤ 0} as π a diﬀerentiable function by setting φ(0, y) = 2 for y > 0. But we cannot extend φ to all of D = R2 \{0} even as a continuous function since φ(x, y) is of course just the angle spanned by the line through (x, y) and (0, 0) and the x-axis which picks up the summand 2π as (x, y) moves around the unit circle.