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The Curl of a Vector Field. Stokes' Theorem Ex- Presses the Integral Of

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The Curl of a Vector Field. Stokes' Theorem Ex- Presses the Integral Of The Curl of a Vector Field. Stokes’ Theorem ex- A related operation, called the divergence will be in- presses the integral of a vector field F around a closed troduced later. curve as a surface integral of another vector field, called the curl of F. This vector field is constructed Each of these acts as a derivative, and there is a version in the proof of the theorem. Once we have it, we in- of the fundamental theorem that evaluates an integral vent the notation ∇×F in order to remember how to of this derivative. compute it. In this notation ∇ stands for the vector The fundamental theorem for line integrals says operator that the difference of a scalar function f (x, y, z) at ∂ ∂ ∂ ∇ , , two points is the integral of f (the gradient of f ) ∂x ∂y ∂z along any path joining the two points. From this it follows that the integral of ∇ f is independent of path ∇ and expressions containing are interpreted by first and that the integral of ∇ f along any closed path is pretending that it is an ordinary vector, so that vector zero. operations will introduce terms that abut a compo- nent from the first factor with one from the second Stokes’ theorem says that the integral of a vector field factor. For numerical vectors, this is interpreted as F(x, y, z) along a closed path is equal to the integral multiplication, but for ∇ the interpretation is to let of ∇×F (the curl of F) along any surface having the the component of ∇ operate on the component from given path as its positively oriented boundary. From the other factor. this it follows that the integral of ∇×F is zero on any closed surface. The gradient also uses this interpretation of ∇ to con- struct a vector field from a scalar function. The curl The divergence theorem (to be discussed later) re- takes one vector field to another using a construction lates the integral of a vector field over a closed surface modeled by the cross product. to the integral of its divergence over the three dimen- 16.5.1 16.5.2 sional region bounded by the surface. Although the #1. hxy, yz, zxi. details are different, all three theorems have similar #5. hex sin y, ex cos y, zi. statements. D E x , y , 1 Special second derivatives. It has already been noted #7. z z z . that For any of these whose curl is zero, express as a gra- ∇×(∇ f ) = 0. dient. Formally, this is just the equality of mixed partials, =∇ Finding a vector field from its curl. The first four but it is tied to Stokes’ Theorem. If F f , the line exercises of Section 16.8 have the form: use Stokes’ integral of F along any curve is the difference of the Theorem to evaluate values of f at the endpoints. For a closed curve, this ZZ is always zero. Stokes’ Theorem then says that the ∇×F dS. surface integral of its curl is zero for every surface, so S it is not surprising that the curl itself is zero. In this form, there isn’t much to the exercise. This Stokes’ theorem also says that the integral of the curl way of stating the exercise gives F, so its curl need of a vector field over a closed surface is zero. If we try never be computed since you must evaluate the line to write a given vector field G as the curl of another integral in order to feel that you have used Stokes’ vector field F, we will meet an obstruction to com- Theorem. There is still something to be done: you pleting the computation in the form of a combination need to produce a parameterization of the positively of derivatives of components of G that must be zero. oriented boundary from a description of the surface, Exercises 16.5. Find curl of the following vector but the statement of the exercise obscures its real con- fields. tent. 16.5.3 16.5.4 It would be more interesting to start from a vector field Looking at the components of the curl, we have G that is of the form G =∇×F, and determine F. Although this is more complicated than the process −Yz = P for writing a conservative vector field F as F =∇f , = (∗) it has a similar flavor. Xz Q − = . We now describe a solution of the equation Yx X y R ∇×F = G If the first of the equations in (∗) is solved for Y (up to a function of x and y), and this result put into the for F when G is given. third equation, we are left with the task of finding It is useful to simplify the problem as much as possible X from its derivatives with respect to y and z. This before beginning. We know that F is only defined up was exactly what we did in recognizing conservative to a term of the form ∇ f , and there is no difficulty (in vector fields as gradients. The only requirement is that principle, although it does require integration) finding the equations Xzy = X yz must hold. These equations a function f for which f3(x, y, z) is any expression. are In particular, any solution F could be replaced by F − Xzy = Q y ∇ f , where f3(x, y, z) is equal to the third component X yz = Yxz − Rz. of F. This means that we need only look for Finally, F = hX, Y, 0i. Yxz = Yzx =−Px , Let so G = hP, Q, Ri. Q y =−Px − Rz. 16.5.5 16.5.6 2 Note that this condition depends only on G and not (since Yx = 0). Then X = xz /2 + f (x, y) and on the particular choice of Y . f2(x, y) =−xy. The solution can be completed be- cause the expression that is supposed to be f (x, y) If this is satisfied, then we can find X for any Y , which 2 really is independent of z. Integrating this, gives completes the determination of F. f (x, y) =−xy2/2 as one solution. Thus, This shows that G = hP, Q, Ri satisfies G =∇×F D E + + = 1 if and only if Px Q y Rz 0. This looks like it F = xz2 − xy2, −yz2, 0 . should be denoted ∇·G. 2 The whole story is a little more subtle. It is easy to If we subtract ∇(x2z2/4) = xz2/2, 0, x2z/2 from verify that ∇×(∇ f ) = 0 and ∇·(∇×F) = 0, but the this, we have the more symmetric solution D E constructive part of these equivalences assumed that 1 the functions obtained in the solution were defined F =− xy2, yz2, zx2 . everywhere. For objects defined only in part of space, 2 the necessary condition my hold but it may not be Additional Exercises. Write the given vector field G possible to perform the construction. as G =∇×F, if possible. An example. From exercise 1 in Section 16.5, we (A) G = h0, 0, 1i. know that∇·G = 0 when G = hyz, xz, xyi. The = 2, 2, first equation of (∗) becomes Yz =−yz, so we take (B) G x y 0 . =− 2/ Y yz 2. This leaves (C) G = h0, 0, xeyi. Xz = xz Why is it called “curl”? Consider the vector field X y =−xy R = h−y, x, 0i 16.5.7 16.5.8 that points in the positive direction along each circle to r 2, so this effect is not overwhelming. However with center at the origin, with a magnitude propor- when considered on a fixed scale, the magnitude of tional to the radius. Such a motion corresponds to our the curl gives a noticeable rotation. To be free of such image of pure rotation. An easy computation shows effects, a vector field must have zero curl. that ∇×R = h0, 0, 2i, identifying the axis of rotation everywhere. A phys- ical model uses a small paddle-wheel moving in the flow R. When the axis of the wheel is lined up with the z-axis, the difference in speeds at the ends of the paddle causes to paddle spin relative to coordinates with fixed directions as its center of mass moves with the flow. This behavior is seen throughout the flow — not just near the center of rotation. The rotation is detected in the vector field R itself by forming the line integral of R along closed curves. Stokes’ theorem tells us that such an integral is given by the integral of ∇×R over a surface bounded by the curve. In particular, the value of the integral may be estimated by the product of the area of the surface and the component of ∇×R perpendicular to the surface. For a small circle of radius r, the area is proportional 16.5.9 16.5.10.
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