= x i+ y j+ k, (1) k. (2) y. curlf = F. (3)

Lecture 14 url of a vector field Let us now take the vector product of the 3D del operator, = x i+ y j+ k, (1) z with a vector field F in R 3 : F = = = ( x i+ y i+ ) z k (F 1 i+f 2 j+f 3 k) i j k x y z F 1 F 2 F 3 ( F3 y F ) ( 2 F1 i+ z z F ) ( 3 F2 j+ x x F ) 1 k. (2) y This is known as the curl of the vector field F. In some books, it is also written as curl F, i.e., curlf = F. (3) It is a vector quantity, as should be the case when the vector product of two vectors is taken. Note that we can also define the curl of vector fields in R 2 by setting F 3 = 0. A curl for two-dimensional vector fields may also be defined according to the above formula. In this case, the vector fields will assume the form, F(x,y) = F 1 (x,y)i+f 2 (x,y)j. (4) We may still use Eq. (2) to compute the curl of F. Since F 3 = 0 and F 1 and F 2 have no z-dependence, the i and j components of F in Eq. (2) vanish to give ( F2 F = x F ) 1 k. (5) y Note that in this special case of planar vector fields, the vector F points in the z-direction (either plus or minus). This will be important in our further discussion of Green s Theorem in the plane. 1

Finally, the curl of a vector field F : R n R n can be defined for higher dimensions, i.e., n > 3, using concepts from differential geometry. (It would be the differential of a 1-form.) We shall also be discussing the physical significance of the curl of a vector field very shortly. Examples: 1. = 0 for any constant vector (x,y,z) = 1 i+ 2 bfj + 3 k. 2. If F = r = xi+yj+zk, then F 1 = x, F 2 = y and F 3 = z so that r = = x y x y ( z y y z i j k 3. Now consider F = yi+xj+0k. Then F = = i j k x y z y x 0 ( 0 y x z ) i+ z z ) i+ ( ( y) ( x z z ) ( y j+ x x x ) k = 0. y z 0 ) ( x j+ x x ( y) ) k = 2k. y We have already encountered this vector field. A view from the positive z-axis looking down onto the xy-plane is shown below. A top view of the vector field F = yi+xj+0k 2

In the xy-plane, it can represent the velocity field of a rotating disk in this case, angular frequency ω = 1. The above vector field could represent the velocity field of a rotating cylinder. Note that curl F = 2k at all points in the plane, not just at the origin, where the axis of rotation is situated. This implies that the vector field is rotational at all points in the plane. We ll return to this idea later in the course. 4. Once again recall the importance of the class of vector fields F(r) = K r3r in physics. When K = GMm, we have the gravitational force exerted on a mass m at r by a point mass M at the origin of a coordinate system. Whe K = Qq/(4πǫ 0 ), we have the electrostatic force exerted on a charge q at r due to a point mass Q at the origin. For convenience, we shall omit the multiplicative factor K. Let us first express this field in terms of artesian coordinates: F(r) = 1 r 3r = 1 (x 2 +y 2 +z 2 ) 3/2[xi+yj+zk], (6) F = i j k x y z x y z (x 2 +y 2 +z 2 ) 3/2 (x 2 +y 2 +z 2 ) 3/2 (x 2 +y 2 +z 2 ) 3/2 We ll leave this as an exercise for the reader! (It s actually not that tedious.) The net result is (7) Gradient fields/conservative forces and the curl Let us now return to the definition of the curl of a vector field F F = = = 1 r3r = 0, (x,y,z) (0,0,0). (8) ( x i+ y i+ ) z k (F 1 i+f 2 j+f 3 k) i j k x y z F 1 F 2 F 3 ( F3 y F ) ( 2 F1 i+ z z F ) ( 3 F2 j+ x x F ) 1 k. (9) y Let us also review the conditions that had to be satisfied for a 1 vector field F to be a gradient or conservative vector field, i.e., that F = f or F = V: : F 1 y = F 2 x, F 1 z = F 3 x, F 2 z = F 3 y. (10) 3

If these three relations are satisfied, then the three entries of the vector F are zero. In other words, F = 0 implies that F is gradient/conservative. In the case of a two-dimensional vector field F : R 2 R 2, only the first of the above conditions needs to be satisfied. The relations in Eq. (10) may also be expressed in the following way. For any scalar field f : R 3 R, ( f) = 0. (11) In other words, the curl of the gradient field F = f is zero: Gradient or conservative fields are irrotational. You will be asked to prove this result in an assignment. Here is another important result that is left as an exercise: div ( curl F) = ( F) = 0. (12) Revisiting Green s Theorem in the plane with the curl Recall Green s Theorem for line integrals of vector fields in the plane: Green s Theorem: Let D be a bounded subset of R 2 with boundary = D that is a piecewise 1 simple closed curve oriented counter-clockwise. If F(x,y) = (F 1 (x,y),f 2 (x,y)) is of class 1 on D D, then F dx = D ( F2 x F ) 1 da. (13) y From Eq. (9, we see that the integrand of the 2D integration on the right is the ˆk- or z-component of the curl of F, which we may write as ( F) z. As such, we can rewrite Eq. (13) as F dx = ( F) z da or (curl F) z da. (14) D This particular way of expressing Green s Theorem will be important in our future discussion of Stokes Theorem, which is the 3D version of Green s Theorem. Equivalently, Green s Theorem is the 2D version of Stokes Theorem. The fact that the integrand on the RHS of Eq. (14) is the ˆk-component of the curl of F is due to the fact that the ˆk vector is the unit normal vector to the xy-plane over which the integration is being performed. This will make more sense when we study Stokes Theorem. D 4

Vorticity and the physical interpretation of the curl of a planar vector field (which explains why F is called the curl ) (Relevant section of AMATH ourse Notes: Section 2.5) In fluid mechanics, if the following planar vector field, v(x,y) = (v 1 (x,y),v 2 (x,y)) = v 1 (x,y)i+v 2 (x,y)j (15) represents the velocity vector field of a fluid in R 2 (for example, the motion of a thin film of fluid), then the vorticity of the fluid at a point (x,y) is defined as Ω(x,y) = v 2 x v 1 y. (16) It is a measure of the rotation of the fluid in the vicinity of the point (x,y). (A discussion of vorticity in terms of paddle wheels being carried by a fluid is presented in Section 2.5 of the ourse Notes.) In fact, the right side of Eq. (16) has already been seen for more general vector fields in the context of 1. conservative fields (Ω(x,y) = 0 over a region D) 2. Green s Theorem in the plane (Ω(x,y) is the integrand of a double integral that may be used to compute the total circulation of a vector field). The vorticity function Ω(x,y) is the z-component of the curl of the planar vector v, i.e., Ω(x,y) = [curlv(x,y)] z = [ v(x,y)] z. (17) As such, we may conjecture that if v(x,y,z) is a three-dimensional velocity vector field, then the three components of the vector, curlv(x,y,z) = v(x,y,z), (18) represent the vorticities of the fluid in the three major planes, i.e., xy, yz, xz. This conjecture, indeed, is correct and will be discussed later in the course. Let us now return to the two-dimensional case, and suppose that F(x,y) is a general vector field in R 2, i.e., not necessarily the velocity field v for a fluid. For convenience, we ll use Ω(x,y) once again to define the vorticity of F, i.e., Ω(x,y) = [curlf(x,y)] z = [ F(x,y)] z = F 2 x F 1 y. (19) 5

Let us assume that the vector field F(x,y) is 1 over the neighbourhood D of a point p = (p 1,p 2 ) of the plane. This means that the first-order partial derivatives of F are continuous functions of (x, y) which, in turn, implies that the vorticity function Ω(x, y) is continuous over the region D. This will be important in our discussion below. Now let ǫ denote the circle of radius ǫ > 0 centered at p and D ǫ its interior region so that ǫ = D ǫ, the boundary of D ǫ. The idea is to let ǫ tend to zero. We assume that for some ǫ = ǫ 0 > 0, and therefore all ǫ values smaller than ǫ, the the circles ǫ lie inside the region D over which F is 1. The situation is sketched below. D ǫ. ǫ ǫ p = (p 1,p 2) From Green s Theorem (which we can apply, since the vector field F satisfies the hypotheses of the theorem), the circulation integral of F over the curve ǫ is [ F2 F dx = ǫ D ǫ x F ] 1 da y = Ω(x,y)dA D ǫ = Ω(x ǫ,y ǫ )A(D ǫ) for some point (x ǫ,y ǫ ) D ǫ. = Ω(x ǫ,y ǫ)πǫ 2. (20) Here, A(D ǫ ) denotes the area of the interior region D ǫ. The existence of a point (x ǫ,y ǫ )) D ǫ for which the second last equality holds follows from the Mean Value Theorem for (double) integrals (since the integrand is continuous): Mean Value Theorem for Integrals: Let f(x, y) be defined over a bounded region D R 2. The mean value of f over D is defined as f D = 1 f(x,y)da. (21) A(D) D If f is continuous on D, then there exists at least one point (x,y ) D for which f D = f(x,y ). (22) 6

This implies that D f(x,y)da = f(x,y )A(D). (23) Back to our main discussion, the notation (x ǫ,y ǫ) in Eq. (20) indicates the dependence of this point on ǫ. Now divide both sides of the final line of Eq. (20) by the area of D ǫ, to yield Ω(x ǫ,yǫ) = 1 πǫ 2 F dx. (24) ǫ In the limit ǫ 0, the region D ǫ enclosed by circle ǫ shrinks toward the point p = (p 1,p 2 ), the center of the circles ǫ. This implies that (x ǫ,y ǫ) (p 1,p 2 ) = p as ǫ 0. (25) It then follows that the limits as ǫ 0 of both sides of Eq. (24) exist: lim ǫ 0 Ω(x ǫ,y ǫ ) = lim ǫ 0 1 πǫ 2 From the continuity of the function Ω(x,y) on D, we may rewrite the LHS as 1 Ω(p 1,p 2 ) = lim ǫ 0 πǫ 2 ǫ F dx. (26) ǫ F dx. (27) In summary, Vorticity at a point p = limiting circulation per unit area at p. (28) But let s now recall that the vorticity is the z-component of the curl of F so that 1 [curlf(p 1,p 2 )] z = Ω(p 1,p 2 ) = lim ǫ 0 πǫ 2 where ǫ is a circle of radius ǫ centered at p. ǫ F dx, (29) 7

Lecture 15 Line integrals of vector fields over closed curves revisited Review: irculation of a vector field around a closed curve in R 2 (as a prelude to Total outward flux ) In this section, we review briefly the idea that the line integral of a planar vector field F around a simple closed curve in R 2, denoted as F dx, (1) may be used to define the circulation of the field over curve. As usual, the convention is that the integration along is performed in the counterclockwise direction so that the region D enclosed by lies always to the left of as we move along the curve. Here, we make no assumption that Green s Theorem in the plane can be employed to compute the circulation integral. As such, we must return to the formal definition. Assuming that we can parametrize the closed curve as x(t) = g(t), a t b, with g(a) = g(b), the line integral is normally computed as follows, b F dx = F(g(t))) g (t) dt. (2) a In a previous lecture, we rewrote the above integral as follows, b F dx = F(g(t)) g (t) dt (3) = = = a b a b a F(g(t)) F(g(t)) ˆT(t) ds fds, g (t) g (t) g (t) dt where the scalar-valued function, f(g(t)) = F(g(t)) ˆT(t), (4) is the projection of F in the direction of the unit tangent vector to the curve at g(t), as sketched in the figure below. Starting at any point P on the curve, the orientation of the tangent vector ˆT will change as we travel along. In one traversal of, the net rotation of the tangent vector is 2π. This is quite clear 1

ˆT(t 2 ) F(x(t 2 )) ˆT(t 1 ) F(x(t 3 ) F(x(t 1 )) ˆT(t 3 ) ˆT(t 4 ) F(x(t 4 )) when is a circle. The line integral in (3) sums up the projection of the vector field F(g(t)) onto the unit tangent vector ˆT(t) to the curve. As such, we say that the line integral in (3) is the circulation of the vector field F around the closed curve. Recall that inthespecial case thatfis a 1 vector fieldover theregion D enclosed by simplecurve, Green s Theorem in the plane may be used to express the circulation integral as a two dimensional integral over D, F dx = D [ F2 x F ] 1 da = [curlf] z da. (5) y D Total outward flux of a vector field across a closed curve in R 2 We now return to the line integral in Eq. (3) above and replace the unit tangent vector ˆT with the unit outward normal ˆN to the curve. This is the unit vector at a point g(t) on the curve which is perpendicular to the unit tangent vector ˆT(t) and therefore to the curve and which points outward from the region D enclosed by, as shown below: ˆn(t 2 ) F(x(t 2 )) F(x(t 3 ) F(x(t 1 )) ˆn(t 1 ) ˆn(t 3 ) F(x(t 4 )) ˆn(t 4 ) 2

The result is the line integral that we denote as b F ˆN ds = F(g(t)) ˆN(t)ds = a b a F(g(t)) ˆN(t) g (t) dt. (6) learly, this line integral adds up the projection of the vector field F onto the outward normal along the curve. The result is the net or total outward flux of F across the closed curve. Note: The word flux comes from the Latin word fluxus, meaning flow. In the theory of threedimensional transport phenomena (heat transfer, mass transfer and fluid dynamics), flux is defined as the rate of flow of a property per unit area, with dimensions [quantity, e.g., mass][time] 1 [area] 1. (7) We now present a specific example of a vector field F that arises in many applications involving flux. Here, however, we present the two-dimensional version in which the flow takes place across a line, as opposed to an area. Later, we shall present the three-dimensional case. At a point (x,y) in the plane, define where F(x,y) = ρ(x,y)v(x,y), (8) 1. ρ(x, y) represents the planar (2D) density of a fluid (mass/unit area). Dimension: 2. v(x, y) represents the velocity of the fluid (length/time). Dimension: Therefore, Dimension of F : L T. M L 2. M L 2 L T = M LT. (9) Now let be a simple closed curve in the plane which encloses a region D. The dimensionality of the total outward flux integral, is F ˆNds, (10) (Dimension of F) (Dimension of ds) = M LT L = M T. (11) In this special case for F, the outward flux integral measures the rate of mass exiting region D per unit time. It is exiting region D because it is flowing/moving across the boundary curve 3

in an outward direction ˆN that is normal to the curve. In the plane R 2, the practical calculation of the outward flux integral is not difficult the only complication is that you have to determine the outward unit normal vector ˆN(t) to the curve. This is easily done from a knowledge of the velocity vector v(t) = g (t). You firstconstruct the unit tangent vector as follows: ˆT(t) = v(t) v(t) = g (t) g (t) = (T 1(t),T 2 (t)). (12) Note that T 2 1 +T2 2 = 1. We now have to find the unit vectors which are normal to (T 1,T 2 ). There is a very easy way to determine the two vectors that are normal to a given vector, say (a,b) in the plane. You simply exchange the two components to produce the vector (b,a) and then negate one of the components, e.g., ( b,a) or (b, a). In fact, one of these vectors is the negative of the other one, i.e. (b, a) = ( b,a). It is easy to see that these vectors are orthogonal to (a,b): (a,b) (b, a) = ab ba = 0 (a,b) ( b,a) = ab+ab = 0. (13) Let us now return to the unit tangent vector ˆT(t). Using the method described above, there are two unit vectors that are perpendicular to ˆT(t). We ll denote them as ˆN 1 = (T 2 (t), T 1 (t)), ˆN2 = (T 2 (t), T 1 (t)). (14) We choose the vector that points outward from the region D enclosed by the simple closed curve. Special case: onsider the circle of radius R and centered at (0,0), which we shall denote as R. The parametrization of R and its velocity are well known, g(t) = (Rcost,Rsint) g (t) = ( Rsint,Rcost), 0 t 2π. (15) Thus the speed is given by g (t) = R so that the element of arclength is The unit tangent vector is then ˆT(t) = ds = g (t) dt = Rdt. (16) 1 g (t) g (t) = 1 ( Rsint,Rcost) = ( sint,cost). (17) R 4

Two unit vectors that are perpendicular to ˆT(t) are ˆN 1 (t) = (cost,sint), ˆN2 (t) = (cost,sint). (18) Both of these vectors are normal to the curve R. The vector that points outward is (cost,sint). Thus we set ˆN(t) = (cost,sint). (19) This result was expected: In the case of a circle, ˆN(t) is simply the unit vector that points outward from the center, namely, ˆx(t) = x(t). The situation is sketched above. x(t) ˆT = ( sint,cost) ˆx ˆN = (cost,sint) O R We now compute the outward flux of some vector fields through these circles R. Example 1: F(x,y) = Kî. We may imagine this vector field to represent a fluid that is travelling on the surface of a table with constant velocity K in the x-direction. y F = Kî O R x R 1. Evaluate F on curve: F(g(t)) = (K,0). 2. Integrand: F(g(t)) ˆN(t) = (K,0) (cost,sint) = Kcost. 3. Element of arclength: ds = g (t) dt = Rdt. 5

Thus F ˆN ds = R 2π 0 F(g(t)) ˆN(t) g (t) dt = 2π 0 KRcost dt = 0. (20) The net outward flux of F = Kî through the circle R is zero. This result was expected, since the amount of fluid entering the circular region enclosed by curve R from the left is balanced by the amount leaving the region on the right (assuming that K is positive). Example 2: F(x,y) = Kxî+Kyĵ. y F = Kr O R x R We expect the outward flux to be positive since this vector field points outward in the direction of the outward normal vector to R. 1. Evaluate F on curve: F(g(t)) = (KRcost,KRsint). 2. Integrand: F(g(t)) ˆN(t) = (KRcost,KRsint) (cost,sint) = KR. 3. Element of arclength: ds = g (t) dt = Rdt. Thus F ˆN ds = R 2π 0 F(g(t)) ˆN(t) g (t) dt = 2π 0 KR 2 dt = 2πKR 2. (21) The net outward flux of F = Kĝ through the circle R is 2πKR 2. As expected, this flux is positive, representing net outward flow. Also note that the result is dependent upon the radius R of the circle R. It seems that the bigger the circle, the more stuff is flowing through it. 6

Example 3: F(x,y) = Kyî+Kxĵ. y O R x R We expect the outward flux to be zero since this vector field points in the direction of the tangent vector to curve R. 1. Evaluate F on curve: F(g(t)) = ( KRsint,KRcost). 2. Integrand: F(g(t)) ˆN(t) = ( KRsint,KRcost) (cost,sint) = 0. 3. Element of arclength: ds = g (t) dt = Rdt. Thus F ˆN ds = R 2π 0 F(g(t)) ˆN(t) g (t) dt = 2π 0 The net outward flux of F = Kyî+Kxĵ through the circle R is 0, as expected. 0 dt = 0. (22) Example 4: The vector field F(x,y) = Kxi + Kyj of Example 2, but we now compute its total outward flux through the triangle with vertices (1,1), (1,2) and (5,1), as sketched below. y 2 1 0 2 3 1 1 2 3 4 5 x We considertheboundarycurve to beaunionofthethreesidesof thetriangle, i.e., = 1 2 3. Then F ˆNds = F ˆNds+ F ˆNds+ F ˆNds. (23) 1 2 3 7

The contributions from each segment i will have to be computed separately. Admittedly, this is a tedious calculation, but it is instructive to see an example that is somewhat more complicated than simply computing line integrals over circles. urve 1 : There are many ways to parametrize this line segment. As long as we employ a parameter that increases as we move from the starting point to the endpoint, things are fine. One possible parametrization of curve 1 is g(t) = (x(t),y(t)) = (t,1), 1 t 5, (24) with associated velocity vector, g (t) = (1,0). (25) 1. Outward unit normal vector ˆN: Quite easy in this case. The unit vector (0, 1). 2. Evaluate F on curve: F(g(t)) = (Kx(t),Ky(t)) = (Kt,K). 3. Integrand: F(g(t)) ˆN(t) = (Kt,K) (0, 1) = K. 4. Element of arclength: ds = g (t) dt = dt. Thus F ˆN ds = 1 5 1 ( K)dt = 4K. (26) The net outward flux of F through 1 is 4K. It is expected to be negative since the field vectors of F point inward across 1. urve 2 : Once again, there are many ways to parametrize this line segment. This curve lies on the line y = 1 4 x + 9 4. One possible parametrization of curve 2 is to let x(t) = 5 t for 0 t 4, so that y = 1 4 (5 t)+ 9 4 = 1 t+1. (27) 4 Thus we have g(t) = (x(t),y(t)) = (5 t, 14 ) t+1, 0 t 4, (28) with associated velocity vector, g (t) = ( 1, 1 ). (29) 4 8

1. Outward unit normal vector ˆN: A vector normal to 1 will have slope 4. One such vector is N = (1,4) We normalize this vector to produce ˆN = 1 17 (4,1). 2. Evaluate F on curve: F(g(t)) = (Kx(t),Ky(t)) = K(5 t, 1 4 t+1). 3. Integrand: F(g(t)) ˆN(t) = K(5 t, 1 4 t+1) 1 1 17 (1,4)) = K 17 (5 t+t+4) = 9K 1 17. 4. Element of arclength: ds = g (t) dt = 1+ 1 16 = 4 1 17dt. Thus 4 F ˆN ds = 9K 1 2 0 17 1 9 4 17dt = 4 4 K dt = 9K. (30) 0 The net outward flux of F through 2 is 9K. It is expected to be positive since the field vectors of F point outward across 2. urve 3 : Let us employ the following parametrization, g(t) = (x(t),y(t)) = (1,2 t), 0 t 1. (31) with associated velocity vector, g (t) = (0, 1). (32) 1. Outward unit normal vector ˆN: This one is also easy. ˆN = ( 1,0). 2. Evaluate F on curve: F(g(t)) = (Kx(t),Ky(t)) = K(1,2 t). 3. Integrand: F(g(t)) ˆN(t) = K(1,2 t) ( 1,0) = K. 4. Element of arclength: ds = g (t) dt = dt. Thus 1 F ˆN ds = Kdt = K. (33) 3 0 The net outward flux of F through 3 is K. It is expected to be negative since the field vectors of F point inward across 1. From Eq. (23), we add up the contributions from these three curves to determine the net outward flux of F through the closed curve : F ˆNds = ( 4+9 1)K = 4K. (34) Needless to say, that was quite a tedious calculation. 9

The Divergence Theorem in the Plane The Divergence Theorem in the Plane is the two-dimensional version of the more general threedimensional version, one of the most important results in Vector alculus. It could be viewed as the total outward flux analogue of Green s Theorem in the plane which deals with total circulation. Divergence Theorem in the Plane: Let D be a bounded subset of R 2 whose boundary D is a piecewise 1 simple closed curve oriented counter-clockwise. If F(x,y) = (F 1 (x,y),f 2 (x,y)) is a 1 vector field defined over D D, then F ˆNds = = D D [ F1 x + F ] 2 da y FdA. (35) 1. The integral on the LHS is a line integral over the boundary curve D =. This integral measures the net outward flux of the vector field F through the closed curve. 2. The integral on the RHS is a 2D integral over the region D. Before we prove this theorem, let us return to the three examples of total outward flux computed over the circle R of radius R and centered at (0,0). We let D R denote the circular region enclosed by R. 1. F(x,y) = Ki. Earlier, we computed the total outward flux of F through R to be zero. The divergence of this vector field is Therefore, which agrees with our computed result. F = x (K)+ (0) = 0. (36) y F ˆNds = FdA = 0dA = 0, (37) R D R D R Note: Because the divergence of this field is zero everywhere, it follows that for any simple closed curve in the plane enclosing a region D, the total outward flux of F through is zero, i.e., F ˆNds = D FdA = D 0dA = 0, (38) 10

2. F(x,y) = K[xi+yj]. Earlier, we computed thetotal outward fluxof F through R to be2πkr 2. The divergence of this vector field is Therefore, F = x (Kx)+ (Ky) = K +K = 2K. (39) y F ˆNds = FdA = 2KdA = 2KArea(D R ) = 2KπR 2, (40) R D R D R which agrees with our computed result. Note: Because the divergence of this vector field is constant, i.e., 2K, it follows that for any simple closed curve in the plane enclosing a region D, the total outward flux of F through is given by F ˆNds = D FdA = D 2KdA = 2KArea(D). (41) 3. F(x,y) = K[ yi+xj]. Earlier, we computed the total outward flux of F through R to be zero. The divergence of this vector field is Therefore, F = x ( Ky)+ (Kx) = 0. (42) y F ˆNds = which agrees with our computed result. D FdA = D 0dA = 0, (43) Note: As in Example 1, the divergence of this vector field is zero everywhere. It follows that the total outward flux of F through any simple closed curve (containing a region D) is zero. Finally, we return to the fourth example investigated earlier: 4. The vector field of Example 2, F(x,y) = K[xi+yj], but through the boundary of the triangle with vertices (1,1), (5,1) and (1,2), sketched earlier. We actually covered the more general case of this vector field in the note at the end of Example 2. From Eq. (41) and the fact that the area of the triangular region D enclosed by the curve is A = 2, F ˆNds = D FdA = D 2KdA = 2KArea(D) = 2K 2 = 4K, (44) which agrees with the result computed earlier. This computation, however, was much easier. 11

Proof of Divergence Theorem in the Plane: We ll actually use Green s Theorem in the Plane. Assume that curve admits a parametrization of the form, g(t) = (g 1 (t),g 2 (t)), a t b. (45) The unit tangent vector to at a point g(t) is then given by ˆT(t) = g (t) g (t) = 1 g (t) (g 1 (t),g 2 (t)) = (T 1(t),T 2 (t)). (46) D ˆT ˆN There are two unit vectors in the plane that are perpendicular to ˆT = (T 1,T 2 ), namely, ˆN ± = ±(T 2, T 1 ). (47) Theoutwardnormalturnsouttobe ˆN + = (T 2, T 1 ), assketchedinthediagrambelow. (Ofcourse, the sketch shows the particular situation in which T 1 and T 2 are both positive. But the other possibilities will yield the same result.) ˆT T 1 T 2 T 1 ˆN T 2 Therefore, the outward normal vector ˆN(t) is given by ˆN(t) = 1 g (t) (g 2 (t), g 1 (t)). (48) 12

We now compute the total outward flux of F through curve using this outward normal vector: F ˆNds = = = = = b a b a b a b a b where we have defined the vector field, a F(g(t)) ˆN(t) g (t) dt (F 1 (g(t)),f 2 (g(t)) [( F 2 )g 1 +(F 1)g 2 ]dt ( F 2,F 1 ) (g 1,g 2 )dt 1 g (t) (g 2 (t), g 1 (t)) g (t) dt H(g(t)) g (t)dt, (49) H(x,y) = F 2 (x,y)i+f 1 (x,y)j = H 1 (x,y)i+h 2 (x,y)j. (50) But the final line in Eq. (49) is, by definition, the circulation integral of the vector field H over curve, i.e., F ˆNds = H dx. (51) We can use Green s Theorem to compute this circulation integral (since the 1 requirements of the components of H are satisfied), which completes the proof. F ˆNds = = = H dx [ H2 D x H ] 1 da y [ F1 x + F ] 2 da, (52) y D 13

Physical interpretation of the divergence of a planar vector field In this section, we provide a physical interpretation of the divergence of a planar vector field F(x, y) at a point p = (p 1,p 2 ). Our approach will be similar to that used to interpret the curl of a vector field at a point. First recall the Divergence Theorem in the Plane: Let be a simple closed curve enclosing a region D in the plane and let ˆN denote the unit outward normal to. Also let F : R 2 R 2 be a vector field which is 1 on the set D, i.e., F has continuous first derivatives at all points of D. Then, F ˆNds = D [ F1 x + F ] 2 da = FdA. (53) y D The leftmost term of the equation is the total outward flux of F through the curve. The middle and right terms of the equation are the integration of the divergence of F over the two-dimensional region D. Now let p = (p 1,p 2 ) R 2 be a point of interest and assume that the vector field F(x,y) is 1 in a suitably large neighbourhood of p. As before, for an ǫ > 0 and ǫ be the circle of radius ǫ centered at p. Furthermore, let D ǫ be the region enclosed by ǫ. The situation is sketched below. D ǫ. ǫ ǫ p = (p 1,p 2) From the Divergence Theorem, the total outward flux of F through the curve ǫ (counterclockwise integration) is given by ǫ F ˆNds = D ǫ F(x)dA = F(q ǫ )A(D ǫ ) for some point q ǫ D ǫ = F(q ǫ )πǫ 2. (54) Here, we have once again used the Mean Value Theorem for Integrals, since the integrand F is continuous on D ǫ. 14

Once again, we divide both sides by the area A(D ǫ ) = πǫ 2 to obtain F(q ǫ ) 1 πǫ 2 F ˆNds. (55) ǫ Now take the limit ǫ 0, which implies that the circle ǫ is being shrunk toward the point p. As such, the points q ǫ must approach p in the limit. We have the result, In other words, 1 F(p) = lim ǫ 0 πǫ 2 F ˆNds. (56) ǫ The divergence of F at point p is the limiting total outward flux of F per unit area at p. This idea will carry over to R 3 in a straightforward way. For a three-dimensional vector field F, F(p) = lim 1 ǫ 0 4 3 πǫ3 F ˆNds. S ǫ (57) The integral on the right is the total outward flux through a spherical surface S ǫ of radius ǫ which encloses a spherical region of volume V ǫ = 4 3 πǫ3. The divergence of the 3D vector field F at a point p R 3 is the limiting total outward flux of F per unit volume at p. 15