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Unformatted text preview: Section 20.1 – The Curl of a Vector Field Conceptual Description of Circulation. Let F~ (x, y, z) be a vector field, and imagine placing a tiny paddlewheel (like the one shown to the right) in the vector field at the point (x, y, z) so that the handle points in the direction of the axis that you want to measure the circulation of F~ around. Your eye Notes. angular velocity • The the circulation. of the paddlewheel measures the strength of (x,y,z) • The circulation is positive if the rotation is counterclockwise and negative if the rotation is clockwise (as you look down the handle of the paddlewheel). Preliminary Example. Let F~ be the vector field shown to the right. Decide whether the circulation of the vector field at (0, 0, 0) around each of the following axes would be positive, negative, or zero: (a) ~k (b) −~k (c) ~j (a) Since the handle of the paddlewheel points straight up, the circulation is positive. y 0 -2 2 2 z0 -2 (b) Since the handle of the paddlewheel points straight down, the circulation is negative. 0 2 x -2 (c) Since the handle of the paddlewheel points straight to the right, that is, along the positive y-axis, the flow of the vector field cannot "push" the paddles. Therefore, the circulation is zero. Definition. The circulation density of a smooth vector field F~ at P = (x, y, z) around the direction of a unit vector ~n is defined as R ~ r C F · d~ ~ , circ~n F (x, y, z) = lim Area → 0 Area inside C where C is a circle centered at P and perpendicular to ~n. Note. The orientation of the circle C is determined by the right- n P C P C n ~n : handle of paddlewheel hand rule: If you place your right thumb in the direction of ~n, your fingers curl in the direction of C. Comments: • The line integral R C F~ · d~r gives the circulation of F~ around ~n. • The quantity circ~n F~ measures the circulation per unit area around ~n. Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 1 2 Example 1. Let F~ (x, y, z) = −y~i + x~j + 0~k. The cross-section of this vector field for z = 0 is shown to the right. Calculate each of the following. 1 y (a) circ~k F~ (0, 0, 0) C1 0 (b) circ−~k F~ (0, 0, 0) -1 (c) circ~j F~ (0, 0, 0) -1 0 x 1 (a) Let C1 be a circle of radius a, centered at (0, 0) and oriented as shown to the right. Observe that, along C1 , we have F~ · d~r = = kF~ k kd~rk cos 0◦ p x2 + y 2 kd~rk = a kd~rk. Therefore, we have ←− k since F~ and d~r are parallel ←− 2 a C1 p since x2 + y 2 = a on C1 circ~k F~ (0, 0, 0) = lim a→0 = lim a→0 = lim a→0 R F~ · d~r C1 a ←− area inside C1 πa2 R C1 kd~rk πa2 a(2πa) πa2 = 2. R (b) In this case, C2 F~ · d~r = −2πa2 (the negative of the corresponding integral from part (a)), because C2 is oriented in the opposite direction. Therefore, −2πa2 = −2. a→0 πa2 a circ−~k F~ (0, 0, 0) = lim (c) RIn this case, F~ and d~r are orthogonal everywhere along C3 . Therefore, ~ r = 0, so we obtain C3 F · d~ circ~j F~ (0, 0, 0) = lim a→0 2 R F~ · d~r C3 πa2 = 0. C2 −k a C3 j Developed by Jerry Morris Example 2. To the right, you are given the graph of a vector field -2 F~ (x, y, z). For approximately what value of ~n does circn~ F~ (0, 0, 0) appear to be the greatest? Explain. y 0 2 ÖÓ n 2 Choosing ~n = ~k would produce the greatest circulation because it appears that placing the paddlewheel so that the handle points straight up would result in the fastest counterclockwise rotation. z0 C -2 0 2 x -2 Geometric Definition of Curl. The curl of a smooth vector field F~ , written curl F~ , is the vector field with the following properties: 1. The direction of curl F~ (x, y, z) is the direction of ~n for which circ~n F~ (x, y, z) is the greatest. 2. The magnitude of curl F~ (x, y, z) is the circulation density of F~ around that direction. If the circulation density is zero around every direction, then we define the curl to be ~0. Coordinate Definition of Curl. If F~ = F1~i + F2~j + F3~k, then curl F~ = =       ∂F3 ∂F2 ~ ∂F3 ~ ∂F1 ~ ∂F1 ∂F2 k − − − i+ j+ ∂y ∂z ∂z ∂x ∂x ∂y ~ ~i ~ ∂ j∂ k∂ ∂x ∂y ∂z F F F 1 = 2 3 ∇ × F~ Notes: 1. The geometric and coordinate definitions of curl are equivalent (see text, pp. 957-958). 2. Recall the "Curl Test": Under "nice" conditions, a vector field F~ is conservative if and only if curl F~ = ~0 everywhere. See page ?? in the workbook for a more precise statement of this result. Examples and Exercises 1. Consider the two vector fields shown below. For each field, draw in a possible vector representing curl F~ (0, 0, 0). F~ (x, y, z) F~ (x, y, z) -2 2 y 0 2 -2 ÖÓ curl F 2 2 z0 z0 -2 -2 0 2 x y 0 -2 Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 ÖÓ curl F 0 2 x -2 3 2. Below, you are given a formula for three vector fields and a picture of the cross section of that field that lies in the plane z = 0. F~2 (x, y, z) = |y|~i F~3 (x, y, z) = −y ~i + x ~j 2 2 1 1 1 0 0 0 -1 y 2 y y F~1 (x, y, z) = ~i + 2 ~j -1 -1 0 x 1 2 -1 -1 0 x 1 2 -1 0 x 1 2 (a) For each vector field, use graphical reasoning to describe the vectors curl F~ (0, 1, 0) and curl F~ (0, −1, 0). In particular, are they zero or nonzero? If they are nonzero, do they point in the ~k or the −~k direction? (i) For the vector field F~1 , we have curl F~1 (0, 1, 0) = curl F~1 (0, −1, 0) = ~0, because objects placed at these two points in F~1 would experience no rotation. (ii) The vector curl F~2 (0, 1, 0) should point in the "−~k" direction because, for a square placed at (0, 1, 0) in F~2 , there is more left to right flow at the top of the square than at the bottom, resulting in a clockwise rotation (see above diagram). By a similar argument, we see that curl F~2 (0, −1, 0) should point in the "~k" direction. (iii) At (0, 1, 0), there is more right to left flow at the top of a square than at the bottom, resulting in a counterclockwise rotation. At (0, −1, 0), there is more left to right flow at the bottom of a square than at the top, also resulting in a counterclockwise rotation. Therefore, curl F~ (0, 1, 0) and curl F~ (0, −1, 0) both point in the "~k" direction. (b) Confirm your answers to part (a) by calculating curl F~ (0, 1, 0) and curl F~ (0, −1, 0) for each vector field. (i) Since F~1 is constant, its partial derivatives are constant everywhere, so curl F~1 = ~0 everywhere, including at (0, 1, 0) and (0, −1, 0). (ii) First, to aid in taking derivatives, we observe that |y| = p y 2 . We therefore have y 1 y curl F~2 (x, y, z) = 0~i + 0 ~j − (y 2 )−1/2 (2y) ~k = − p ~k = − ~k. 2 |y| y2 Therefore, curl F~2 (0, 1, 0) = −~k and curl F~2 (0, −1, 0) = ~k. (iii) We have curl F~3 (x, y, z) = 0~i + 0 ~j + (1 − (−1)) ~k = 2~k, so curl F~3 = 2~k everywhere, including at (0, 1, 0) and (0, −1, 0). 4 Developed by Jerry Morris 3. Using Green’s Theorem and the definition of curl, it can be shown that for any smooth vector field F~ and any unit vector ~n, we have curl F~ · ~n = circ~n F~ , (∗) a fact that you may use freely from here on out. Let S represent a small, two-dimensional solid square or disk centered at the point (x, y, z) and having area vector ~ Let C be the curve along the boundary of S, oriented ∆A. ~ according to the right hand rule (see to correspond to ∆A diagram to the right). Use equation (*) above to show that C S (∗∗) ~ curl F~ (x, y, z) · ∆A ≈ Z ∆A ∆A S C F~ · d~r. C We have ~ curl F~ · ∆A = curl F~ · (~n ∆A) ←− where ~n is a unit normal vector to S and ∆A is the scalar area of S = (curl F~ · ~n) ∆A ←− by properties of dot products = (circ~n F~ ) ∆A ←− by Equation (∗) above R C ≈ = Z F~ · d~r ∆A ! ∆A ←− by the definition of circulation F~ · d~r. C Note: The smaller ∆A is, the better the above approximation will be. Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 5 4. Suppose that F~ is a smooth vector field and that curl F~ = −x~i + z~k. Estimate the circulation, around a circle of radius 0.01 in each of the following situations. R C F~ · d~r, (a) Centered at (2, 0, 0), parallel to the yz-plane, and oriented counterclockwise when viewed from the positive x-axis. (b) Centered at (0, 0, 3), parallel to the xy-plane, and oriented counterclockwise when viewed from the positive z-axis. (c) Centered at (0, 0, 3), contained in the yz-plane, and oriented counterclockwise when viewed from the positive x-direction. Note: For each of the following calculations, we will use Formula (∗∗) from the previous exercise. (a) We have z Z F~ · d~r ≈ ~ curl F~ (2, 0, 0) · ∆A = −2~i · π(0.01)2 ~i = −2π(0.01)2 ≈ −0.000628. C ∆A x C y (b) We have Z F~ · d~r ≈ ~ curl F~ (0, 0, 3) · ∆A = 3 ~k · π(0.01)2 ~k = 3π(0.01)2 ≈ 0.000942. C C y x (c) We have Z z ∆A z F~ · d~r ≈ ~ curl F~ (0, 0, 3) · ∆A = 3 ~k · π(0.01)2 ~j = 0. C ∆A C x 6 Developed by Jerry Morris y 5. Use the “curl test” to determine whether each of the following vector fields is conservative. If it is conservative, find a function f such that F~ = ∇f. (a) F~ = yexy ~i + xexy ~j + (2z + y) ~k (b) F~ = 2xyz ~i + x2 z ~j + (x2 y + 1) ~k (a) Calculating, we have curl F~ =  = (1 − 0)~i + (0 − 0) ~j + ∂F3 ∂F2 − ∂y ∂z  ~i +  = ~i. ∂F3 ∂F1 − ∂z ∂x  ~j +  ∂F1 ∂F2 − ∂x ∂y  ~k  (xyexy + exy ) − (xyexy + exy ) ~k Since curl F~ 6= ~0, we see that F~ is not conservative. Remark: The above calculation could have been shortened by noticing that the ~i component of curl F~ was nonzero, and concluding that curl F~ 6= ~0 without having to calculate the ~j or the ~k component of curl F~ . (b) Calculating, we have curl F~ =  = (x2 − x2 )~i + (2xy − 2xy) ~j + (2xz − 2xz) ~k ∂F3 ∂F2 − ∂y ∂z  ~i +  ∂F3 ∂F1 − ∂z ∂x  ~j +  ∂F1 ∂F2 − ∂x ∂y  ~k = ~0. Therefore, since curl F~ = ~0, the vector field F~ is conservative. We will now find a potential function, f, for F~ : fx (x, y, z) = 2xyz =⇒ f (x, y, z) = x2 yz + P (y, z) fy (x, y, z) = x2 z =⇒ f (x, y, z) = x2 yz + Q(x, z) fz (x, y, z) = x2 y + 1 =⇒ f (x, y, z) = x2 yz + z + R(x, y) Since each of the above three representations for f must agree, we see that f (x, y, z) = x2 yz + z is our potential function. This can be checked easily by verifying that F~ = ∇f. Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 7 Section 20.2 – Stokes’ Theorem Preliminary Example. Let S be the surface shown below, oriented upward, and let C be the curve running along the boundary of S, oriented counterclockwise. Note that S has been divided into M small “patches”. Si = ith surface patch (oriented upward) Ci ~i ∆A = = curve along boundary of Si area vector of Si ∆Ai Si C We will now calculate the sum of the circulation of F~ around each of the M patches. Note that the orientation of each curve Ci has been chosen to correspond to the orientation of the surface, S, according to the right-hand rule. We therefore have S (1) M Z X i=1 F~ · d~r Z = Ci F~ · d~r ←− because of "cancellation" along the "inside" edges (labeled green in the above diagram) C However, by the result of Exercise 3 in the previous section, we also have (2) M Z X i=1 F~ · d~r Ci sample point on Si ↓ M X ~ ~i (curl F )(Pi ) · ∆A ≈ ←− by Equation (∗∗) on page 5 i=1 Therefore, by combining Equations (1) and (2) above, we see that Z F~ · d~r ≈ C M X ~i , (curl F~ )(Pi ) · ∆A i=1 and this approximation becomes more accurate as the number of patches, M, increases. take the limit as M → ∞, we obtain Stokes’ Theorem: If we Stokes’ Theorem. Let S be a smooth oriented surface with piecewise smooth, oriented boundary C, and let F~ be a smooth vector field on an open region containing S and C. Then Z Z ~ ~ curl F~ · dA F · d~r = C S The orientation of C is determined from the orientation of S according to the right hand rule. Notes. 1. The integral 2. The integral R C R S F~ · d~r is a line ~ is a curl F~ · dA flux integral around the integral over "boundary" of a surface S a surface S . . 3. Stokes’ Theorem is mainly useful to replace a tedious line integral with an easier flux integral, or vice versa. 8 Developed by Jerry Morris The Boundary of a Surface Definition. The boundary of a surface, S, is the curve (or curves) C running around the edge of S (like the hem on clothing). Once an orientation for S has been chosen, the corresponding orientation for C is determined by the right-hand rule. Examples. For each of the following surfaces, draw in or describe both possible orientations. Then, for each surface orientation, draw in the induced orientation of the boundary curve(s). 1. Cylinder (without top and bottom) n C n C "outward" orientation "inward" orientation 2. Hemisphere (with open bottom) n n rC "upward" orientation r C "downward" orientation 3. Punctured Yield Sign C out of page Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 C into page 9 Exercises and Examples 1. Let F~ (x, y, z) = x2 z ~i + y 2 exz ~j + z 2 ey ~k, and let S be the portion of the paraboloid z = 9 − x2 − y 2 that lies R ~ above the xy-plane, oriented upward. Calculate S curl F~ · dA. Since the boundary, C, of the surface S appears to be a circle in the xy-plane, which is easy to parameterize, we will use Stokes’ Theorem as follows: Z Z ~ = curl F~ · dA F~ · d~r S z S C Note that C is oriented counterclockwise, as viewed from the positive z-axis, so that its orientation corresponds with the upward orientation of S. We must now find and parameterize C: 3 x y Ck 3 Calculation of Boundary Curve C: Since the boundary curve consists of the intersection of S with the plane z = 0, we calculate as follows: z z = 9 − x2 − y 2 = 0 =⇒ 9 − x2 − y 2 = 0 =⇒ x2 + y 2 = 9 ←− circle, radius 3 Next, we parameterize C as follows: x = y = z = Calculation of Integral: Z 3 cos t 3 sin t 0 ~r(t) = 3 cos t~i + 3 sin t ~j + 0 ~k, 0 ≤ t ≤ 2π Continuing the integral calculation from above, we have ~ curl F~ · dA = S Z F~ · d~r ←− by Stokes’ Theorem C = Z t=2π (x2 z ~i + y 2 exz ~j + z 2 ey ~k) · ~r ′ (t) dt t=0 = Z 2π (0~i + (3 sin t)2 e0 ~j + 0 ~k) · (−3 sin t~i + 3 cos t ~j) dt 0 = Z 2π 27 sin2 t cos t dt 0 = 10 2π 9 sin3 t 0 = 9(sin 2π)3 − 9(sin 0)3 = 0. Developed by Jerry Morris 2. Let F~ (x, y, z) = −2y~i + 2x~j. Use Stokes’ Theorem to find R C F~ · d~r, where C is a circle... (a) parallel to the yz-plane, of radius a, centered at a point on the x-axis, and oriented counterclockwise as viewed from the positive x-axis. S = disk forming interior of C, oriented in the positive x-direction z Since C forms the boundary of S, we have Z F~ · d~r C 4~k ~n dA Z z }| { z}|{ ~ = curl F~ · dA C n S = Z 4~k · ~i dA xt y S S = Z 0 dA S = 0. (b) parallel to the xy-plane, of radius a, centered at a point on the z-axis, and oriented counterclockwise as viewed from above. S = disk forming interior of C, oriented in the positive z-direction tz Since C forms the boundary of S, we have n C Z C F~ · d~r ~n dA 4~k Z z }| { z}|{ ~ ~ curl F · dA = S = Z 4~k · ~k dA x S t y S = 4 Z dA S = 4 (πa2 ) = 4πa2 . Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 11 C2 3. Let F~ (x, y, z) = −y ~i + x ~j + zex+y ~k, and let E be the solid region that lies above the plane z = 0, below the plane z = 1, and inside the half of the cylinder x2 + y 2 = 1 for which x ≤ 0. Let S be the surface of E, EXCLUDING THE SEMICIRCULAR TOP PANEL that lies in the plane z = 1 (see diagram to the right). Calculate R ~ where S is given the outward orientation. curl F~ · dA, S C1 z x ÖÓ n y R ~ would be Since S is comprised of three separate surfaces, it is likely that S curl F~ · dA unpleasant to evaluate directly as a flux integral. We will therefore attempt to use Stokes’ Theorem. Note that the boundary curve (highlighted above) can be broken into two pieces: a line segment C1 and a semicircle C2 , oriented as indicated in the above diagram. We therefore have the following: Z Z Z ~ = F~ · d~r F~ · d~r + curl F~ · dA S Calculation of F~ d~r R C1 F~ · d~r: Along C1 , we have the following: 0 ↓ = −y ~i + x ~j + zex+y ~k ←− parallel to ~j Calculation of R C2 F~ · d~r: x = y = z = C2 C1 Since F~ · d~r = 0 everywhere along C1 , we have Z F~ · d~r = 0. C1 Along C2 , we have the following: cos t − sin t 1 3π π ≤t≤ 2 2 ~r(t) = cos t~i − sin t ~j + ~k, Therefore, continuing with the integral calculation from above, we have Z Z Z ~ = F~ · d~r F~ · d~r + curl F~ · dA S C2 C1 = Z F~ · d~r ←− C2 = Z t= 3π 2 Z t= 3π 2 Z 3π 2 t= π 2 = since R C1 F~ · d~r = 0 (−y ~i + x ~j + zex+y ~k) · (− sin t~i − cos t ~j) dt (y sin t − x cos t) dt t= π 2 = = Z (− sin2 t − cos2 t) dt π 2 3π 2 (−1) dt π 2 = −π. 12 Developed by Jerry Morris 4. Let F~ (x, y, z) = (2y + x2 ez )~i + (4x + ey ) ~j + cos(x2 yz) ~k, and let S be the portion of the sphere x2 + y 2 + z 2 = 4 for which z ≥ 0, oriented upward. (a) Describe and draw a flat surface T that has the same boundary curve as S. Also state the proper orientation for T so that it corresponds to the orientation on the boundary curve induced by S. R ~ must equal (b) Assuming the orientations for S and T specified above, explain why S curl F~ · dA R ~ ~ curl F · dA. T R ~ curl F~ · dA. (c) Evaluate S (a) Let T be the portion of the plane z = 0 that lies inside the sphere x2 + y 2 + z 2 = 4 (see diagram to the right). Then T must be oriented upward to match the orientation of the boundary curve, C. t S T C (b) Because the surfaces S and T both have C as their boundary curve, we have the following: Z Z ~ curl F~ · dA ←− Stokes’ Theorem applied to S F~ · d~r = C Z S F~ · d~r = C Z ~ curl F~ · dA ←− Z ~ = curl F~ · dA Z Stokes’ Theorem applied to T T Therefore, we must have ~ curl F~ · dA. T S R ~ and use part (b). (c) Since T is a simpler surface than S, we will evaluate T curl F~ · dA We begin by making some observations about our integral. Note that the normal vector to T is ~n = ~k, so only the ~k component of curl F~ will be relevant to our calculations. Therefore, we will not explicitly calculate the ~i and ~j components of curl F~ ; we will simply write "P " and "Q" in their places: ~n curl F~ = ~k ←− unit normal to T = P ~i + Q ~j + (4 − 2) ~k Calculating, we have Z ~ curl F~ · dA = Z ~ curl F~ · dA ) ←− curl F~ · ~n dA = 2 dA by part (b) T S = Z curl F~ · ~n dA T = Z 2 dA ←− by the above calculations T = 2 (Surface Area of T ) = 2(4π) = 8π. Note: This strategy could also have been applied to the previous exercise. Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2013 13 5. Let F~ = F1 ~i + F2 ~j + F3 ~k be a smooth vector field, and let S be a piecewise smooth surface that completely encloses a solid region in R3 ; in other words, S is a closed surface. Let S have the outward orientation, and suppose that S has been divided into M small patches Si , each having boundary curve Ci and area vector ~ i , where the orientations of Si and Ci are chosen to correspond to the outward orientation of S (see ∆A diagram below). (a) Explain why M Z X i=1 F~ · d~r must equal zero. Ci Since S is totally enclosed, it has no boundary curve (contrary to the surface in the proof of Stokes’ Theorem). Therefore, in calculating the above sum, every patch edge is internal and is traversed in both directions one time, resulting in cancellation. Thus, this sum must equal zero. Ci ∆ Ai Si (b) Let Pi be a point chosen on the surface Si for each value of i. Use the fact that Z ~ ~ F~ · d~r (curl F )(Pi ) · ∆Ai ≈ Ci and your answer to part (a) to calculate the value of We have Z ~ ≈ curl F~ · dA S R S M X ~ curl F~ · dA. ~i (curl F~ )(Pi ) · ∆A i=1 ≈ M Z X i=1 = F~ · d~r Ci 0. In the limit as M → ∞, we therefore obtain Z ~ = 0. curl F~ · dA S 14 Developed by Jerry Morris (c) Calculate div (curl F~ ). First, recall that  curl F~ = ∂F2 ∂F3 − ∂y ∂z   ~i + ∂F3 ∂F1 − ∂z ∂x  ~j +  ∂F1 ∂F2 − ∂x ∂y  ~k. Therefore, we have div (curl F~ ) = = = ∂ ∂x  ∂F3 ∂F2 − ∂y ∂z   (F3 )yx − (F2 )zx +  (F1 )zy − (F1 )yz + = 0 + 0 + 0 = 0. (d) Use the Divergence Theorem to calculate ←− R S    ∂ ∂F2 ∂F1 ∂F3 ∂F1 + − − ∂z ∂x ∂z ∂x ∂y   (F1 )zy − (F3 )xy + (F2 )xz − (F1 )yz   (F2 )xz − (F2 )zx + (F3 )yx − (F3 )xy ∂ ∂y +  since mixed partial der...
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