chap17-et-student-solutions
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chap17-et-student-solutions

Course: MATH 2057, Fall 2008

School: LSU

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FUNDAMENTA L 18 THEOREMS OF VE C TOR ANALYSIS 18.1 Greens Theorem Preliminary Questions 1. Which vector eld F is being integrated in the line integral SOLUTION (ET Section 17.1) x 2 d y e y d x? The line integral can be rewritten as e y d x + x 2 d y. This is the line integral of F = e y , x 2 along the curve. 2. Draw a domain in the shape of an ellipse and indicate with an arrow the boundary orientation of...

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L FUNDAMENTA 18 THEOREMS OF VE C TOR ANALYSIS 18.1 Greens Theorem Preliminary Questions 1. Which vector eld F is being integrated in the line integral SOLUTION (ET Section 17.1) x 2 d y e y d x? The line integral can be rewritten as e y d x + x 2 d y. This is the line integral of F = e y , x 2 along the curve. 2. Draw a domain in the shape of an ellipse and indicate with an arrow the boundary orientation of the boundary curve. Do the same for the annulus (the region between two concentric circles). SOLUTION C. The orientation on C is counterclockwise, meaning that the region enclosed by C lies to the left in traversing C For the annulus, the inner boundary is oriented clockwise and the outer boundary is oriented counterclockwise. The region between the circles lies to the left while traversing each circle. 3. The circulation of a gradient vector eld around a closed curve is zero. Is this fact consistent with Greens Theorem? Explain. SOLUTION Greens Theorem asserts that F ds = P dx + Q dy = Q P x y dA (1) C C D If F is a gradient vector eld, the cross partials are equal, that is, Q P = y x Q P =0 x y (2) Combining (1) and (2) we obtain C F ds = 0. That is, Greens Theorem implies that the integral of a gradient vector eld around a simple closed curve is zero. 4. Which of the following vector elds possess the following property: For every simple closed curve C, equal to the area enclosed by C? (a) F = y, 0 (b) F = x, y (c) F = sin(x 2 ), x + e y SOLUTION 2 C F ds is By Greens Theorem, F ds = Q P x y dx dy (1) C D S E C T I O N 18.1 C D Greens Theorem (ET Section 17.1) 637 We compute the curl of each one of the given elds. (a) Here, P = y and Q = 0, hence Q P = 0 (1) = 1. Therefore, by (1), x y C F ds = D 1 d x d y = Area(D) (b) We have P = x and Q = y, therefore Q P = 0 0 = 0. By (1) we get x y C F ds = D 0 d x d y = 0 = Area(D) 2 (c) In this vector eld we have P = sin(x 2 ) and Q = x + e y . Therefore, P Q = 1 0 = 1. x y By (1) we obtain C F ds = D 1 d x d y = Area(D). Exercises 1. Verify Greens Theorem for the line integral SOLUTION C x y d x + y d y, where C is the unit circle, oriented counterclockwise. Step 1. Evaluate the line integral. We use the parametrization () = cos , sin , 0 2 of the unit circle. Then d x = sin d, and x y d x + y d y = cos sin ( sin d) + sin cos d = cos sin2 + sin cos d The line integral is thus xy dx + y dy = = 2 0 2 0 d y = cos d C cos sin2 + sin cos d cos sin2 d + 2 0 y C D x sin cos d = sin3 2 cos 2 2 =0 30 4 0 (1) Step 2. Evaluate the double integral. Since P = x y and Q = y, we have P Q = 0 x = x x y We compute the double integral in Greens Theorem: Q P x y dx dy = x d x d y = x dx dy D D D 638 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) The integral of x over the disk D is zero, since by symmetry the positive and negative values of x cancel each other. Therefore, D P Q x y dx dy = 0 (2) Step 3. Compare. The line integral in (1) is equal to the double integral in (2), as stated in Greens Theorem. In Exercises 311, use Greens Theorem to evaluate the line integral. Orient the curve counterclockwise unless otherwise 2 F ds, where F = y + sin x 2 , x 2 + e y and C is the circle of radius 4 centered at the origin. Let indicated. I = C (a) Which is 2 easier: evaluating I directly or using Greens Theorem? 3. y 2 d x + x d y, where C is the boundary of the unit square 0 x 1, 0 y 1 (b) Evaluate I using the easier method. C SOLUTION y 1 C 1 D x We have P = y 2 and Q = x 2 , therefore P Q = 2x 2y x y Using Greens Theorem we obtain C y2 d x + x 2 d y = Q D x P dA = y D (2x 2y) d x d y = 2 D x dx dy 2 D y dx dy By symmetry, the positive and negative values of x cancel each other in the rst integral, so this integral is zero. The second double integral is zero by similar reasoning. Therefore, C y2 d x + x 2 d y = 0 0 = 0 5. C x 2 y 2x+y d x, where C y the unit circle centered at the origin is e d x + e d y, where C is the triangle with vertices (0, 0), (1, 0), and (1, 1) C SOLUTION y C D x In this function P = x 2 y and Q = 0. Therefore, P Q = 0 x 2 = x 2 x y We obtain the following integral: I= x2 y dx = Q P x y dA = x 2 d A C D D We convert the integral to polar coordinates. This gives I= 2 0 0 1 r 2 cos2 r dr d = 2 0 0 1 r 3 cos2 dr d S E C T I O N 18.1 Greens Theorem (ET Section 17.1) 639 = 2 0 cos2 d 1 0 r 3 dr = sin 2 2 + 2 4 =0 r4 1 4 r =0 = 1 4 = 4 7. C F ds, where F = x 2 , x 2 and C consists of the arcs y = x 2 and y = x for 0 x 1 F ds, where F = x + y, x 2 y and C is the boundary of the region enclosed by y = x 2 and y = x for C SOLUTION 0x By 1. Greens Theorem, I= F ds = y C y=x D y = x2 0 1 x C D Q P x y dA We have P = Q = x 2 , therefore P Q = 2x 0 = 2x x y Hence, I= = 2x d A = 1 0 x x2 D 2x d y d x = 1 x 2x y 0 y=x 2 dx = 1 0 2x(x x 2 ) d x = 1 0 (2x 2 2x 3 ) d x 2 3 1 41 1 21 xx == 3 20 32 6 9. The line integral of F = x 3 , 4x around the boundary of the parallelogram in Figure 15 (note the orientation) (ln x + y) d x x 2 d y, where C is the rectangle with vertices (1, 1), (3, 1), (1, 4), and (3, 4) C y (2, 2) (4, 2) (2, 0) x FIGURE 15 SOLUTION We have P = x 3 and Q = 4x, therefore P Q =40=4 x y Hence, Greens Theorem implies x 3 d x + 4x d y = P Q x y dA = 4dA = 4 d A = 4 Area(D) = 4 4 = 16 C D D D 11. C 2 x y d line integral of y, = e x+y ,is xy path in Figure 16 The x + (x + x) d F where C e the around the boundary of the parallelogram in Figure 15 y (0, 1) (1, 0) (1, 0) x FIGURE 16 640 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S SOLUTION y (0, 1) C D (1, 0) (ET CHAPTER 17) (1, 0) x In the given function, P = x y and Q = x 2 + x. Therefore, P Q = 2x + 1 x = x + 1 x y By Greens Theorem we obtain the following integral: x y d x + (x 2 + x) d y = Q P x y dA = (x + 1) d A = x dA + 1dA C D D D D By symmetry, the positive and negative values of x cancel each other, causing the rst integral to be zero. Thus, C x y d x + (x 2 + x) d y = 0 + D d A = Area(D) = 21 = 1. 2 In Exercises 1316, use Eq. (5) to calculate the area of the given region. Let C R be the circle of radius R centered at the origin. Use Greens Theorem to determine F ds, where F is C2 13. The circle of radius 3 centered at the origin 2 2 2 2 F a vector By such that SOLUTION eld Eq. (5), we have ds = 9 and curlz (F) = x + y for 1 x + y 4. C1 A= 1 x dy y dx 2C We parametrize the circle by x = 3 cos , y = 3 sin , hence, x d y y d x = 3 cos 3 cos d 3 sin (3 sin ) d = (9 cos2 + 9 sin2 ) d = 9 d Therefore, A= 1 1 2 9 x dy y dx = 9 d = 2 = 9. 2C 20 2 15. The region between the x-axis and the cycloid parametrized by c(t) = (t sin t, 1 cos t) for 0 t 2 (Figure The triangle with vertices (0, 0), (1, 0), and (1, 1) 17) y 1 x 2 FIGURE 17 Cycloid. SOLUTION By Eq. (5), the area is the following integral: A= 1 x dy y dx 2C where C is the closed curve determined by the segment O A and the cycloid . y 1 O A = (2 , 0) x S E C T I O N 18.1 Greens Theorem (ET Section 17.1) 641 Therefore, A= 1 1 x dy y dx + 2 OA 2 x dy y dx (1) We compute the two integrals. The segment O A is parametrized by t, 0 , t = 0 to t = 2. Hence, x = t and y = 0. Therefore, x d y y d x = t 0 dt 0 dt = 0 OA x dy y dx = 0 (2) On we have x = t sin t and y = 1 cos t, therefore x d y y d x = (t sin t) sin t dt (1 cos t)(1 cos t) dt = (t sin t sin2 t 1 + 2 cos t cos2 t) dt = (t sin t + 2 cos t 2) dt Hence, x dy y dx = 0 2 (t sin t + 2 cos t 2) dt = 2 0 2 0 (2 2 cos t t sin t) dt 2 0 = 2t 2 sin t + t cos t sin t Substituting (2) and (3) in (1) we get A= = 2t 3 sin t + t cos t = 6 (3) 1 1 0 + 6 = 3. 2 2 17. Let x 3 + y 3 = 3x y be the folium of Descartes (Figure 18). The region between the graph of y = x 2 and the x-axis for 0 x 2 y 2 2 2 x 2 FIGURE 18 Folium of Descartes. (a) Show that the folium has a parametrization in terms of t = y/x given by x= (b) Show that x dy y dx = Hint: By the Quotient Rule, x2 d (c) Find the area of the loop of the folium. SOLUTION 2 (a) We show that x = 3t 3 , y = 3t 3 satisfy the equation x 3 + y 3 3x y = 0 of the folium: 3t , 1 + t3 y= 3t 2 1 + t3 ( < t < ) (t = 1) 9t 2 dt (1 + t 3 )2 y x = x dy y dx 1+t 1+t x 3 + y 3 3x y = 3 3t + 3 1+t 3t 2 1 + t3 3 3 3t 3t 2 3 1 + t3 1+t 27t 3 1 + t 3 (1 + t 3 ) 3 (1 + t 3 ) = 27t 3 + 27t 6 3 (1 + t 3 ) 27t 3 (1 + t 3 ) 3 (1 + t 3 ) = = 0 (1 + t 3 ) 3 =0 642 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) 2 This proves that the curve parametrized by x = 3t 3 , y = 3t 3 lies on the folium of Descartes. This parametrization 1+t 1+t parametrizes the whole folium since the two equations can be solved for t in terms of x and y. That is, x= 3t 1 + t3 3t 2 y= 1 + t3 t= y x A glance at the graph of the folium shows that any line y = t x, with slope t, intersects the folium exactly once. Thus, there is a one-to-one relationship between the values of t and the points on the graph. y (b) We differentiate the two sides of t = x with respect to t. Using the Quotient Rule gives 1= or x This equality can be written in the form x dy y dx = 9t 2 (1 + t 3 ) 2 x dy y dx dt dt x2 2 3t 3 1+t dx dy y = x2 = dt dt dt (c) We use the formula for the area enclosed by a closed curve and the result of part (b) to nd the required area. That is, A= 9t 2 1 1 x dy y dx = dt 2C 2 0 (1 + t 3 )2 From our earlier discussion on the parametrization of the folium, we see that the loop is traced when the parameter t is increasing along the interval 0 t < . We compute the improper integral using the substitution u = 1 + t 3 , du = 3t 2 dt. This gives A= = 3 R 1+R 3 3 du 9t 2 1 3 1 1+R 1 lim lim lim dt = = 2 R 0 (1 + t 3 )2 2 R 1 2 R u u=1 u2 3 1 lim 1 2 R 1 + R3 = 3 3 (1 0) = 2 2 y 2 t= C x 2 2 2 t=0 19. Show that if C is a simple closed curve, then Follow the procedure of Exercise 17 to nd the area of the loop of the lemniscate curve with equation (x 2 + 2 )2 = x y (Figure 19). y y d x = x dy C C and both integrals are equal to the area enclosed by C. SOLUTION P = y and Q = x, we have Q = 1 and P = 1. Therefore, the cross partials are equal and therefore F is conservative. x y By the formula for the area enclosed by a simple closed curve, the area enclosed by C is A= 1 1 1 x dy y dx = x dy + y d x 2C 2C 2C We show that C y d x + x d y = 0 by showing that the vector eld F = y, x is conservative. Indeed, since Using the equality obtained above, we have A= 1 1 x dy + x dy = x dy = y d x. 2C 2C C C For the vector elds (A)(D) in Figure 20, state whether the curl at the origin appears to be positive, negative, or zero. S E C T I O N 18.1 Greens Theorem (ET Section 17.1) 643 21. Let F = 2xe y , x + x 2 e y and let C be the quarter-circle path from A to B in Figure 21. Evaluate I = follows: (a) Find a function (x, y) such that F = G + , where G = 0, x . (b) Show that the line integrals of G along the segments O A and O B are zero. (c) Use Greens Theorem to show that I = (B) (A) + 4 and evaluate I . y B = (0, 4) C F ds as O A = (4, 0) x FIGURE 21 SOLUTION (a) We need to nd a potential function (x, y) for the difference F G = 2xe y , x + x 2 e y 0, x = 2xe y , x 2 e y We let (x, y) = x 2 e y . (b) We use the parametrizations AO : t, 0 , 0 t 4 and O B : 0, t , 0 t 4 to evaluate the integrals of G = 0, x . We get G ds = G ds = 4 0 4 0 y B = (0, 4) OA 0, t 1, 0 dt = 0, 0 0, 1 dt = 4 0 4 0 0 dt = 0 0 dt = 0 OB O A = (4, 0) x (c) Since F G = , we have C (F G) ds = (B) (A) = C F ds C G ds = I C G ds That is, I = (B) (A) + To compute the line integral on the right-hand side, we rewrite it as C C G ds (1) G ds = B O+O A+C G ds BO G ds OA G ds Using part (b) we may write C G ds = B O+O A+C G ds (2) We now use Greens Theorem. Since G = 0, x , we have P = 0 and Q = x, hence Q P = 1 0 = 1. Thus, x y G ds = 1 d A = Area(D) = B O+O A+C D 42 = 4 4 (3) 644 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) Combining (1), (2), and (3), we obtain I = (B) (A) + 4 Since (x, y) = x 2 e y , we conclude that I = (0, 4) (4, 0) + 4 = 0 42 e0 + 4 = 4 16. y B = (0, 4) C D O A = (4, 0) x 23. Evaluate I =the (sin xintegral x +F(3x + 3 , 4x for the the path from AABCD in Figure 22. Hint: Usesave method of Compute line + y) d of = x y) d y along nonclosed path to B in Figure 23. Hint: To the work, use C Greens Exercise 22. Theorem to relate this line integral to the line integral along the vertical path from B to A. y D = (0, 6) (2, 4) (2, 2) A = (0, 0) x FIGURE 23 SOLUTION y D = (0, 6) C D B = (2, 2) A = (0, 0) x C = (2, 4) Let F = sin x + y, 3x + y , hence P = sin x + y and Q = 3x + y. We denote by C1 the closed path determined by C and the segment D A. Then by Greens Theorem, P dx + Q dy = Q P x y dA = (3 1) d A = 2 d A = 2 Area(D) (1) C1 D D D The area of D is the area of the trapezoid ABC D, that is, Area(D) = BC + AD h (2 + 6) 2 = = 8. 2 2 y D = (0, 6) C = (2, 4) h B = (2, 2) A = (0, 0) x Combining with (1) we get C1 P d x + Q d y = 2 8 = 16 Using properties of line integrals, we have C P dx + Q dy + DA P d x + Q d y = 16 (2) S E C T I O N 18.1 Greens Theorem (ET Section 17.1) 645 We compute the line integral over D A, using the parametrization D A : x = 0, y = t, t varies from 6 to 0. We get P dx + Q dy = = 0 6 0 6 DA F(0, t) 0 d sin 0 + t, 3 0 + t 0, 1 dt 0, t dt = dt 6 0 6 t, t 0, 1 dt = t dt = t2 0 = 18 2 t=6 We substitute in (2) and solve for the required integral: C P d x + Q d y 18 = 16 or C P d x + Q d y = 34. 25. Let F be the velocity eld. Estimate the circulation of F around a circle of radius R = 0.05 with center P, assuming Estimate the circulation of a vector eld F around a circle of radius R = 0.1, assuming that curlz (F) takes the that curlz (F)(P) = 3. In which direction would a small paddle placed at P rotate? How fast would it rotate (in radians value 4 at the center of the circle. per second) if F is expressed in meters per second? SOLUTION We use the following estimation: F ds curl(F)(P)Area(D) C D P 0.05 C (1) We are given that curl(F)(P) = 3. Also, the area of the disk of radius R = 0.05 is 0.052 = 0.0025. Therefore, we obtain the following estimation: C F ds 3 0.0025 0.024. Since the curl is negative, the paddle would rotate in the clockwise direction. Using the formula |curl(F)| = 2, we see that the angular speed is = 1.5 radians per second. 27. Referring to Figure 25, suppose that Referring to Figure 24, suppose that that curlz (F) = 3 in D. C2 F ds = 12. Use Greens Theorem to determine F ds, assuming C1 F C2 = 3, ds F ds = 4 C3 Use Greens Theorem to determine the circulation of F around C1 , assuming that curlz (F) = 9 on the shaded region. C1 D 1 5 C3 C2 1 FIGURE 25 We must calculate C1 F ds. We use Greens Theorem for the region D between the three circles C1 , C2 , and C3 . Because of orientation, the line integrals C2 F ds = C2 F ds and C3 F ds = C3 F ds must be used in applying Greens Theorem. That is, SOLUTION C1 F ds C2 F ds C3 F ds = D curl(F) d A We substitute the given information to obtain C1 F ds 3 4 = D 9dA = 9 D 1 d A = 9 Area(D) (1) 646 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) The area of D is computed as the difference of areas of discs. That is, Area(D) = 52 12 12 = 23 We substitute in (1) and compute the desired circulation: C1 F ds 7 = 9 23 or C1 F ds = 214. 29. Use the result of Exercise 28 to compute the areas of the polygons in Figure 26. Check your result for the area of the triangle Area of a Polygon Greens Theorem leads to a convenient formula for the area of a polygon. in (A) using geometry. (a) Let C be the line segment joining (x 1 , y1 ) to (x 2 , y2 ). Show that y y 5 4 3 the 2 1 (3, 5) 1 1 (x y d x + x d y = 5 1 y2 x 2 y1 ) 2C 24 (2, 3) (5, 3) (b) Prove that the area of (x n+1 , yn+1 ) = (x 1 , y1 )] to y polygon with vertices (x 1 , 2 1 ), (x 2 , y2 ), . . . , (x n , yn ) is equal [where we set (2, 1) 1 2 3 4 (5, 1) 51 (1, 1) 1 x (3, 2) 3 3 (1, 3) n (A) (xi yi+1 xi+1 yi ) (B) 2 i=1 FIGURE 26 3 2 1 x 1 2 4 5 SOLUTION (a) The vertices of the triangle are (x 1 , y1 ) = (x 4 , y4 ) = (2, 1), y 5 4 3 2 1 1 (x 2 , y2 ) = (5, 1), (x 3 , y3 ) = (2, 3) (2, 3) (2, 1) 2 3 4 (5, 1) 5 x Using the formula obtained in Exercise 28, the area of the triangle is the following sum: A= 1 (x 1 y2 x 2 y1 ) + (x 2 y3 x 3 y2 ) + (x 3 y1 x 1 y3 ) 2 1 1 = (2 1 5 1) + (5 3 2 1) + (2 1 2 3) = (3 + 13 4) = 3 2 2 We verify our result using the formula for the area of a triangle: A= (b) The vertices of the polygon are (x 1 , y1 ) = (x 6 , y6 ) = (1, 1) (x 2 , y2 ) = (1, 3) (x 3 , y3 ) = (3, 2) (x 4 , y4 ) = (5, 3) (x 5 , y5 ) = (3, 5) y (3, 5) 5 4 (1, 3) 3 2 (3, 2) (1, 1) 1 1 2 3 4 1 1 bh = (5 2) (3 1) = 3 2 2 (5, 3) 3 2 1 x 5 S E C T I O N 18.1 Greens Theorem (ET Section 17.1) 647 Using the formula in part (a), the area of the polygon is the following sum: A= 1 (x 1 y2 x 2 y1 ) + (x 2 y3 x 3 y2 ) + (x 3 y4 x 4 y) + (x 4 y5 x 5 y4 ) + (x 5 y1 x 1 y5 ) 2 1 (1 3 1 1) + (1 2 3 3) + (3 3 5 2) + 5 5 (3) 3 + 3 1 (1) 5 = 2 1 = (4 7 1 + 34 + 2) = 12 2 Further Insights and Challenges In Exercises 3031, let F be the vortex vector eld [dened for (x, y) = (0, 0)]: F= y x , x 2 + y2 x 2 + y2 31. Prove that that curlz (F)(P) closed curve whose 0). F ds = 2 (Figure 27). Hint: (a) Show if C is a simple = 0 for all P = (0, interior contains the origin, then C Apply Greens that the circulation of F around the and C R R of radius so centered atC R is contained in C. 2 for all R. (b) Show Theorem to the domain between C circle C where R is R small that the origin is equal to y C x CR FIGURE 27 SOLUTION Let R > 0 be sufciently small so that the circle C R is contained in C. y C x CR D Let D denote the region between C R and C. We apply Greens Theorem to the region D. The curve C is oriented counterclockwise and C R is oriented clockwise. We have C F ds + CR F ds = D curl(F) d A (1) From Exercise 30(b) we know that C R F ds = 2. Since D does not contain the origin, we have by part (a) of Exercise 30, curl (F) = 0 on D. Substituting in (1) we obtain C F ds 2 = D 0dA = 0 or C F ds = 2. In Exercises 3234, the conjugate of a vector eld F = P, Q is the vector eld F = Q, P . 33. The normal component of F at a point P on a simple closed C is the quantity F(P) n(P), where n(P) is the Explain the following statement: F of F vector eld obtained by rotating the vectors the counterclockwise outward-pointing unit normal vector. The uxis the across curve C is dened as the line integral of of Fnormal component through an angle of . 2 F ds. around C (Figure 28). Show that the ux across C is equal to C 648 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) P n(P) FIGURE 28 The ux of F is the integral of the normal component F n around the curve. SOLUTION We must show that if C is a simple closed curve, then C (F n)ds = C F ds By the denition of the vector line integral, the line integral of the vector eld F over C is C F ds = C (F T) ds (1) In Exercise 32 we showed that F is a rotation of F by counterclockwise. Therefore, the angle between F and the 2 tangent T is equal to the angle between F and the normal n. Also, F = F . Hence, F T = F Fn= F Since the dot products are equal, we conclude that C T cos = F cos n cos = F cos (F T) ds = C (F n)ds (2) Combining (1) and (2) we obtain C F ds = T q C F* q (F n)ds. n F In Exercises 3538, the Laplaceoperator is dened by Q P + . Use Greens Theorem to prove that for any simple closed curve C, Dene div(F) = x y 2 2 = +2 9 2 x y Flux across C = div(F) d A D 35. Let F = . Show that curlz (F ) = , where F is the conjugate vector eld (dened in Exercises 3234). where For a region enclosed P, Q , the conjugate vector eld is F = the P . By the Theorem discussed SOLUTIOND is the vector eld F = by C. This is a two-dimensional version ofQ,Divergence given information, in Section 18.3. , F = = F = , x y y x We compute the curl of F : curl(F ) = x x y y = 2 2 + 2= 2 x y 37. Let P = (a, b) and let C(r ) be the circle of radius r centered at P. The average value of a continuous function on Let n denote the outward-pointing unit normal vector to a simple closed C. The normal derivative of a function C(r ) is dened as the integral , is the directional derivative Dn () = n. Prove the formula , denoted 2 n 1 (a + r cos , b + r sin ) d I (r ) = 2 0 ds = dA C n D (a) Show that where D is the domain enclosed+ r C. Hint: Let F =) = Show that cos , b +rT, where T is the unit tangent vector (a by cos , b + r sin . (a + r n = F sin ) n r pointing in the counterclockwise direction along C, and apply Greens Theorem. S E C T I O N 18.1 Greens Theorem (ET Section 17.1) 649 (b) Use differentiation under the integral sign to prove that 1 d I (r ) = ds dr 2r C(r) n (c) Use Exercise 36 to conclude that d 1 I (r ) = dr 2r where D(r ) is the interior of C(r ). SOLUTION In this solution, r (a + r cos , b + r sin ) denotes the partial derivative r computed at (a + r cos , b + r sin ), whereas r (a + r cos , b + r sin ) is the derivative of the composite function. D(r) dA (a) Since n = n, we rst express the gradient vector in terms of polar coordinates. We use the Chain Rule and the derivatives: x = We get sin , r y = cos , r r x = cos , r y = sin x = r r x + x = r cos + y = r r y + y = r sin + Hence, = r cos We use the following parametrization for C(r ): C(r ) : c() = a + r cos , b + r sin , The unit normal vector is n = cos , sin . We compute the dot product: sin r (1) cos r sin cos , r sin + r r 0 2 sin cos = n = r cos , r sin + cos , sin n r r = r cos2 That is, (a + r cos , b + r sin ) = r (a + r cos , b + r sin ) n (b) We compute the following derivative using the Chain Rule and (1): (a + r cos , b + r sin ) = x (a + r cos ) + y (b + r sin ) r r r = r cos = r cos2 sin r cos + r sin + cos r sin (2) sin cos cos sin + r sin2 + = r cos2 + sin2 = r r r sin cos cos sin + r sin2 + r r = r (a + r cos , b + r sin ) We now differentiate I (r ) under the integral sign, and use (3) and (2) to obtain 2 2 1 1 d I (r ) = (a + r cos , b + r sin ) d = r (a + r cos , b + r sin ) d dr 2 0 r 2 0 (3) = 2 1 (a + r cos , b + r sin ) d 2 0 n (4) 650 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) On the other hand, since c () = r sin , r cos , we have 2 2 ds = (a + r cos , b + r sin ) c () d = (a + r cos , b + r sin )r d n n n C(r) 0 0 =r 2 0 n (a + r cos , b + r sin ) d (5) Combining (4) and (5) we get d 1 I (r ) = ds. dr 2r C(r) n (c) We combine the result of part (b) and Exercise 36 to conclude d 1 1 I (r ) = ds = dr 2r C(r) n 2r D(r) d A. In Exercises 3940, let) D be (r ) region bounded m(r ) simple closed curve C. A function (x, y) on D (whose on C(r ). Prove that m(r I the M(r ), where by a and M(r ) are the minimum and maximum values of secondorder partial derivatives exist and are continuous) I (r ) = (P). Then use the continuity of to prove that lim is called harmonic if = 0, where is the Laplace operator r 0 dened in Eq. (9). 39. Use the results of Exercises 37 and 38 to prove the mean-value property of harmonic functions: If is harmonic, then I (r ) = (P) for all r . SOLUTION In Exercise 37 we showed that d 1 I (r ) = dr 2r D dA If is harmonic, = 0. Therefore the right-hand side of the equality is zero, and we get d I (r ) = 0 dr We conclude that I (r ) is constant, that is, I (r ) has the same value for all r . The constant value is determined by the limit limr 0 I (r ) = (P) obtained in Exercise 38. That is, I (r ) = (P) for all r . Show that f (x, y) = x 2 y 2 is harmonic. Verify the mean-value property for f (x, y) directly [expand f (a + r cos , b + r sin ) as a function of and compute I (r )]. Show that x 2 + y 2 is not harmonic and does not satisfy 18.2 Stokes Theorem (ET Section 17.2) the mean-value property. Preliminary Questions 1. Indicate with an arrow the boundary orientation of the boundary curves of the surfaces in Figure 13, oriented by the outward-pointing normal vectors. n (A) n (B) FIGURE 13 SOLUTION The indicated orientation is dened so that if the normal vector is moving along the boundary curve, the surface lies to the left. Since the surfaces are oriented by the outward-pointing normal vectors, the induced orientation is as shown in the gure: n (A) n (B) 2. Let F = curl(A). Which of the following are related by Stokes Theorem? (a) The circulation of A and ux of F. (b) The circulation of F and ux of A. S E C T I O N 18.2 SOLUTION Stokes Theorem (ET Section 17.2) 651 Stokes Theorem states that the circulation of A is equal to the ux of F. The correct answer is (b). A vector eld A such that F = curl(A) is a vector potential for F. 3. What is the denition of a vector potential? SOLUTION 4. Which of the following statements is correct? (a) The ux of curl(F) through every oriented surface is zero. (b) The ux of curl(F) through every closed, oriented surface is zero. SOLUTION Statement (b) is the correct statement. The ux of curl(F) through an oriented surface is not necessarily zero, unless the surface is closed. 5. Which condition on F guarantees that the ux through S1 is equal to the ux through S2 for any two oriented surfaces S1 and S2 with the same oriented boundary? SOLUTION If F has a vector potential A, then by a corollary of Stokes Theorem, F ds = A ds S C Therefore, if two oriented surfaces S1 and S2 have the same oriented boundary curve, C, then S1 F ds = C A ds and S2 F ds = C A ds Hence, S1 F ds = S2 F ds Exercises In Exercises 14, calculate curl(F). 1. F = z y 2 , x + z 3 , y + x 2 SOLUTION We have i j y x + z3 k z y + x2 = (1 3z 2 )i (2x 1)j + (1 + 2y)k = 1 3z 2 , 1 2x, 1 + 2y curl(F) = x z y2 3. F = e y , sin x, cos x yyz F = We have SOLUTION x , z , x i curl(F) = x ey j y sin x k z cos x = 0i ( sin x)j + (cos x e y )k = 0, sin x, cos x e y In Exercises 58, x verify Stokes Theorem for the given vector eld and surface, oriented with an upward-pointing normal. y , ,0 F= 2 x x,+ y 2 z x 2 + y 2 5. F = 2x y, y + , the surface z = 1 x 2 y 2 for x 2 + y 2 1 SOLUTION We must show that F ds = curl (F) dS C S 652 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) Step 1. Compute the line integral around the boundary curve. The boundary curve C is the unit circle oriented in the counterclockwise direction. We parametrize C by (t) = (cos t, sin t, 0), Then, F (t) = 2 cos t sin t, cos t, sin t 0 t 2 (t) = sin t, cos t, 0 F (t) (t) = 2 cos t sin t, cos t, sin t sin t, cos t, 0 = 2 cos t sin2 t + cos2 t We obtain the following integral: F ds = 2 0 C 2 cos t sin2 t + cos2 t dt = t sin 2t 2 2 sin3 t ++ = 3 2 40 (1) Step 2. Compute the ux of the curl through the surface. We parametrize the surface by (, t) = t cos , t sin , 1 t 2 , We compute the normal vector: T = = t sin , t cos , 0 = cos , sin , 2t Tt = t i t sin cos j t cos sin k 0 2t = (2t 2 cos )i (2t 2 sin )j t (sin2 + cos2 )k 0 t 1, 0 2 T Tt = = (2t 2 cos )i (2t 2 sin )j tk Since the normal is always supposed to be pointing upward, the z-coordinate of the normal vector must be positive. Therefore, the normal vector is n = 2t 2 cos , 2t 2 sin , t We compute the curl: i x 2x y j y x k z y+z curl(F) = = i + (1 2x)k = 1, 0, 1 2x We compute the curl in terms of the parameters: curl(F) = 1, 0, 1 2t cos Hence, curl(F) n = 1, 0, 1 2t cos 2t 2 cos , 2t 2 sin , t = 2t 2 cos + t 2t 2 cos = t The surface integral is thus curl(F) dS = 2 0 0 1 S t dt d = 2 1 0 t dt = 2 t2 1 = 20 (2) The values of the integrals in (1) and (2) are equal, as stated in Stokes Theorem. 7. F = e yz , 0, 0 , the square with vertices (1, 0, 1), (1, 1, 1), (0, 1, 1), and (0, 0, 1) y x F = yz, 0, x , the portion of the plane + + z = 1 where x, y, z 0 SOLUTION 2 3 Step 1. Compute the integral around the boundary curve. The boundary consists of four segments C1 , C2 , C3 , and C4 shown in the gure: S E C T I O N 18.2 z (0, 0, 1) C1 (1, 0, 1) C2 (1, 1, 1) y x S C3 C4 Stokes Theorem (ET Section 17.2) 653 (0, 1, 1) We parametrize the segments by C1 : 1 (t) = (t, 0, 1), C2 : 2 (t) = (1, t, 1), 0t 1 0t 1 0t 1 0t 1 C3 : 3 (t) = (1 t, 1, 1), C4 : 4 (t) = (0, 1 t, 1), We compute the following values: F 1 (t) = e yz , 0, 0 = e1 , 0, 0 F 2 (t) = e yz , 0, 0 = et1 , 0, 0 F 3 (t) = e yz , 0, 0 = 1, 0, 0 F 4 (t) = e yz , 0, 0 = et1 , 0, 0 Hence, F 1 (t) 1 (t) = e1 , 0, 0 1, 0, 0 = e1 F 2 (t) 2 (t) = et1 , 0, 0 0, 1, 0 = 0 F 3 (t) 3 (t) = 1, 0, 0 1, 0, 0 = 1 F 4 (t) 4 (t) = et1 , 0, 0 0, 1, 0 = 0 We obtain the following integral: 4 C F ds = i=1 Ci F ds = 1 0 e1 dt + 0 + 1 0 (1) dt + 0 = e1 1 Step 2. Compute the curl. i x e yz j y 0 k z 0 curl(F) = = e yz j e yz k = 0, e yz , e yz Step 3. Compute the ux of the curl through the surface. We parametrize the surface by (x, y) = (x, y, 1), 0 x, y 1 The upward pointing normal is n = 0, 0, 1 . We express curl(F) in terms of the parameters x and y: curl(F) ( (x, y)) = 0, e y1 , e y1 Hence, curl(F) n = 0, e y1 , e y1 0, 0, 1 = e y1 The surface integral is thus curl(F) dS = e y1 d A = 1 0 0 1 S D e y1 d y d x = 1 0 e y1 d y = e y1 1 0 654 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) = 1 + e1 = e1 1 We see that the integrals in (1) and (2) are equal. In Exercises 910, use Stokes Theorem to compute the 2 of curl(F) through the given surface. ux F = y, 2x, x + z , the upper hemisphere x + y 2 + z 2 = 1, z 0 9. F = z, y, x , SOLUTION (1) the hemisphere x 2 + y 2 + z 2 = 1, x 0 By Stokes Theorem, F ds = curl (F) dS C S We compute the line integral. The boundary curve, which is the unit circle in the (y, z) plane, is parametrized by (t) = (0, cos t, sin t), We compute the following values: F (t) = z, y, x = sin t, cos t, 0 0 t 2 (t) = 0, sin t, cos t 1 F (t) (t) = sin t, cos t, 0 0, sin t, cos t = sin t cos t = sin 2t 2 Hence, F ds = 2 0 C F (t) (t) dt = 2 0 cos 2t 2 1 sin 2t dt = =0 2 40 Then also S curl(F) dS = 0 11. Let S be the surface of the cylinder (not including the top and bottom) of radius 2 for 1 z 6, oriented with F = x 2 + y 2 , x (Figure the outward-pointing normal+ z 2 , 0 ,14). part of the cone z 2 = x 2 + y 2 such that 2 z 4 (a) Indicate with an arrow the orientation of S (the top and bottom circles). (b) Verify Stokes Theorem for S and F = yz 2 , 0, 0 . z 6 1 x y FIGURE 14 SOLUTION (a) The induced orientation is dened so that as the normal vector travels along the boundary curve, the surface lies to its left. Therefore, the boundary circles on top and bottom have opposite orientations, which are shown in the gure. z 6 C1 n 1 x C2 y (b) We verify Stokes Theorem for S and F = yz 2 , 0, 0 . Step 1. Compute the integral around the boundary circles. We use the following parametrizations: C1 : 1 (t) = (2 cos t, 2 sin t, 6), C2 : 2 (t) = (2 cos t, 2 sin t, 1), t from 2 to 0 t from 0 to 2 S E C T I O N 18.2 Stokes Theorem (ET Section 17.2) 655 We compute the following values: F 1 (t) = yz 2 , 0, 0 = 72 sin t, 0, 0 , 1 (t) = 2 sin t, 2 cos t, 0 F 1 (t) 1 (t) = 72 sin t, 0, 0 2 sin t, 2 cos t, 0 = 144 sin2 t F 2 (t) = yz 2 , 0, 0 = 2 sin t, 0, 0 , 2 (t) = 2 sin t, 2 cos t, 0 F 2 (t) 2 (t) = 2 sin t, 0, 0 2 sin t, 2 cos t, 0 = 4 sin2 t The line integral is thus F ds = = Step 2. Compute the curl i x yz 2 j y 0 k z 0 F ds + F ds = 0 2 C C1 C2 (144 sin2 t) dt + 2 0 (4 sin2 t) dt 70 sin 2t 2 = 140 2 0 2 0 140 sin2 t dt = 140 2 1 cos 2t 0 2 dt = 70 2 curl(F) = = (2yz)j z 2 k = 0, 2yz, z 2 Step 3. Compute the ux of the curl through the surface. We parametrize S by (, z) = (2 cos , 2 sin , z), 0 2, 1z6 In Example 2 of Chapter 17.4 it is shown that the outward pointing normal is n = 2 cos , 2 sin , 0 We compute the dot product: curl(F) ( (, z)) n = 0, 4z sin , z 2 2 cos , 2 sin , 0 = 8z sin2 We obtain the following integral (and use the integral we computed before): curl(F) dS = 6 1 0 2 S 8z sin2 d dz = 6 2 8z dz 1 0 sin2 d = 4z 2 6 1 = 140 The line integral and the ux have the same value. This veries Stokes Theorem. 13. LetLet Sthe ux portion ofy , 2xe x , z 2 through the upper hemisphere S of the radius R depicted in Figure 15. Use I be be the of F = e the plane z = x contained in the half-cylinder of unit sphere. Stokes e y , 2xe 2 0 . Find vector eld A such that curl(A) 2z around the boundary of S (a half-ellipse) in the (a) Let G =Theorem xto, calculateathe circulation of F = z, x, y += G. counterclockwise direction when viewed from above. Hint:is zero. that curl(F) is orthogonal to theof A around S. (b) Use Stokes Theorem to show that the ux of G through S Show Hint: Calculate the circulation normal vector to the plane. (c) Calculate I . Hint: Use (b) to show that I is equal to the ux of 0, 0, z 2 through S. SOLUTION 2 (a) We search for a vector eld A so that G = curl(A). That is, 2 A3 A2 A1 A3 A2 A1 , , = e y , 2xe x , 0 y z z x x y We note that the third coordinate of this curl vector must be zero; this can be satised if A1 = 0 and A2 = 0. With this 2 2 in mind, we let A = 0, 0, e y e x . The vector eld A = 0, 0, e y e x satises this equality. Indeed, A3 A2 = ey , y z 2 A3 A1 = 2xe x , z x A1 A2 =0 x y 656 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S 2 (ET CHAPTER 17) (b) We found that G = curl(A), where A = 0, 0, e y e x . We compute the ux of G through S. By Stokes Theorem, G dS = curl(A) dS = A ds S S C The boundary C is the circle x 2 + y 2 = 1, parametrized by (t) = (cos t, sin t, 0), z 0 t 2 y C x We compute the following values: A (t) = 0, 0, e y e x 2 2 = 0, 0, esin t ecos t (t) = sin t, cos t, 0 2 A (t) (t) = 0, 0, esin t ecos t sin t, cos t, 0 = 0 Therefore, A ds = 2 0 C 2 0 dt = 0 (c) We rewrite the vector eld F = e y , 2xe x , z 2 as F = e y , 2xe x , z 2 = e y , 2xe x , 0 + 0, 0, z 2 = curl(A) + 0, 0, z 2 Therefore, S 2 2 F dS = S curl(A) dS + S 0, 0, z 2 dS (1) In part (b) we showed that the rst integral on the right-hand side is zero. Therefore, S F dS = S 0, 0, z 2 dS (2) The upper hemisphere is parametrized by (, ) = (cos sin , sin sin , cos ), with the outward pointing normal n = sin cos sin , sin sin , cos See Example 4, Section 17.4. We have 0, 0, cos2 n = sin cos3 Therefore, 0, 0, z 2 dS = 2 0 0 0 2, 0 . 2 /2 S sin cos3 d d = 2 /2 0 sin cos3 d = 2 Combining with (2) we obtain the solution cos4 /2 = (0 1) = 4 2 2 0 S F dS = . 2 S E C T I O N 18.2 Stokes Theorem (ET Section 17.2) 657 15. Let A be the vector potential and B the magnetic eld of the innite solenoid of radius R in Example 5. Use Stokes Theorem toF = 0, z, 1 and let S be the spherical cap x 2 + y 2 + z 2 1, where z 1 . Evaluate Let compute: F dS directly 2 S (a) The ux of B through Then verify thaty-plane of radius r < R = (0, x, x z) and evaluate the surface integral again as a surface integral. a circle in the x F = curl(A), where A (b) The circulation of A around the boundary C of a surface lying outside the solenoid using Stokes Theorem. SOLUTION (a) In Example 5 it is shown that B = curl(A), where 1 2 R B y , x ,0 2 r2 r2 A= 1 B y, x, 0 2 if if r>R (1) r<R Therefore, using Stokes Theorem, we have (S is the disk of radius r in the x y-plane) S B dS = S curl(A) dS = y S A ds (2) S r x We parametrize the circle C = S by c(t) = r cos t, r sin t, 0 , 0 t 2. Then c (t) = r sin t, r cos t, 0 By (1) for r < R, A (c(t)) = Hence, A (c(t)) c (t) = Now, by (2) we get B dS = A dS = 2 1 1 r 2 B dt = r 2 B 2 0 1 B r sin t, r cos t, 0 2 1 1 1 B r sin t, r cos t, 0 r sin t, r cos t, 0 = B r 2 sin2 t + r 2 cos2 t = r 2 B 2 2 2 S S 0 2 2 dt = r 2 B (b) Outside the solenoid B is the zero eld, hence B = 0 on every domain lying outside the solenoid. Therefore, Stokes Theorem implies that S A dS = S curl(A) dS = S B dS = S 0 dS = 0. 17. (a) (b) A uniform magnetic eld B has constant strength b teslas in the z-direction [i.e., B = 0, 0, b ]. The magnetic eld B due to a small current loop (which we place at the origin) is called a magnetic dipole (Figure Verify that = (x 21+ y 2 r is 2 )vector potential for = curl(A), where y, 0 . 16). Let A = 2 B + z a 1/2 . For large, B B, where r = x, Calculate the ux of B through the rectangle with vertices A, B, C, and D in Figure 17. yx A = 3, 3,0 (a) Let C be a horizontal circle of radius R located far from the origin with center on z-axis. Show that A is tangent to C. (b) Use Stokes Theorem to calculate the ux of B through C. FIGURE 17 SOLUTION 658 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) (a) We compute the vector A = 1 B r. Since B = bk and r = xi + yj, we have 2 A= 1 1 1 by bx 1 B r = bk (xi + yj) = b(xk i + yk j) = b(xj yi) = , ,0 2 2 2 2 22 We now show that curl(A) = B. We compute the curl of A: i x by 2 j y bx 2 k z 0 curl (A) = = 0, 0, b b + = 0, 0, b = B 2 2 Therefore, A is a vector potential for B. (b) By Stokes Theorem the ux of B through the surface S is S B dS = S curl(A) dS = C A ds (1) We compute the line integral. The unit circle C is parametrized by c(t) = cos t, sin t, 0 , In part (a) we found that A = by , bx , 0 . Hence, 22 A (c(t)) c (t) = b b sin t b cos t b b b , , 0 sin t, cos t, 0 = sin2 t + cos2 t + 0 = sin2 t + cos2 t = 2 2 2 2 2 2 0 t < 2 We obtain the following line integral: A ds = 2 b 0 C 2 dt = b 2 = b 2 (2) We combine (1) and (2) to conclude that the ux of B through S is S B dS = b 19. Let F = z 2 , 2zx, 4y x 2 and let C be a simple closed curve in the plane x + y + z = 4 that encloses a region of Let F = x 2 y, x, 0 . Referring to Figure 17, let C be the closed path ABC D. Use Stokes Theorem to evaluate F ds, where C is oriented in the counterclockwise direction (when viewed from above area 16 (Figure 18). Calculate F ds in two ways. First, regard C as the boundary of the rectangle with vertices A, B, C, and D. Then treat C as C C the plane). the boundary of the wedge-shaped box with open top. z 4 x+y+z=4 C 4 4 x y FIGURE 18 SOLUTION We denote by S the region enclosed by C. Then by Stokes Theorem, F ds = S c curl(F) ds (1) We compute the curl of F = z 2 , 2zx, 4y x 2 : i x z 2 j y 2zx k z 4y x 2 curl(F) = = 4 2x, 2x 2z, 2z The plane x + y + z = 4 has the parametrization (x, y) = x, y, 4 x y S E C T I O N 18.2 Stokes Theorem (ET Section 17.2) 659 Hence, = 1, 0, 1 0, 1, 1 = (i k)(j k) = k + j + i = 1, 1, 1 x y The normal determined by the induced orientation is n = 1, 1, 1 Let D be the parameter domain in the parametrization (x, y) = (x, y, 4 x y) of S; that is, D will be the base triangle in the x y plane that lies underneath the pyramid in the picture. To compute the surface integral in (1) we compute the values curl(F) ( (x, y)) = 4 2x, 2x 2(4 x y), 2(4 x y) = 4 2x, 8 + 4x + 2y, 8 2x 2y curl(F) n = 4 2x, 8 + 4x + 2y, 8 2x 2y 1, 1, 1 = 4 2x 8 + 4x + 2y + 8 2x 2y = 4 Therefore, using (1) and (2) we obtain C F ds = S curl(F) dS = D 4 d A = 4Area(D) = 4 16 = 32 2 y z x 21. Let C be the triangular boundary of the portion of the plane + + = 1 lying in the octant x, y, z 0. Use Let F = y 2 , x 2 , z 2 . Show that a b c Stokes Theorem to nd positive constants a, b, c such that the line integral of F = y 2 , 2z + x, 2y 2 around C is zero. Hint: Choose constants so that curl(F) is orthogonal toF ds = vector. the normal F ds SOLUTION C1 C2 for any two closed curves lying on a cylinder whose central axis is the z-axis (Figure 19). z c C b a x y We compute the curl: i x y2 j y 2z + x k z 2y 2 curl(F) = = (4y 2)i + (1 2y)k = 4y 2, 0, 1 2y 11 The outward-pointing normal to the plane is n = a , b , 1 . Hence, c curl(F) n = 4y 2, 0, 1 2y 111 4y 2 1 2y ,, + = abc a c Note that if we choose a = 2 and c = 1, then curl(F) n = 0 (here, b can be any positive number). Thus, using Stokes Theorem, we have C F ds = D curl(F) n d A = 0 23. You know two things about a vector eld F: The curl of a vector eld F at the origin is v0 = 3, 1, 4 . Estimate the circulation around the small parallelogram (a) F has a vector potential A = 0,A is1unknown). 0, 0, 1 . spanned by the vectors A (but 1 , 2 and B = 2 3 (b) F(x, y, 0) = 0, 0, 1 for all (x, y). Determine the ux of F through the surface S in Figure 20. 660 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) z S 1 Unit circle x y FIGURE 20 SOLUTION Since F has a vector potentialthat is, F is the curl of a vector eldthe ux of F through a surface depends only on the boundary curve C. Now, the surface S and the unit disc S1 in the x y-plane share the same boundary C. Therefore, S z 1 F dS = S1 F dS (1) D x 1 We compute the ux of F through S1 , using the parametrization S1 : (r, ) = (r cos , r sin , 0), n = 0, 0, 1 By the given information, we have F ( (r, )) = F(r cos , r sin , 0) = 0, 0, 1 Hence, F ( (r, )) n = 0, 0, 1 0, 0, 1 = 1 We obtain the following integral: F dS = 2 0 0 1 0 r 1, 0 2 S1 F ( (r, )) n dr d = 2 0 0 1 1 dr d = 2 Combining with (1) we obtain S F dS = 2 25. Use Eq. (8) to prove that if a is a constant vector, then curl(a) = a. Find the ux of F through the surface S in Figure 20, assuming that F has a vector potential and F(x, y, 0) = SOLUTION 0, By Eq. (8) we have cos x, 0 . curl(a) = curl(a) + a Since a is a constant vector, all the partial derivatives of the components of a are zero, hence the curl of a is the zero vector: curl(a) = 0 Thus we obtain curl(a) = a A vector eld F is called radial if it is of the form F = () x, y, z for some function (), where = x 2 + y 2 + z 2 . Show that the curl of a radial vector eld is zero. Hint: It is enough to show that one component of S E C T I O N 18.2 Stokes Theorem (ET Section 17.2) 661 27. Verify the identity curl(()) = 0 SOLUTION 7 We have = , , x y z We compute each component of curl () . The rst component is y The second component of curl(()) is z And the third component is x We conclude that curl(()) = 0 29. Assume that the second partial derivatives of and exist and are continuous. Use(7) and (8) to prove that Prove the Product Rule S z z y = 2 2 =0 yz zy x x z = 2 2 =0 z x xz y y x = 2 2 =0 xy y x () ds = curl(F) + Fds () () curl(F) = S where S is a smooth surface with boundary S. SOLUTION By Stokes Theorem, we have S () ds = S curl() dS We now use Eq.(8) to evaluate the curl of . That is, S () ds = = S S curl() + dS curl() dS + () () dS (1) S Now, since the gradient eld is conservative, this eld satises the cross-partials condition. In other words, curl() = 0 Combining with(1) we obtain S () ds = S 0 dS + S () () dS = S () () dS Further Explain carefully why Greens Theorem is a special case of Stokes Theorem. Insights and Challenges 31. Complete the proof of Theorem 1 by proving the equality C F3 (x, y, z)k ds = S curl(F3 (x, y, z)k) dS where S is the graph of a function z = f (x, y) over a domain D in the x y-plane whose boundary is a simple closed curve. SOLUTION Let (x(t), y(t)), a t b be a parametrization of the boundary curve C0 of the domain D. 662 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S z (x, y, f(x, y)) (ET CHAPTER 17) n (x, y) x D C0 y The boundary curve C of S projects on C0 and has the parametrization (t) = (x(t), y(t), f (x(t), y(t))) , Let F = 0, 0, F3 (x, y, z) We must show that C at b F ds = S curl(F) dS (1) We rst compute the surface integral, using the parametrization S: The normal vector is n= = 1, 0, f x (x, y) 0, 1, f y (x, y) = (i + f x (x, y)k) j + f y (x, y)k x y (x, y) = (x, y, f (x, y)) = f y (x, y)j f x (x, y)i + k = f x (x, y), f y (x, y), 1 We compute the curl of F: i x 0 j y 0 k z F3 (x, y, z) F (x, y, z) F3 (x, y, z) , 3 ,0 y x curl(F) = = Hence, curl(F) ( (x, y)) n = F3 F3 (x, y, f (x, y)) (x, y, f (x, y)) , 0 f x (x, y), f y (x, y), 1 y x F3 (x, y, f (x, y)) F3 (x, y, f (x, y)) f x (x, y) + f y (x, y) y x = The surface integral is thus curl(F) dS = S D F3 (x, y, f (x, y)) F3 (x, y, f (x, y)) f x (x, y) + f y (x, y) y x dx dy (2) We now evaluate the line integral in (1). We have F (t) (t) = 0, 0, F3 x(t), y(t), f x(t), y(t) = F3 x(t), y(t), f x(t), y(t) Using the Chain Rule gives d f x(t), y(t) = f x x(t), y(t) x (t) + f y x(t), y(t) y (t) dt Substituting in (3), we conclude that the line integral is F ds = b a x (t), y (t), d f x(t), y(t) dt (3) d f x(t), y(t) dt C F3 x(t), y(t), f x(t), y(t) f x x(t), y(t) x (t) + f y x(t), y(t) y (t) dt (4) S E C T I O N 18.3 Divergence Theorem (ET Section 17.3) 663 We consider the following vector eld: G(x, y) = F3 x, y, f (x, y) f x (x, y), F3 x, y, f (x, y) f y (x, y) Then the integral in (4) is the line integral of the planar vector eld G over C0 . That is, C F ds = C0 G ds Therefore, we may apply Greens Theorem and write F ds = G ds = F x, y, f (x, y) f y (x, y) F x, y, f (x, y) f x (x, y) x 3 y 3 dx dy (5) C C0 D We use the Product Rule to evaluate the integrand: F3 F3 x, y, f (x, y) f x (x, y) F3 x, y, f (x, y) f x y (x, y) (x, y, f (x, y)) f y (x, y) + F3 x, y, f (x, y) f yx (x, y) x y = F3 F3 x, y, f (x, y) f y (x, y) x, y, f (x, y) f x (x, y) x y Substituting in (5) gives F ds = F3 (x, y, f (x, y)) F3 (x, y, f (x, y)) f y (x, y) f x (x, y) x y dx dy (6) C D Equations(2) and(6) give the same result, hence C F ds = S curl(F) ds for F = 0, 0, F3 (x, y, z) Let F be a continuously differentiable vector eld in R3 , Q a point, and S a plane containing Q with unit normal vector e. Let Cr be a circle of radius r centered at Q in S and let Sr be the disk enclosed by Cr . Assume Sr is 18.3 Divergence Theorem e. (ET Section 17.3) oriented with unit normal vector (a) Let m(r ) and M(r ) be the minimum and maximum values of curl(F(P)) e for P Sr . Prove that Preliminary Questions 1 1. What is the ux of F = 1, 0, 0 through a closed surface? m(r ) 2 curl(F) dS M(r ) r Sr P Q R SOLUTION The divergence of F = 1, 0, 0 is div(F) = x + y + z = 0, therefore the Divergence Theorem (b) that thethat of F through a closed surface S is Prove ux implies S 1 e = lim = F dScurl(F(Q)) div(F) d V r 2 = r 0 W Cr W F dsd V = 0 0 This proves that curl(F(Q)) e is the circulation per unit area in the plane S. 2. Justify the following statement: The ux of F = x 3 , y 3 , z 3 through every closed surface is positive. SOLUTION The divergence of F = x 3 , y 3 , z 3 is div(F) = 3x 2 + 3y 2 + 3z 2 Therefore, by the Divergence Theorem, the ux of F through a closed surface S is S F dS = W (3x 2 + 3y 2 + 3z 2 ) d V Since the integrand is positive for all (x, y, z) = (0, 0, 0), the triple integral, hence also the ux, is positive. 3. Which of the following expressions are meaningful (where F is a vector eld and is a function)? Of those that are meaningful, which are automatically zero? (a) div() (b) curl() (c) curl() (d) div(curl(F)) (e) curl(div(F)) (f) (div(F)) SOLUTION 664 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) (a) The divergence is dened on vector elds. The gradient is a vector eld, hence div() is dened. It is not automatically zero since for = x 2 + y 2 + z 2 we have div() = div 2x, 2y, 2z = 2 + 2 + 2 = 6 = 0 (b) The curl acts on vector valued functions, and is such a function. Therefore, curl() is dened. Since the gradient eld is conservative, the cross partials of are equal, or equivalently, curl() is the zero vector. (c) The curl is dened on vector elds rather than on scalar functions. Therefore, curl() is undened. Obviously, curl() is also undened. The (d) curl is dened on the vector eld F and the divergence is dened on the vector eld curl(F). Therefore the expression div (curl(F)) is meaningful. We show that this vector is automatically zero: div (curl (F)) = div = = = x F3 F2 F1 F3 F2 F1 , , y z z x x y F3 F2 y z + y F1 F3 z x + z F2 F1 x y 2 F2 2 F1 2 F3 2 F2 2 F1 2 F3 + + xy xz yz y x z x zy 2 F3 2 F3 xy y x + 2 F2 2 F2 z x xz + 2 F1 2 F1 yz zy =0+0+0=0 (e) The curl acts on vector valued functions, whereas div(F) is a scalar function. Therefore the expression curl (div(F)) has no meaning. (f) div(F) is a scalar function, hence (divF) is meaningful. It is not necessarily the zero vector as shown in the following example: F = x 2, y2, z2 div (F) = 2x + 2y + 2z (divF) = 2, 2, 2 = 0, 0, 0 4. Which of the following statements is correct (where F is a continuously differentiable vector eld dened everywhere)? (a) The ux of curl(F) through all surfaces is zero. (b) If F = , then the ux of F through all surfaces is zero. (c) The ux of curl(F) through all closed surfaces is zero. SOLUTION (a) This statement holds only for conservative elds. If F is not conservative, there exist closed curves such that C F ds = 0, hence by Stokes Theorem S curl(F) dS = 0. (b) This statement is false. Consider the unit sphere S in the three-dimensional space and the function (x, y, z) = x 2 + y 2 + z 2 . Then F = = 2x, 2y, 2z and div (F) = 2 + 2 + 2 = 6. Using the Divergence Theorem, we have (W is the unit ball in R 3 ) S F dS = W div(F) d V = W 6 dV = 6 W d V = 6 Vol(W) (c) This statement is correct, as stated in the corollary of Stokes Theorem in section 18.2. 5. How does the Divergence Theorem imply that the ux of F = x 2 , y e z , y 2zx through a closed surface is equal to the enclosed volume? SOLUTION By the Divergence Theorem, the ux is F dS = div(F) d V = (2x + 1 2x) d V = 1 d V = Volume(W) S W W W Therefore the statement is true. S E C T I O N 18.3 Divergence Theorem (ET Section 17.3) 665 Exercises In Exercises 14, compute the divergence of the vector eld. 1. F = x y, yz, y 2 x 3 SOLUTION The divergence of F is div(F) = (x y) + (yz) + (y 2 x 3 ) = y + z + 0 = y + z x y z 3. F = x 2zx 2 , z x y, z 2 x 2 xi + yj + zk SOLUTION div(F) = (x 2zx 2 ) + (z x y) + (z 2 x 2 ) = (1 4zx) + (x) + (2zx 2 ) = 1 4zx x + 2zx 2 x y z In Exercises 58, verify thekDivergence Theorem for the vector eld and region. sin(x + z)i ye x z 5. F = z, x, y and the box [0, 4] [0, 2] [0, 3] SOLUTION Let S be the surface of the box and R the region enclosed by S. z 3 2 y 4 x We rst compute the surface integral in the Divergence Theorem: S F dS = R div(F) d V (1) We denote by Si , i = 1, . . . , 6, the faces of the box, starting at the face on the x z-plane and moving counterclockwise, then moving to the bottom and the top. We use parametrizations S1 : S2 : S3 : S4 : S5 : S6 : 1 (x, z) = (x, 0, z), 0 x 4, 0 y 2, 0 x 4, 0 y 2, 0 x 4, 0 x 4, 0z3 0z3 0z3 0z3 0y2 0y2 n = 0, 1, 0 2 (y, z) = (0, y, z), n = 1, 0, 0 3 (x, z) = (x, 2, z), n = 0, 1, 0 4 (y, z) = (4, y, z), n = 1, 0, 0 5 (x, y) = (x, y, 0), n = 0, 0, 1 6 (x, y) = (x, y, 3), n = 0, 0, 1 Then, F dS = = 3 0 3 0 0 0 4 4 S1 F ( 1 (x, z)) 0, 1, 0 d x dz = x d x dz = 3 x 2 4 = 24 20 3 0 0 4 z, x, 0 0, 1, 0 d x dz 666 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) 3 0 0 2 S2 F dS = = F dS = = F dS = = F dS = = F dS = = 3 0 3 0 3 0 3 0 3 0 3 0 2 0 2 0 2 0 2 0 0 0 0 0 0 0 0 0 0 0 2 F ( 2 (y, z)) 1, 0, 0 d y dz = z d y dz = 2 z 2 3 = 9 20 z, 0, y 1, 0, 0 d y dz 2 4 S3 F ( 3 (x, z)) 0, 1, 0 d x dz = x d x dz = 3 x2 4 = 24 20 3 0 0 4 z, x, 2 0, 1, 0 d x dz 4 2 S4 F ( 4 (y, z)) 1, 0, 0 d y dz = z d y dz = 2 z2 3 =9 20 3 0 0 2 z, 4, y 1, 0, 0 d y dz 2 4 S5 F ( 5 (x, y)) 0, 0, 1 d x d y = y d x d y = 4 y 2 2 = 8 20 2 0 0 4 2 0 0 4 0, x, y 0, 0, 1 d x d y 4 4 S6 F ( 6 (x, y)) n d x d y = y dx dy = 4 y2 2 =8 20 3, x, y 0, 0, 1 d x d y 4 We add the integrals to obtain the surface integral 6 S F dS = i=1 Si F dS = 24 9 + 24 + 9 8 + 8 = 0 (2) We now evaluate the triple integral in (1). We compute the divergence of F = z, x, y : div(F) = Hence, R (z) + (x) + (y) = 0 x y z div(F) d V = R 0 dV = 0 (3) The equality of the integrals in (2) and (3) veries the Divergence Theorem. 2 2 7. F = 2x, 3z, 3y and the region x2 + y2 1, 0 z 2 F = y, x, z and the region x + y + z 2 4 SOLUTION z 2 1 y x Let S be the surface of the cylinder and R the region enclosed by S. We compute the two sides of the Divergence Theorem: S F dS = R div(F) d V (1) We rst calculate the surface integral. S E C T I O N 18.3 Divergence Theorem (ET Section 17.3) 667 Step 1. Integral over the side of the cylinder. The side of the cylinder is parametrized by (, z) = (cos , sin , z), n = cos , sin , 0 Then, F ( (, z)) n = 2 cos , 3z, 3 sin cos , sin , 0 = 2 cos2 + 3z sin We obtain the integral F dS = 2 0 0 2 0 2, 0z2 side 2 cos2 + 3z sin d dz = 4 + 0 = 4 2 0 cos2 d + 2 2 3z dz 0 0 sin d =4 sin 2 2 + 2 40 Step 2. Integral over the top of the cylinder. The top of the cylinder is parametrized by (x, y) = (x, y, 2) with parameter domain D = (x, y) : x 2 + y 2 1 . The upward pointing normal is n = Tx T y = 1, 0, 0 0, 1, 0 = i j = k = 0, 0, 1 Also, F ( (x, y)) n = 2x, 6, 3y 0, 0, 1 = 3y Hence, F dS = D top 3y d A = 0 The last integral is zero due to symmetry. y D x 1 Step 3. Integral over the bottom of the cylinder. We parametrize the bottom by (x, y) = (x, y, 0), The downward pointing normal is n = 0, 0, 1 . Then F ( (x, y)) n = 2x, 0, 3y 0, 0, 1 = 3y We obtain the following integral, which is zero due to symmetry: F dS = D (x, y) D bottom 3y d A = 0 Adding the integrals we get S F dS = side F dS + top F dS + bottom F dS = 4 + 0 + 0 = 4 (2) Step 4. Compare with integral of divergence. div(F) = div 2x, 3z, 3y = (2x) + (3z) + (3y) = 2 x y z 668 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) R div (F) d V = R 2 dV = 2 R d V = 2 Vol(R) = 2 2 = 4 (3) The equality of (2) and (3) veries the Divergence Theorem. F = x, 0, use the Divergence + y 2 to evaluate the surface integral In Exercises 916, 0 and the region x 2Theorem z 4 9. F = x, y, z , S is the sphere x 2 + y 2 + z 2 = 1. SOLUTION S F dS. We compute the divergence of F = x, y, z : div(F) = (x) + (y) + (z) = 1 + 1 + 1 = 3 x y z Using the Divergence Theorem and the volume of the sphere, we obtain S F dS = W div(F) d V = W 3 dV = 3 W 1 d V = 3 Volume (W) 4 = 3 13 = 4 3 11. F = x 3 , 0, z 3 , S is the sphere x 2 2 y 2 2 z 2 2 4. ++= F = y, z, x , S is the sphere x + y + z = 1. SOLUTION We compute the divergence of F = x 3 , 0, z 3 : div(F) = 3 (x ) + (0) + (z 3 ) = 3x 2 + 3z 2 = 3(x 2 + z 2 ) x y z Using the Divergence Theorem we obtain (W is the region inside the sphere) S F dS = W div(F) d V = W 3(x 2 + z 2 ) d V We convert the integral to spherical coordinates. We have x 2 + z 2 = 2 cos2 sin2 + 2 cos2 = 2 cos2 sin2 + 2 (1 sin2 ) = 2 sin2 (1 cos2 ) + 2 = 2 sin2 sin2 + 2 = 2 (1 sin2 sin2 ) We obtain the following integral: F dS = 3 =3 =3 = 6 2 0 2 0 2 0 0 0 0 0 2 S 2 (1 sin2 sin2 ) 2 sin d d d 4 (sin sin3 sin2 )d d d 4 sin d d d 3 2 0 2 0 2 0 0 0 2 0 2 0 2 4 sin3 sin2 d d d 0 0 sin d =0 4 d 3 3 sin2 d sin3 d 2 0 4 d 5 2 5 =0 = 6 cos = 12 5 2 5 =0 sin2 cos sin 2 2 2 cos 2 4 =0 3 3 =0 256 32 4 32 3 = 5 35 5 2 13. F = = yx, y, y ,, S is the boundary of the region contained in z cylinder x 2 + y 2 = 4 between the planes z = x F x, , z + z S is the boundary of the unit cube 0 x, y, the 1. and z = 8. SOLUTION Let W be the region enclosed by S. S E C T I O N 18.3 z Divergence Theorem (ET Section 17.3) 669 x y We compute the divergence of F = x, y 2 , z + y : div(F) = 2 (x) + (y ) + (z + y) = 1 + 2y + 1 = 2 + 2y. x y z By the Divergence Theorem we have S F dS = W div(F) d V = W (2 + 2y) d V We compute the triple integral. Denoting by D the disk x 2 + y 2 4 in the x y-plane, we have F dS = 8 S Dx (2 + 2y) dz d x d y = D (2 + 2y)z 8 z=x dx dy = D (2 + 2y)(8 x) d x d y We convert the integral to polar coordinates: F dS = = = = 2 0 2 0 2 0 2 0 0 0 2 2 S (2 + 2r sin )(8 r cos )r dr d 16r + 2r 2 (8 sin cos ) r 3 sin 2 dr d 2 2 r4 sin 2 8r 2 + r 3 (8 sin cos ) d 3 4 r =0 32 + 16 (8 sin cos ) 4 sin 2 3 d = 64 + 2 128 2 16 2 sin d cos d 4 sin 2 d = 64 30 30 0 15. F = x + y, z, z x , S is the boundary of the region between the paraboloid z = 9 x 2 y 2 and the x y-plane. 2 F = x 2 z 2 , e z cos x, y 3 , S is the boundary of the region bounded by x + 2y + 4z = 12 and the coordinate SOLUTIONin the rst octant. divergence of F = x + y, z, z x , planes We compute the div(F) = (x + y) + (z) + (z x) = 1 + 0 + 1 = 2. x y z z 9 3 x y Using the Divergence Theorem we have S F dS = W div(F) d V = W 2 dV We compute the triple integral: F dS = 2 dV = 9x 2 y 2 S W D0 2 dz d x d y = 9x 2 y 2 D 2z 0 dx dy 670 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) = W 2(9 x 2 y 2 ) d x d y y x 2 + y2 = 9 D 3 x We convert the integral to polar coordinates: x = r cos , S y = r sin , 0 r 3, 3 0 0 2 9r 2 r4 3 2 40 = 81 F dS = 2 0 0 3 2 9 r 2 r dr d = 4 (9r r 3 ) dr = 4 17. Let W be the region in Figure 16 bounded by the cylinder x 2 + y 2 = 9, the plane z = x + 1, and the x y-plane. Use 2 F = e z , sin(x 2 z), x 2 + 9y 2 S is the = z, x 2 + y 2 z 8 x 2 y 2 . the Divergence Theorem to compute the, ux of Fregion x, y + 2z through the boundary of W. z x y FIGURE 16 SOLUTION We compute the divergence of F = z, x, y + 2z : div(F) = (z) + (x) + (y + 2z) = 2 x y z By the Divergence Theorem we have S F dS = W div(F) d V = W 2 dV To compute the triple integral, we identify the projection D of the region on the x y-plane. D is the region in the x y plane enclosed by the circle x 2 + y 2 = 9 and the line 0 = x + 1 or x = 1. We obtain the following integral: F dS = 2 dV = x+1 x+1 S W D0 2 dz d x d y = D 2z z=0 dx dy = D (2x + 2) d x d y We compute the double integral as the difference of two integrals: the integral over the disk D2 of radius 3, and the integral over the part D1 of the disk, shown in the gure. y D1 3 x S E C T I O N 18.3 Divergence Theorem (ET Section 17.3) 671 We obtain S F dS = = D2 D2 (2x + 2) d x d y 2x d x d y + y D1 (2x + 2) d x d y (2x + 2) d x d y D2 2 dx dy D1 D2 3 x The rst integral is zero due to symmetry. The second integral is twice the area of D2 , that is, 2 32 = 18. Therefore, S F dS = 18 D1 (2x + 2) d x d y We compute the double integral over the upper part of D1 . Due to symmetry, this integral is equal to half of the integral over D1 . y q 1 3 x We describe the region in polar coordinates: 1.91 , Then D1 1 r 3 cos (2x + 2) d x d y = 2 = = = 3 1.91 1/ cos (2r cos + 2)r dr d 3 4r 3 cos + 2r 2 d 1 3 1.91 r = cos 3 1.91 1/ cos (4r 2 cos + 4r ) dr d = 4 cos 2 3 cos3 cos2 1 2 3 cos2 1.91 36 cos + 18 + 36 cos + 18 d 2 tan 3 1.91 1.91 d = 36 sin + 18 2 tan 1.91 = 13.67 3 = 18 36 sin 1.91 + 18 1.91 so we have S F dS = 18 D1 (2x + 2) d x d y = 70.22. 19. Volume a constant c for which the velocity x, y, z . Prove that if W is a region R3 with a smooth boundary S, then Find as a Surface Integral Let F = eld 1 F = (cx y)i += z)j + (3x + 4cz)k Volume(W) (y F dS 3 S of a uid is incompressible [meaning that div(F) = 0]. 9 672 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S SOLUTION (ET CHAPTER 17) Using the volume as a triple integral we have Volume(W) = 1 dV (1) W We compute the surface integral of F over S, using the Divergence Theorem. Since div(F) = x (x) + y (y) + z (z) = 3, we get S F dS = W div(F) d V = W 3 dV = 3 W 1 dV (2) We combine (1) and (2) to obtain S F dS = 3 volume(W) or Volume(W) = 1 3 S F dS 21. Show that (a cos sin , b sin sin , c cos ) is a parametrization of the ellipsoid Use Eq. (9) to calculate the volume of the unit ball as a surface integral over the unit sphere. y2 z2 x2 + + =1 a b c Then use Eq. (9) to calculate the volume of the ellipsoid as a surface integral over its boundary. SOLUTION For the given parametrization, x = a cos sin , y = b sin sin , z = c cos (1) We show that it satises the equation of the ellipsoid x2 y2 z2 + + = a b c a cos sin 2 + a b sin sin 2 + b c cos 2 c = cos2 sin2 + sin2 sin2 + cos2 = sin2 (cos2 + sin2 ) + cos2 = sin2 + cos2 = 1 Conversely, for each (x, y, z) on the ellipsoid, there exists and so that (1) holds. Therefore whole ellipsoid. Let W be the interior of the ellipsoid S. Then by Eq. (9): Volume(W) = 1 3 S (, ) parametrizes the F dS, F = x, y, z We compute the surface integral, using the given parametrization. We rst compute the normal vector: = a sin sin , b cos sin , 0 = a cos cos , b sin cos , c sin = ab sin2 sin cos k ac sin sin2 j ab cos2 sin cos k bc cos sin2 i = bc cos sin2 , ac sin sin2 , ab sin cos Hence, the outward pointing normal is n = bc cos sin2 , ac sin sin2 , ab sin cos F (, ) n = a cos sin , b sin sin , c cos bc cos sin2 , ac sin sin2 , ab sin cos = abc cos2 sin3 + abc sin2 sin3 + abc sin cos2 = abc sin3 (cos2 + sin2 ) + abc sin cos2 = abc sin3 + abc sin cos2 = abc sin3 + abc sin (1 sin2 ) S E C T I O N 18.3 Divergence Theorem (ET Section 17.3) 673 = abc sin We obtain the following integral: Volume(W) = = 1 3 F dS = 1 2 abc sin d d 30 0 = 4abc 3 S 2abc 2abc sin d = cos 3 3 0 0 23. Find and prove a Product Rule expressing div( f F) in terms of div(F) and f . Prove the identity SOLUTION Let F = P, Q, R . We compute div( f F): div(curl(F)) = 0 ( f P) + ( f Q) + ( f R) div( f F) = div f P, f Q, f R = x y z Applying the product rule for scalar functions we obtain div( f F) = =f f f P + P+ x x P Q R + + x y z f Q f + Q+ y y + f R f + R z z f f f P+ Q+ R = f div(F) + F f x y z We thus proved the following identity: div( f F) = f div(F) + F f 25. Prove that div( f g) = 0. Prove the identity SOLUTION We compute the cross product: div(F G) = curl(F) G F curl(G) i j k Then prove that the cross product = two f y , f z gxvectorgz = isf x f y f z f g of f x , irrotational , g y , elds incompressible [F is called irrotational if curl(F) = 0 and incompressible if div(F) = 0]. g x g y gz = f y gz f z g y , f z g x f x gz , f x g y f y g x We now compute the divergence of this vector. Using the Product Rule for scalar functions and the equality of the mixed partials, we obtain div( f g) = ( f y gz f z g y ) + ( f z g x f x gz ) + ( f x g y f y g x ) x y z f yz gx f y gx z = ( f yx f x y )gz + (gzx gx z ) f y + ( f x z f zx )g y + (gx y g yx ) f z + ( f zy f yz )gx + (g yz gzy ) f x = 0 In Exercises 2628, let denote the Laplace operator dened by = f yx gz + f y gzx f zx g y f z g yx + f zy gx + f z gx y f x y gz f x gzy + f x z g y + f x g yz = 2 2 2 + 2+ 2 x2 y z A function satisfying = 0 is called harmonic. Prove the identity Show that = div() for any function . curl(curl(F)) Show that is harmonic if and only if div() = 0. = (div(F)) F Show that if F is the gradient of a harmonic function, then curl(F) = 0 and div(F) = 0. where F denotes F1 , F2 , F3 . 1 (d) Show F = x z, yz, (x 2 y 2 ) is the gradient of a harmonic function. What is the ux of F through a closed 2 surface? 27. (a) (b) (c) SOLUTION 674 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) (a) We compute the divergence of : div() = div , , x y z = 2 2 2 + 2+ 2= x2 y z (b) In part (a) we showed that = div(). Therefore = 0 if and only if div() = 0. That is, is harmonic if and only if is divergence free. (c) We are given that F = , where = 0. In part (b) we showed that div(F) = div() = 0 We now show that curl(F) = 0. We have i x x j y y k z z curl(F) = curl() = curl x , y , z = = zy yz , x z zx , yx x y = 0, 0, 0 = 0 The last equality is due to the equality of the mixed partials. 2 2 2 2 (d) We rst show that F = x z, yz, x y is the gradient of a harmonic function. We let = x2 z y2 z such that 2 F = . Indeed, = We show that is harmonic, that is, x 2 y2 , , = x z, yz, x y z 2 =F = 0. We compute the partial derivatives: = xz x = yz y 2 =z x2 2 = z y 2 2 =0 z 2 x 2 y2 = z 2 Therefore, = 2 2 2 + 2 + 2 = zz+0=0 2 x y z Since F is the gradient of a harmonic function, we know by part (c) that div(F) = 0. Therefore, by the Divergence Theorem, the ux of F through a closed surface is zero: S F dS = W div(F) d V = W 0 dV = 0 z 29. The electric n er , due to a unit electric dipole oriented 2 + z 2 )1/2 , and er is E 1 x, y,3z ,is the unit = (x 2 vector.+ where radial + y 2 Let F = eld where n is any number, = (x 2 + y in the k direction = = (a) Calculate Let e z 2 )1/2 (Figure 17).div(F). = 1 x, y, z . r (b) Use the Divergence Theorem to calculate the ux of F through the surface of a sphere of radius R centered at (a) Show that E = 3 k 3z4 er . the origin. For which values of n is this ux independent of R? (b) Calculate the ux of E through a sphere centered at the origin. (c) Prove that (n ) = nn1 er . (c) Calculate div(E). (d) Use (c) to show that F is a gradient vector eld for n = 1. Then show that F = 1 er is also a gradient vector Can we use the Divergence Theorem to compute the ux of E through a sphere centered at the origin? (d) eld by computing the gradient of ln . (e) What is the value of C F ds, where C is a closedzcurve? (f) Find the values of n for which the function = n is harmonic. x FIGURE 17 The dipole vector eld restricted to the x z-plane. S E C T I O N 18.3 SOLUTION Divergence Theorem (ET Section 17.3) 675 (a) We rst compute the partial derivatives of : 1 x 1/2 = (x 2 + y 2 + z 2 ) 2x = x 2 1 y 1/2 = (x 2 + y 2 + z 2 ) 2y = y 2 1 z 1/2 = (x 2 + y 2 + z 2 ) 2z = z 2 We compute the partial derivatives of z3 , using the Chain Rule and the partial derivatives in (1): (1) x y z Therefore, z 3 z 3 z 3 =z =z = x 3zx 3 ( ) = z (3)4 = 3z 4 = 5 = 3z5 x x x y 3 ( ) = z (3)4 = 3z 4 = 3z5 y y y z (z 3 ) = 1 3 + z (3)4 = 3 3z 4 = 3 3z 2 5 z z E= z 3 = 3z5 xi 3z5 yj + (3 3z 2 5 )k = 3 k 3z4 1 (xi + yj + zk) = 3 k 3z4 er (b) To compute the ux 2, 0 : S E dS we use the parametrization (, ) = (R cos sin , R sin sin , R cos ), 0 n = R 2 sin er We compute E (, ) n. Since = R on S, we get E (, ) n = R 3 k 3z R 4 er R 2 sin er = R 1 sin k er 3z R 2 sin = R 1 sin k R 1 (xi + yj + zk) 3z R 2 sin = R 2 z sin 3z R 2 sin = 2z R 2 sin = 2R cos R 2 sin = R 1 sin 2 Hence, E dS = 2 0 0 S R 1 sin 2 d d = 2 sin 2d = cos 2 =0 R0 R =0 (c) We use part (a) to write the vector E componentwise: E = 3 k 3z4 er = 3 k 3z4 1 x, y, z = 3z5 x, 3z5 y, 3z 2 5 + 3 To nd div(E) we compute the following derivatives, using (1) and the laws of differentiation. This gives (3z5 x) = 3z (5 x) = 3z 56 x + 5 1 x x x = 3z 56 x Similarly, (3z5 y) = 3z7 (5y 2 2 ) y and (3z 2 5 + 3 ) = 6z5 3z 2 (5)6 34 z z z x + 5 = 3z7 (5x 2 2 ) 676 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) = 6z5 + 15z 2 6 Hence, z z 34 = 3z7 (5z 2 32 ) div(E) = 3z7 (5x 2 2 + 5y 2 2 + 5z 2 32 ) = 15z7 (x 2 + y 2 + z 2 2 ) = 15z7 (2 2 ) = 0 (d) Since E is not dened at the origin, which is inside the ball W, we cannot use the Divergence Theorem to compute the ux of E through the sphere. 31. LetLet E be the dS, where due to a long, uniformly charged rod of radius R with charge density per unit length I= F electric eld (Figure 18).SBy symmetry, we may assume that E is everywhere perpendicular to the rod and its magnitude E(r ) depends only on the distance r to the rod (strictly speaking, thisywould hold only if the rod were innite, but it is 2yz xz x F= , , 2 2 ) = 2 for r > R. Hint: Apply Gausss Law to a cylinder nearly true if the rod is long enough). Show that E(r 2 0 r of x 2 + R + z of unit length boundary of a region rod. ( = radius y 2 and 2 ) and S is thewith its axis along the W. (a) Check that F is divergence-free. (b) Show that I = 0 if S is a sphere centered at the origin. Note that the Divergence Theorem cannot be used. Why not? (c) Give an argument showing that I = 0 for all S. SOLUTION (a) To nd div(F), we rst compute the partial derivatives of = 2x x = =, 2 + y2 + z2 x 2x We compute the partial derivatives: x y z 2yz 2 xz 2 xy 2 = 2yz x 2 + y2 + z2: 2z z = = 2 + y2 + z2 z 2x 2y y = =, 2 + y2 + z2 y 2x x 2 ( ) = 2yz (2)3 = 4yz 3 = 4x yz4 x x y 2 ( ) = x z (2)3 = 2zx 3 = 2x yz4 y y z 2 ( ) = x y (2)3 = 2x y3 = 2x yz4 z z = x z = x y The divergence of F is the sum of these partials. That is, div(F) = 4x yz4 + 2x yz4 + 2x yz4 = 0 We conclude that F is divergence-free. (b) We compute the ux of F over S, using the following parametrization: S: (, ) = (R cos sin , R sin sin , R cos ), where er = 1 x, y, z 0 2, 0 n = R 2 sin er We compute the dot product: Fn= Therefore, F 2yz xz xy , 2 , 2 x, y, z 1 R 2 sin = (2x yz x yz x yz)3 R 2 sin = 0 2 (, ) n = 0, so we have F dS = 2 0 0 S F (, ) nd d = 2 0 0 0 d d = 0 The Divergence Theorem cannot be used since F is not dened at the origin, which is inside the ball with the boundary S. (c) In part (b) we showed that F n = 0 for all values of R. Therefore, I = 0 over all spheres centered at the origin. S E C T I O N 18.3 Divergence Theorem (ET Section 17.3) 677 Further Insights and Challenges 33. Assume that is harmonic. Show that div() = 3 2 and conclude that Let S be the boundary surface of a region W in R and let Den denote the directional derivative of , where en is the outward unit normal vector. Let be the Laplace operator dened earlier. Den d S = 2 d V (a) Use the Divergence Theorem to prove S W SOLUTION In Exercise 23 we proved the following Product Rule: Den d S = S div( f F) = f F + f div (F) (b) Show that if is a harmonic function (dened in Exercise 27), then We use this rule for f = and F = to obtain Den d S = 0 div () = + div () = 2 + div () S By Exercise 27 part (a), div() = Also, since is harmonic, (1) W dV (2) =0 Combining (1), (2), and (3), we obtain div() = 2 + 0 = 2 Now, by the Theorem on evaluating directional derivatives, Den = en Hence, S (3) (4) Den d S = S ( en ) d S (5) By the denition of the vector surface integral we have S dS = S ( en ) d S (6) Combining (5) and (6) and using the Divergence Theorem and equality (4), we get S Den d S = S dS = W div() d V = W 2 d V 35. Show that F = 2y 1, 3z 2 , 2x y has a vector potential and nd one. Let F = P, Q, R be a vector eld dened on R3 such that div(F) = 0. Use the following steps to show that F has a vector potential. SOLUTION Since div(F) = x (2y 1) + y (3z 2 ) + z (2x y) = 0, we know by Exercise 34 that F has a vector potentialLet which is 0, g . Show that (a) A, A = f, g f A = f, 0, g g f , , curl(A) = y z x z y y R(x, t, z) dt + Q(x, y0 , t) dt f (x, y, z) = y0 z0 (b) Fix any value y0 and show that if we dene g(x, y, z) = y, z) P(x, t, z) R(x, t, z) dt + (x, z) dt f (x, = y0 y0 y Hence, P(x, y, z) = 2y 1, Q(x, y, z) = 3z 2 , and R(x, y, z) = 2x y. We choose z 0 = y0 = 0 and nd f and g: g(x, y, z) = P(x, t, z) dt + (x, z) y z y z y0 f (x, y, z) = 2xt dt + 3t 2 dt = xt 2 + t3 = x y 2 + z 3 y y (1) where and are any functions of x and z, then g/y = P and t=0f /y = R. 0 0 t=0 (c) It remains for us to show that and can be chosen so Q = f /z g/ x. Verify that the following choice y y works (for any choice of z 0 ): g(x, y, z) = (2t 1) dt = t 2 t = y2 y 0 z z0 t=0 Substituting in (1) we obtain the vector(x, z) = potential Q(x, y0 , t) dt, (x, z) = 0 2 2 A= 3 Hint: You will need to use the relation div(F) = z0. x y , 0, y y Show that F = 2ye z x y, y, yz z has a vector potential and nd one. 678 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) 37. A vector eld with a vector potential has zero ux through every closed surface in its domain. In the text, we observed er that although the inverse-square radial vector eld F = 2 satises div(F) = 0, F cannot have a vector potential on its domain {(x, y, z) = (0, 0, 0)} because the ux of F through a sphere containing the origin is nonzero. (a) Show that the method of Exercise 34 produces a vector potential A such that F = curl(A) on the restricted domain D consisting of R3 with the y-axis removed. (b) Show that F also has a vector potential on the domains obtained by removing either the x-axis or the z-axis from R3 . (c) Does the existence of a vector potential on these restricted domains contradict the fact that the ux of F through a sphere containing the origin is nonzero? SOLUTION (a) We have F(x, y, z) = er = 2 <x,y,z> 3/2 , hence (x 2 +y 2 +z 2 ) P(x, y, z) = Q(x, y, z) = R(x, y, z) = x 3/2 (x 2 + y 2 + z 2 ) y 3/2 (x 2 + y 2 + z 2 ) z 3/2 (x 2 + y 2 + z 2 ) In Exercise 34, we dened the functions (taking y0 = z 0 = 0) f (x, y, z) = g(x, y, z) = y 0 y 0 z (x 2 + t 2 + z 2 ) x dt + 3/2 dt z 0 Q(x, 0, t) dt = y 0 z 3/2 (x 2 + t 2 + z 2 ) dt (x 2 + t 2 + z 2 ) 3/2 These functions are dened for (x, z) = (0, 0), since the points with x = 0 and z = 0 are on the y-axis. (Notice that for any xed (x, z) = (0, 0) the interval of integration do not intersect the y-axis, therefore they are contained in the domain D.) For (x, z) = (0, 0) we have by the Fundamental Theorem of Calculus y g x x = dt = = P(x, y, z) 3/2 y y 0 (x 2 + t 2 + z 2 )3/2 (x 2 + y 2 + z 2 ) y (x 2 + t 2 + z 2 ) z 3 (x 2 + t 2 + z 2 ) f g 2 = 3 z x 0 (x 2 + t 2 + z 2 ) 3/2 1/2 2z dt y (x 2 + t 2 + z 2 )3/2 x 3 (x 2 + t 2 + z 2 )1/2 2x 2 dt 3 0 (x 2 + t 2 + z 2 ) = y (x 2 + t 2 + z 2 )1/2 (x 2 + t 2 + z 2 3z 2 ) dt 2 2 23 0 (x + t + z ) y (x 2 + t 2 + z 2 )1/2 (x 2 + t 2 + z 2 3x 2 ) dt 2 2 23 0 (x + t + z ) y x 2 + t 2 2z 2 + t 2 + z 2 2x 2 dt = 2 2 2 5/2 0 (x + t + z ) = y 0 (x + t + z ) x 2 2t 2 + z 2 dt = 2 2 2 5/2 y 3/2 (x 2 + y 2 + z 2 ) = Q(x, y, z) The last integral can be veried by showing that y and y f z z = dt = = R(x, y, z) y y 0 (x 2 + t 2 + z 2 )3/2 2 + y 2 + z 2 )3/2 (x y 3/2 (x 2 + y 2 + z 2 ) = x 2 2y 2 + z 2 (x 2 + y 2 + z 2 ) 5/2 Chapter Review Exercises 679 We conclude that the vector A = f, 0, g is a vector potential of F in D, since curl(A) = g f g f , , = P, Q, R = F. y z x y (b) Suppose we remove the x-axis. In this case, we let A = 0, f, g g(x, y, z) = f (x, y, z) = x x x0 Q(t, y, z)dt + y y0 P(x 0 , t, z) dt R(t, y, z) dt x0 Using similar procedure to that in Exercise 34, one can show that F = curl(A). In removing the z-axis the proof is similar, with corresponding modications of the functions in Exercise 34. (c) The ball inside any sphere containing the origin must intersect the x, y, and z axes; therefore, F does not have a vector potential in the ball, and the ux of F through the sphere may differ from zero, as in our example. CHAPTER REVIEW EXERCISES 1. Let F(x, y) = x + y 2 , x 2 y and let C be the unit circle, oriented counterclockwise. Evaluate a line integral and using Greens Theorem. C F ds directly as SOLUTION We parametrize the unit circle by c(t) = (cos t, sin t), 0 t 2. Then, c (t) = sin t, cos t and F(c(t)) = (cos t + sin2 t, cos2 t sin t). We compute the dot product: F(c(t)) c (t) = cos t + sin2 t, cos2 t sin t sin t, cos t = ( sin t)(cos t + sin2 t) + cos t (cos2 t sin t) = cos3 t sin3 t 2 sin t cos t The line integral is thus F (c(t)) c (t) dt = = = 2 0 2 0 C cos3 t sin3 t 2 sin t cos t dt cos3 t dt 2 0 sin3 t dt 2 sin 2t dt 0 2 0 2 sin t 2 cos2 t sin t + + 3 30 2 cos t sin2 t cos t + 3 3 + cos 2t 2 =0 20 We now compute the integral using Greens Theorem. We compute the curl of F. Since P = x + y 2 and Q = x 2 y, we have Q P = 2x 2y x y Thus, C F ds = D y (2x 2y) d x d y C D 1 x 680 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) We compute the double integral by converting to polar coordinates. We get F ds = = 2 0 1 0 0 1 C (2r cos 2r sin )r dr d = 2 0 2 0 0 1 2r 2 (cos sin ) dr d sin + cos 2 0 2r 2 dr (cos sin ) d = 2 31 r 30 = 2 (1 1) = 0 3 In Exercises 36, use Greens Theoremrectangle in Figure 1integralR and R be the boundaries of the two triangles, Let R be the boundary of the to evaluate the line and let around the given closed curve. 1 2 all oriented counterclockwise. x y 3 d x + x 3 y d y, where C is the rectangle 1 x 2, 2 y 3, oriented counterclockwise. 3. F ds if F ds = 4 and F ds = 2. (a) Determine C SOLUTION R1 R R2 y 3 (b) What is the value of R F ds if R is oriented clockwise? C D x 1 2 2 Since P = x y 3 , Q = x 3 y the curl of F is P Q = 3x 2 y 3x y 2 x y By Greens Theorem we obtain x y3 d x + x 3 y d y = = = 3 2 3 2 C D (3x 2 y 3x y 2 ) d x d y = 3 2 2 1 (3x 2 y 3x y 2 ) d x d y dy x3 y 3 3x 2 y 2 2 3y 2 dy = (8y 6y 2 ) y 2 2 2 x=1 9y 2 + 9y 2 dy = 3y 3 9y 2 3 81 81 + = + 2 2 2 2 2 (12 + 18) = 30 y 2 d x x 2 d y, where C consists of the arcs y = x 2 and y = x, 0 x 1, oriented clockwise. (3x + 5y cos y) d x + x sin y d y, where C is any closed curve enclosing a region with area 4, oriented counC C SOLUTION We compute the curl of F. terclockwise. 5. y C y= x D y = x2 0 1 x We have P = y 2 and Q = x 2 , hence P Q = 2x 2y x y We now compute the line integral using Greens Theorem. Since the curve is oriented clockwise, we consider the negative of the double integrals. We get y2 d x x 2 d y = (2x 2y) d A = 1 0 x2 x C D (2x 2y) d y d x Chapter Review Exercises 1 0 1 0 x 1 0 681 = = 2x y + y 2 y=x 2 dx = 2x x + x (2x x 2 + x 4 ) d x x4 4x 5/2 x2 1 x5 + + 5 2 5 20 (x 4 2x 3 + 2x 3/2 + x) d x = 3 1141 = + + = 5252 5 7. Let c(t) = t 2 (1 t), t(t 1)2 . y ye x dthe path c(t) where tis the triangle with vertices (1, 0), (0, 4), and (0, 1), oriented counterclockwise. (a) Plot x + xe d y, for 0 C 1. C (b) Calculate the area of the region enclosed by c(t) for 0 t 1. SOLUTION (a) The path c(t) for 0 t 1 is shown in the gure: y 0.1 0 0.1 x Note that the path is traced out clockwise as t goes from 0 to 1. (b) We use the formula for the area enclosed by a closed curve, A= 1 (x d y y d x) 2C We compute the line integral. Since x = t 2 (1 t) and y = t(t 1)2 , we have d x = 2t (1 t) t 2 dt = 2t 3t 2 dt d y = (t 1)2 + t 2(t 1) = (t 1)(3t 1) dt Therefore, x d y y d x = t 2 (1 t) (t 1)(3t 1) dt t(t 1)2 (2t 3t 2 ) dt = t 2 (t 1)2 dt We obtain the following integral (note that the path must be counterclockwise): A= 1 14 1 102 t (t 1)2 dt = (t 2t 3 + t 2 ) dt = 21 20 2 t5 t4 t3 1 + 5 2 30 = 1 60 In Exercises 912, state whether the equation is an identity (valid for all F or ). If it is not, provide an example in which In (a)(d), calculate the curl and divergence of the vector eld. the equation does not hold. 9. F = yi zk (a) curl() = 0 SOLUTION We compute the curl of the vector eld, (c) div(curl(F)) = 0 i j k curl(F) = x y z y 0 z = (z) (0) i y z (b) div() = 0 (d) (div(F)) = 0 (z) (y) j + x z (0) (y) k x y = 0i + 0j 1k = k The divergence of F is div(F) = (y) + (0) + (z) = 0 + 0 1 = 1. x y z F = e x+y , e y+z , x yz 682 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S 2 2 2 11. F = (ex y z ) (ET CHAPTER 17) 2 2 2 2 2 2 In Exercise 8 we proved the identity curl() = 0. Here, = ex y z , and we have curl ex y z 0. To compute div F, we rst write F explicitly: SOLUTION = 2 2 2 2 2 2 2 2 2 2 2 2 F = ex y z = 2xex y z , 2yex y z , 2zex y z = P, Q, R div(F) = Q R P + + x y z 2 2 2 2 2 2 2 2 2 2 2 2 = 2ex y z + 4x 2 ex y z + 2ex y z + 4y 2 ex y z 2 2 2 2 2 2 + 2ex y z + 4z 2 ex y z 2 2 2 = 2ex y z 2(x 2 + y 2 + z 2 ) 3 13. Recall that if F1 , F2 , and F3 are differentiable functions of one variable, then e = 1 x, y, z = x 2 + y 2 + z 2 curl ( F1 (x), F2 (y), F3 (z) ) = 0 Use this to calculate the curl of F = x 2 + y 2 , ln y + z 2 , z 3 sin(z 2 )e z SOLUTION 3 We use the linearity of the curl and the property mentioned in the exercise to compute the curl of F: x 2 + y 2 , ln y + z 2 , z 3 sin z 2 e z 3 curl F = curl = curl x 2 , ln y, z 3 sin(z 2 )e z 3 + curl y2, z2, 0 = 0 + curl y 2 , z 2 , 0 = 22 2 2 (0) z, y (0), z y = 2z, 0, 2y y z z x x y 15. Verify the identities of Exercises 22 and 24 in Section 18.3 for the vector elds F = x z, ye x , yz and G = Give an example of a nonzero vector eld F such that curl(F) = 0 and div(F) = 0. z2, x y3, x 2 y . SOLUTION We rst show div(curl(F)) = 0. Let F = P, Q, R = x z, ye x , yz . We compute the curl of F: i x P j y Q k z R Q P R Q P R , , y z z x x y curlF = = Substituting in the appropriate values for P, Q, R and taking derivatives, we get curlF = z 0, x 0, ye x 0 Thus, div (curl(F)) = (z)x + (x) y + (ye x )z = 0 + 0 + 0 = 0. Likewise, for G = P, Q, R = z 2 , x y 3 x 2 y , we compute the curl of G: i x P j y Q k z R curlG = = Q P R Q P R , , y z z x x y Substituting in the appropriate values for P, Q, R and taking derivatives, we get curlG = x 2 0, 2z 2x y, y 3 0 Thus, div (curl(G)) = (x 2 )x + (2z 2x y) y + (y 3 )z = 2x 2x = 0. We now work on the second identity. For F = x z, ye x , yz and G = z 2 , x y 3 , x 2 y , it is easy to calculate F G = x 2 y 2 e x x y 4 z, yz 3 x 3 yz, x 2 y 3 z yz 2 e x Chapter Review Exercises 683 Thus, div(F G) = (2x y 2 e x + x 2 y 2 e x y 4 z) + (z 3 x 3 z) + (x 2 y 3 2yze x ) On the other hand, from our work above, curlF = z, x, ye x curlG = x 2 , 2z 2x y, y 3 So, we calculate G curl F F curl G = z 2 z + x y 3 x + x 2 y ye x x z x 2 ye x (2z 2x y) yz y 3 = z 3 + x 2 y 3 + x 2 y 2 e x + 2x y 2 e x x 3 z 2yze x y 4 z = (2x y 2 e x + x 2 y 2 e x y 4 z) + (z 3 x 3 z) + (x 2 y 3 2yze x ) = div(F G) 17. Prove that if F is a gradient vector eld, then the ux of curl(F) through a smooth surface S (whether closed or not) Suppose that S1 and S2 are surfaces with the same oriented boundary curve C. Which of the following conditions is equal to zero. guarantees that the ux of F through S1 is equal to the ux of F through S2 ? SOLUTION If F is a gradient vector eld, then F is conservative; therefore the line integral of F over any closed curve (a) F = for some function is zero. Combining with Stokes Theorem yields (b) F = curl(G) for some vector eld G S curl(F) dS = S F ds = 0 2 2 x Verify Stokes F = z 2 x F = and x, 0 and the surface = 4 ellipsoid 2 0, oriented oriented by In Exercises 1920, letTheorem,for+ z, y 2 y, z let S be the upper halfzof the x y , z+ y 2 + z 2 = 1,by outward4 pointing normals. outward-pointing normals. 19. Use Stokes Theorem to compute SOLUTION S curl(F) dS. We compute the curl of F = z 2 , x + z, y 2 : i x z2 j y x +z k z y2 curl(F) = = (2y 1)i (0 2z)j + (1 0)k = 2y 1, 2z, 1 2 Let C denote the boundary of S, that is, the ellipse x4 + y 2 = 1 in the x y-plane, oriented counterclockwise. Then by Stokes Theorem we have S curl(F) dS = C F ds (1) We parametrize C by C : r (t) = (2 cos t, sin t, 0), Then F (r (t)) r (t) = 0, 2 cos t, sin2 t 2 sin t, cos t, 0 = 2 cos2 t Combining with (1) gives curl(F) ds = 2 0 0 t 2 S 2 cos2 t dt = t + sin 2t 2 = 2 20 21. Use Stokes Theorem to evaluate y, z, x ds, where C is the curve in Figure 2. Compute F dS. Hint: Find a vector potential A, that is, a vector eld A such that F = curl (A), and use C Stokes Theorem. (0, 0, 1) S z y2 + z2 = 1 S x (0, 1, 0) y FIGURE 2 684 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S SOLUTION (ET CHAPTER 17) We compute the curl of F = y, z, x : i x y j y z k z x curl(F) = = i j k = 1, 1, 1 By Stokes Theorem, we have C y, z, x ds = S curl(F) dS = S (curl(F) en ) dS Since the boundary C of the quarter circle S is oriented clockwise, the induced orientation on S is normal pointing in the negative x direction. Thus, en = 1, 0, 0 . Hence, curl(F) en = 1, 1, 1 1, 0, 0 = 1. Combining with (1) we get C y, z, x ds = S 1 ds = Area (S) = 4 Verify the Divergence Theorem Theorem to calculate In Exercises 2326, use the Divergence for F = 0, 0, z and the region x 2for y 2 + z 2 = 1. eld and surface. F dS + the given vector S 23. F = x y, yz, x 2 z + z 2 , S is the boundary of the box [0, 1] [2, 4] [1, 5]. SOLUTION z 5 1 2 4 y 1 x We compute the divergence of F = x y, yz, x 2 z + z 2 : div(F) = xy + yz + (x 2 z + z 2 ) = y + z + x 2 + 2z = x 2 + y + 3z x y z The Divergence Theorem gives x y, yz, x 2 z + z 2 dS = = = = 5 1 5 1 5 1 1 2 4 4 0 1 S (x 2 + y + 3z) d x d y dz = d y dz = 51 13 5 1 2 4 x3 3 + (y + 3z)x 1 d y dz x=0 1 + y + 3z 3 y+ 4 12 y + 3zy dz 2 y=1 4 16 + + 12z 3 2 11 + + 3z 32 dz = 5 1 17 + 9z 2 dz 9z 2 5 17z (17 5 + 9 25) (17 + 9) + = 142 = 2 21 2 2 2 25. F = x yz + x y, 12y 2 (1 z) + e x , e x +y , S is the boundary of the solid bounded by the cylinder x 2 + y 2 = 16 2 F = x y, yz, x z + z 2 , S is the boundary of the unit sphere. and the planes z = 0 and z = y 4. Chapter Review Exercises SOLUTION 685 We compute the divergence of F: div(F) = (x yz + x y) + x y y2 (1 z) + e x 2 + x 2 +y 2 (e ) = yz + y + y(1 z) = 2y z Let S denote the surface of the solid W. The Divergence Theorem gives F dS = = div(F) d V = 0 S W W 2y d V = 0 D y4 2y dz d x d y (8y 2y 2 ) d x d y D 2yz z=y4 dx dy = D 2y (0 (y 4)) d x d y = D We convert the integral to polar coordinates: F dS = =8 =0 2 0 0 4 0 4 S (8r cos 2r 2 cos2 )r dr d 2 0 r 2 dr cos d = 4 0 r 3 dr 0 2 2 cos2 d r4 4 40 + sin 2 2 20 44 2 = 128 4 27. Find the volume of a region W if F = sin(yz), x 2 + z 4 , x cos(x y) , S is any smooth closed surface that is the boundary of a region in R3 . 1 x + x y + z, x + 3y y 2 , 4z dS = 16 2 W SOLUTION Let F = x + x y + z, x + 3y 1 y 2 , 4z . We compute the divergence of F: 2 div(F) = (x + x y + z) + x y x + 3y 12 y + (4z) = 1 + y + 3 y + 4 = 8 2 z Using the Divergence Theorem and the given information, we obtain 16 = That is, 16 = 8 Volume (W) or Volume (W) = 2 In Exercises 2932, let F be a vector eld whose curl and 2 divergence at the origin are Show that the circulation of F = x 2 , y 2 , z(x 2 + y ) around any curve C on the surface of the cone z 2 = x 2 + y 2 is equal to zero (Figure 3). curl(F)(0, 0, 0) = 2, 1, 4 , div(F)(0, 0, 0) = 2 29. Estimate SOLUTION S F dS = W div(F) d V = W 8 dV = 8 W 1 d V = 8 Volume (W) C F ds, where C is the circle of radius 0.03 in the x y-plane centered at the origin. We use the estimation F ds (curl(F)(0) en ) Area(R) z en C R C x y 686 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) The unit normal vector to the disk R is en = k = 0, 0, 1 . The area of the disk is Area (R) = 0.032 = 0.0009. Using the given curl at the origin, we have C F ds 2, 1, 4 0, 0, 1 0.0009 = 4 0.0009 0.0113 31. Suppose that F is the velocity eld of a uid and imagine placing a small paddle wheel at the origin. Find the equation of the plane in which the paddle wheel should be placed to make of rotate as quicklyyz-plane centered at the origin. Does F ds, where C is the boundary of the square it side 0.03 in the as possible. Estimate the maximum spin when the circulation of the velocity circulation depend on how square is oriented within the yz-plane? Might the actual eld F around the wheel is maximum. The maximum circulation occurs when en , and the curl of F at the origin (i.e., the vector 2, 1, 4 ) point in it is oriented? the same direction. Therefore, the plane in which the paddle wheel should be placed is the plane through the origin with the normal 2, 1, 4 . This plane has the equation, 2x y + 4z = 0. 33. The velocity eld of a uid (in meters per second) is Estimate the ux of F through the box of side 0.5 in Figure 4. Does the result depend on how the box is oriented relative to the coordinate axes? F = x 2 + y 2 , 0, z 2 Let W be the region between the hemisphere S = (x, y, z) : x 2 + y 2 + z 2 = 1, and the disk D = (x, y, 0) : x 2 + y 2 1 in the x y-plane. (a) Show that no uid ows across D. (b) Use (a) to show that the rate of uid ow across S is equal to spherical coordinates. SOLUTION SOLUTION The paddleon how has the estimate depend wheel the C x, y, z 0 W div(F) d V . Compute this triple integral using (a) To show that no uid ows across D, we show that the normal component of F at each point on D is zero. At each point P = (x, y, 0) on the x y-plane, F(P) = x 2 + y 2 , 0, 02 = x 2 + y 2 , 0, 0 . Moreover, the unit normal vector to the x y-plane is en = (0, 0, 1). Therefore, F(P) en = x 2 + y 2 , 0, 0 0, 0, 1 = 0. Since D is contained in the x y-plane, we conclude that the normal component of F at each point on D is zero. Therefore, no uid ows across D. (b) By the Divergence Theorem and the linearity of the ux we have S F dS + D F dS = W div(F) d V Since the ux through the disk D is zero, we have S F dS = W div(F) d V (1) To compute the triple integral, we rst compute div(F): div(F) = 2 (x + y 2 ) + (0) + (z 2 ) = 2x + 2z = 2(x + z). x y z z W y x Chapter Review Exercises 687 Using spherical coordinate we get div(F) d V = 2 =2 = /2 0 1 0 0 2 0 1 W ( sin cos + cos )2 sin d d /2 3 d 0 sin2 d = 2 0 cos d + 2 /2 0 /2 0 cos sin d /2 1 sin 2 d 0+ 2 0 2 cos 2 2 = (1 1) = 4 2 Combining with (1) we obtain the ux: S F dS = 2 x 35. LetThe velocity eld 2 a uid (in vector eld second) is (x, y) = x + of 2 . The meters per F = (Figure 5) provides a model in the plane of the velocity eld x +y of an incompressible, irrotational uid owing past a cylindrical obstacle (in 2 case, the obstacle is the unit circle this F = (3y 4)i + ey(z+1) j + (x 2 + y )k x 2 + y 2 = 1). (a) Verify that F is irrotational (in cubic meters perirrotational if curl(F) = 0]. (a) Estimate the ow rate [by denition, F is second) through a small surface S around the origin if S encloses a region of volume 0.01 m3 . y (b) Estimate the circulation of F about a circle in the x y-plane of radius r = 0.1 m centered at the origin (oriented 3 counterclockwise when viewed from above). 2 (c) Estimate the circulation of F about a circle in the1 yz-plane of radius r = 0.1 m centered at the origin (oriented counterclockwise when viewed from the positive x-axis). x 3 2 1 1 2 3 1 2 3 FIGURE 5 The vector eld for (x, y) = x + x x 2 + y2 . (b) Verify that F is tangent to the unit circle at each point along the unit circle except (1, 0) and (1, 0) (where F = 0). (c) What is the circulation of F around the unit circle? (d) Calculate the line integral of F along the upper and lower halves of the unit circle separately. SOLUTION (a) In Exercise 8, we proved the identity curl() = 0. Since F is a gradient vector eld, it is irrotational; that is, curl(F) = 0 for (x, y) = (0, 0), where F is dened. (b) We compute F explicitly: F = = y2 x 2 2x y , = 1+ , x y 2 + y 2 )2 2 + y 2 )2 (x (x Now, using x = cos t and y = sin t as a parametrization of the circle, we see that F = 1 + sin2 t cos2 t, 2 cos t sin t = 2 sin2 t, 2 cos t sin t , and so F = 2 sin t sin t, cos t = 2 sin t y, x , which is clearly perpendicular to the radial vector x, y for the circle. (c) We use our expression of F from Part (b): F = = 1 + y2 x 2 2 (x 2 + y 2 ) , 2x y (x 2 + y 2 ) 2 Now, F is not dened at the origin and therefore we cannot use Greens Theorem to compute the line integral along the unit circle. We thus compute the integral directly, using the parametrization c(t) = (cos t, sin t), 0 t 2. 688 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S y (ET CHAPTER 17) C 1 x Then, F (c(t)) c (t) = 1 + sin2 t cos2 t 2 (cos2 t + sin2 t) , 2 cos t sin t (cos2 t + sin2 t) 2 sin t, cos t = 1 + sin2 t cos2 t, 2 cos t, sin t sin t, cos t = 2 sin2 t, 2 cos t sin t sin t, cos t = 2 sin3 t 2 cos2 t sin t = 2 sin t (sin2 t + cos2 t) = 2 sin t Hence, F ds = 2 0 C 2 sin t dt = 0 (d) We denote by C1 and C2 the upper and lower halves of the unit circle. Using part (c) we have C1 F ds + C2 F ds = 0 y C2 F ds = C1 F ds (1) C1 1 C2 x To compute the circulation along C1 , we compute the integral as in part (c), only that the limits of integration are now t = 0 and t = . Using the computations in part (c) we obtain C1 F ds = 0 2 sin2 t dt = 4 Therefore, by (1), C2 F ds = 4. 37. In Section 18.1, we showed that if C is a simple closed curve, oriented counterclockwise, then the line integral is Figure 6 shows the vector eld F = , where (x, y) = ln(x 2 + (y 1)2 ) + ln(x 2 + (y + 1)2 ), which is the velocity eld for the ow of a uidArea enclosedof equal 1 with sources by C = strength y y d x (note that is undened at these two x d at (0, 1) 1 2 that is, curlz (F) = 0 and div(F) = 0 [in computing points). Show that F is both irrotational and incompressible, C div(F), treat F as a vector eld in R3 with a zero z-component]. Is it necessary to compute curlz (F) to conclude that Suppose that C is a path from P to Q that is not closed but has the property that every line through the origin intersects it is zero? C in at most one point, as in Figure 7. Let R be the region enclosed by C and the two radial segments joining P and Q to the origin. Show that the line integral in Eq. (1) is equal to the area of R. Hint: Show that the line integral of F = y, x along the two radial segments is zero and apply Greens Theorem. y C Q R P x FIGURE 7 Chapter Review Exercises SOLUTION y C Q R P Q x 689 Let F = y, x . Then P = y and Q = x, and Q P = 2. By Greens Theorem, we have x y C y d x + x d y + QO y d x + x d y + OP y d x + x d y = R 2dA = 2 R dA Denoting by A the area of the region R, we obtain A= 1 1 1 y d x + x d y + y d x + x d y + y d x + x d y 2C 2 QO 2 OP (1) We parametrize the two segments by Q O : c(t) = (t, t tan ) O P : d(t) = (t, t tan ) Then, F (c(t)) c (t) = t tan , t 1, tan = t tan + t tan = 0 F (d(t)) d (t) = t tan , t 1, tan = t tan + t tan = 0 Therefore, F ds = F ds = 0. c (t) = 1, tan d (t) = 1, tan QO OP Combining with (1) gives A= 1 y d x + x d y. 2C 39. Prove the following generalization of Eq. (1). Let C be a simple closed curve in the plane (Figure 8) Suppose that the curve C in Figure 7 has the polar equation r = f (). (a) Show that c() = ( f () cos , f () S : ) ax a+ by + cz + d = 0parametrization of C. sin is counterclockwise (b) In Section 12.4, we showed that the area of the region R is given by the formula Then the area of the region R enclosed by C is equal to 1 2n C 1 2 (bz Aread x R (cx2 az) f () (ay bx) dz cy) of + = d y + d Use = a, b, of Exercise 37 to give a new proof of this the boundary Evaluate the to integral in vector n). Hint: where n the resultc is the normal to S and C is oriented as formula. Hint:of R (relativeline the normal (1) using c(). Apply Stokes Theorem to F = bz cy, cx az, ay bx . z Plane S n = a, b, c R C x y FIGURE 8 SOLUTION By Stokes Theorem, curl(F) dS = (curl(F) en ) d S = F ds (1) S S C 690 C H A P T E R 18 F U N D A M E N TA L TH E O R E M S O F V E C T O R A N A LY S I S (ET CHAPTER 17) We compute the curl of F: i x bz cy j y cx az k z ay bx curl(F) = = 2ai + 2bj + 2ck = 2 a, b, c The unit normal to the plane ax + by + cz + d = 0 is en = Therefore, curl(F) en = 2 a, b, c = Hence, S a, b, c a 2 + b2 + c2 1 a 2 + b2 + c2 a, b, c 2 a 2 + b2 + c2 (a 2 + b2 + c2 ) = 2 a 2 + b2 + c2 curl(F) dS = S curl(F) en d S = S 2 a 2 + b2 + c2 d S = 2 a 2 + b2 + c2 S 1 dS (2) The sign of S 1 dS is determined by the orientation of S. Since the area is a positive value, we have S 1 ds = Area (S) Therefore, (2) gives curl(F) dS = 2 a 2 + b2 + c2 Area(S) S Combining with (1) we obtain 2 a 2 + b2 + c2 Area(S) = or Area(S) = 1 2 a 2 + b2 + c2 = 1 2n C C F ds (bz cy) d x + (cx az) d y + (ay bx) dz Use the result of Exercise 39 to calculate the area of the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1) as a line integral. Verify your result using geometry.

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LSU - MATH - 4171
MATH 4171SPRING 2006Graph TheoryA supplementGuoli Ding http:/math.lsu.edu/ ding Last Updated: May 3, 2006Copyright c 1996, by Guoli Ding. All Rights Reserved.Table of Contents1. Introduction . . . . . . . . . . . . . . . . . . . . . . . .
LSU - MATH - 4171
MATH 4171 Graph TheorySPRING 2006Homework Set I (Four problems) Due date : Monday 1-30-061. Describe a real world problem (preferable from your own eld) that can be modeled by graphs. Specify vertices, edges, and the incidence relation for your
LSU - MATH - 4171
MATH 4171 Graph Theory Homework Set I - SolutionsSPRING 20061. Describe a real world problem (preferable from your own eld) that can be modeled by graphs. Specify vertices, edges, and the incidence relation for your graph. . 2. Is there a simple
UMBC - CS - 671
CMSC 671 - Fall '01 LECTURE NOTES FOR 9/6/01 Prof. Marie desJardins 4:00 - 4:45 4:45 - 5:15 Problem solving as search / uninformed search More detailed Lisp example (nested-find-odd)- PROBLEM DEFINITION Given an arbitrary nested list, find and retu
LSU - MATH - 4171
MATH 4171 Graph Theory Homework Set II (5 problems) Due date : Monday 2-13-06SPRING 20061. For the graph below, nd the distance from u to h, and a shortest u-h path (with respect to the given numbers on each edge). Show your workd 4 a 1 u 3 2 7
LSU - MATH - 4171
MATH 4171 Graph Theory Homework Set II -SolutionsSPRING 20061. For the graph below, nd the distance from u to h, and a shortest u-h path (with respect to the given numbers on each edge). Show your work.d 4 a 1 u 3 2 7 b 2 6 e 3 1 9 c 1 7 5 5 1 2
LSU - MATH - 4171
MATH 4171 Graph Theory Homework Set III Due date : Monday 3-6-06SPRING 20061. Consider the two arc-disjoint s-t dipaths, sabhgt and sef t, in the following digraph. (a) Find an augmenting path with respect to these two paths; (b) Using the given
LSU - MATH - 4171
MATH 4171 Graph Theory Homework Set III -SolutionsSPRING 20061. Consider the two arc-disjoint s-t dipaths, sabhgt and sef t, in the following digraph. (a) Find an augmenting path with respect to these two paths; (b) Using the given two paths and
LSU - MATH - 4171
MATH 4171 Graph Theory Homework Set IV (4 problems) Due date : Monday 3-20-06SPRING 20061. Let G = (X, Y, E) be an r-regular (r &gt; 0) bipartite graph. Prove that G has a perfect matching.2. Find a maximum matching in the following graph. Prove t
LSU - MATH - 4171
MATH 4171 Graph Theory Homework Set IV -SolutionsSPRING 20061. Let G = (X, Y, E) be an r-regular (r &gt; 0) bipartite graph. Prove that G has a perfect matching. Proof. Let V = X Y . We prove that 1 |M | |V | . 2 |V | , 2which means that G has
UMBC - CS - 671
CMSC 671 Fall 2001Class #3/4 Thursday, September 6 / Tuesday, September 11Todays class Goal-based agents Representing states and operators Example problems Generic state-space search algorithm Specific algorithms Breadth-first search Dep
LSU - MATH - 4171
MATH 4171 Graph Theory Homework Set V (4 problems) Due date : Monday 4-3-06SPRING 20061. Theorem 4.8 states that a necessary condition for a graph G = (V, E) to be Hamiltonian is that G S has at most |S| components, for all S V . Is this condit
LSU - MATH - 4171
MATH 4171 Graph Theory Homework Set V SolutionsSPRING 20061. Theorem 4.8 states that a necessary condition for a graph G = (V, E) to be Hamiltonian is that G S has at most |S| components, for all S V . Is this condition sucient? In other words
LSU - MATH - 4171
MATH 4171 Graph Theory Homework Set VI (3 problems) Due date : Monday 4-17-06SPRING 20061. Let G be a plane graph of minimum degree at least three. Prove that G has a region of size at most ve.2. Decide if the following graphs are planar. You n
LSU - MATH - 4171
MATH 4171 Graph Theory Homework Set VI -SolutionsSPRING 20061. Let G = (V, E) be a plane graph of minimum degree at least three. Prove that G has a region of size at most ve. Proof. Let d1 , d2 , ., d|V | be the degrees of G. Then 2|E| = d1 + d2
LSU - MATH - 4171
MATH 4171 Graph TheorySPRING 2006Homework Set VII (4 problems) Due date : Monday 5-1-061. Prove without using Theorem 8.24 that every planar graph is 6-colorable. 2. For the graph G below, prove that (G) = 4 by showing: (a) (G) 4; and (b) (G)
LSU - MATH - 4171
MATH 4171 Graph Theory Homework Set VII - Solutions1. Prove without using Theorem 8.24 that every planar graph is 6-colorable.SPRING 2006Proof. Let G be a planar simple graph. Then every induced subgraph G of G is also a planar simple graph. By
UMBC - CS - 671
CMSC 671 Fall 2001Class #5 Tuesday, September 18Todays class Heuristic search Best first search Greedy search Beam search A, A* Examples Memory-conserving variations of A* Heuristic functions Iterative improvement methods Hill climbing
UMBC - CS - 671
CMSC 671 Fall 2001Class #6-7 Thursday, September 20 / Tuesday, September 251Todays class Constraint Processing / Constraint Satisfaction Problem (CSP) paradigm Algorithms for CSPs Backtracking (systematic search) Constraint propagation (k-c
UMBC - CS - 671
CMSC 671 Fall 2001Class #7 Tuesday, September 251Todays class Interleaving backtracking and consistency checking Variable-ordering heuristics Value-ordering heuristics Intelligent backtracking2Advanced Constraint TechniquesKumar, Algor
UMBC - CS - 671
CMSC 671 Fall 2001Class #8 Thursday, September 27Todays class Game playing Game trees Minimax Alpha-beta pruning Adding randomness Deep Blue (da chess champeen of da woild!)Game PlayingChapter 5Some material adopted from notes by Charl
UMBC - CS - 671
CMSC 671 Fall 2001Class #9 Tuesday, October 21Todays class Knowledge-based agents Propositional logic2Agents that Reason LogicallyChapter 6Some material adopted from notes by Andreas Geyer-Schulzand Chuck Dyer3A knowledge-based age
UMBC - CS - 671
CMSC 671 Fall 2001Class #10 Thursday, October 41Todays class History of AI Key people Significant events Future of AI Where are we going2History of AIChronology of AI; Russell &amp; Norvig Ch. 26 Alan M. Turing, Computing Machinery and I
UMBC - CS - 671
CMSC 671 Fall 2001Class #11 Tuesday, October 91Todays class Philosophy of AI Can we build intelligent machines? If we do, how will we know theyre intelligent? Should we build intelligent machines? If we do, how should we treat them and h
UMBC - CS - 671
CMSC 671 Fall 2001Class #12/13 Thursday, October 11 / Tuesday, October 16Todays class First-order logic Properties, relations, functions, quantifiers, Terms, sentences, wffs, axioms, theories, proofs, Extensions to first-order logic Logic
UMBC - CS - 671
CMSC 671 Fall 2001Class #13/14 Tuesday, October 16 / Thursday, October 181Todays class Inference in first-order logic Inference rules Forward chaining Backward chaining Resolution Unification Proofs Clausal form Resolution as search2
UMBC - CS - 671
CMSC 671 Fall 2001Class #15 Tuesday, October 231The Breadth of AI: A Lecture on Diversity2What is AI? Intensional definitions: describe the end goals and objectives of the discipline Searle, McCarthy, Ginsberg, etc. definitions from HW #1
UMBC - CS - 671
CMSC 671 Fall 2001Class #16 Thursday, October 251Todays class What is planning? Approaches to planning GPS / STRIPS Situation calculus formalism Partial-order planning2PlanningChapter 11Some material adopted from notes by Andreas Gey
UMBC - CS - 671
CMSC 671 Fall 2001Class #17 Tuesday, October 301Todays class Real-world domains Job shop scheduling Hierarchical decomposition (HTN planning) Increasing expressivity Temporal constraints2Practical PlanningChapter 123Real-world pl
UMBC - CS - 671
CMSC 671 Fall 2001Class #20 Thursday, November 81Todays class Conditional independence Bayesian networks Network structure Conditional probability tables Conditional independence Inference in Bayesian networks2Conditional Independence
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CMSC 671 Fall 2001Class #21 Tuesday, November 131Todays class Conditional independence Bayesian networks Network structure Conditional probability tables Conditional independence Inference in Bayesian networks2Inference in Bayesian Ne
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CMSC 671 Fall 2001Class #22 Thursday, November 151Todays class Evaluation methodologies Knowledge engineering KE process Ontologies and representations Sisyphus-III CommonKADS2Knowledge EngineeringChapters 8, 11.83Creating knowle
UMBC - CS - 671
CMSC 671 Fall 2001Class #23 / 24 Tuesday, November 20 / Tuesday, November 271Todays class Semester endgame Machine learning What is ML? Inductive learning Supervised Unsupervised Decision trees2Upcoming dates Mon 11/27 Th/Fr
UMBC - CS - 671
CMSC 671 Fall 2001Class #25-26 Tuesday, November 27 / Thursday, November 291Todays class Neural networks Bayesian learning2Machine Learning: Neural and BayesianChapter 19Some material adapted from lecture notes by Lise Getoor and Ron Pa
UMBC - CS - 07
1National Science Foundation Symposium on Next Generation of Data Mining and Cyber-Enabled Discovery for Innovation (NGDM07): Final ReportReport Committee: Tim Finin, Jo o Gama, Robert Grossman, Diane Lambert, Huan Liu, a Kun Liu, Olfa Nasraoui, L
UMBC - ID - 111
Swoogle SearchSemantic Web NavigationSWOOGLE 2 Swoogle Search Ontology Dictionary Swoogle Statistics Web Server Human usersOntology Dictionaryservice IR analyzer SWD analyzerWeb Service Intelligent AgentsSwoogle RankingOntoRank: Ranking
UMBC - ID - 112
Convincing Internet Search Engines to Index Semantic Markupmy.owl.doc my.swangled.docOWL Document&lt;A,B,C&gt; &lt;A,E,C&gt; etc.TheSwanglerconvertseach tripleofthedocumenttoa singleindexingtermSwangled Document&lt;A,B,C&gt; &lt;A,E,C&gt; etc. 5V4F56A5XJ LA846CM7RR
Southeastern Bible - MT - 101
MT101 FALL Instructor:CALCULUS II SEMESTER 2003Professor Solomon Friedberg Carney Hall, Room 325 Phone: 617 552-3002. Email: friedber@bc.eduTeaching Assistant: Mr. Ted Zarrabi Carney Hall, Room 358 Phone: 617 552-8205. Email: zarrabi@bc.edu Cla
BC - MT - 101
MT101 FALL Instructor:CALCULUS II SEMESTER 2003Professor Solomon Friedberg Carney Hall, Room 325 Phone: 617 552-3002. Email: friedber@bc.eduTeaching Assistant: Mr. Ted Zarrabi Carney Hall, Room 358 Phone: 617 552-8205. Email: zarrabi@bc.edu Cla
Southeastern Bible - MT - 105
Notes on Improper Integrals MT105, Prof. Friedberg, Fall 2005 In class I mentioned that, by denition, Bf (x) dx = limA Bf (x) dx,Awhere A and B independently approach and +, respectively, provided the limit exists and is nite. If this lim
BC - MT - 105
Notes on Improper Integrals MT105, Prof. Friedberg, Fall 2005 In class I mentioned that, by denition, Bf (x) dx = limA Bf (x) dx,Awhere A and B independently approach and +, respectively, provided the limit exists and is nite. If this lim
Southeastern Bible - MT - 105
First Applications of Integration by Parts Lecture in MT105, 9/21/05, Prof. Friedberg We will give two applications of Integration by Parts. Remember that we have the version for denite integralsb bu(x) v (x) dx = u(b)v(b) u(a)v(a) a au (x) v(
BC - MT - 105
First Applications of Integration by Parts Lecture in MT105, 9/21/05, Prof. Friedberg We will give two applications of Integration by Parts. Remember that we have the version for denite integralsb bu(x) v (x) dx = u(b)v(b) u(a)v(a) a au (x) v(
Southeastern Bible - MT - 816
MATH 816MODERN ALGEBRA IFALL 2004Instructor: Professor Solomon Friedberg Carney Hall, Room 325 617 552-3002 friedber@bc.edu Office Hours: Monday and Friday 10-11, Wedneday 12-1 (tentative). In addition, you are always welcome to make an appoint
BC - MT - 816
MATH 816MODERN ALGEBRA IFALL 2004Instructor: Professor Solomon Friedberg Carney Hall, Room 325 617 552-3002 friedber@bc.edu Office Hours: Monday and Friday 10-11, Wedneday 12-1 (tentative). In addition, you are always welcome to make an appoint
Southeastern Bible - MT - 101
279Program to Calculate Riemann Sums to Evaluate a Definite Integral (TI-81)Select PRGM to get the program menu, move to EDIT to enter a program. When you select a program number, you must rst give it a name (for example, RSUMS) to the right of th
BC - MT - 101
279Program to Calculate Riemann Sums to Evaluate a Definite Integral (TI-81)Select PRGM to get the program menu, move to EDIT to enter a program. When you select a program number, you must rst give it a name (for example, RSUMS) to the right of th
Southeastern Bible - MT - 101
288Program to Calculate Riemann Sums to Evaluate a Definite Integral (TI-82)Select PRGM to get the program menu, move to NEW to enter a new program. You must rst give it a name (for example, RSUMS) when prompted; to nish entering/editing a program
BC - MT - 101
288Program to Calculate Riemann Sums to Evaluate a Definite Integral (TI-82)Select PRGM to get the program menu, move to NEW to enter a new program. You must rst give it a name (for example, RSUMS) when prompted; to nish entering/editing a program
Southeastern Bible - MT - 101
297Program to Calculate Riemann Sums to Evaluate a Definite Integral (TI-85)Select PRGM to get the program menu, then select EDIT to enter a program. When you enter a program, you must rst give it a name (for example, RSUMS). To nish editing, hit
BC - MT - 101
297Program to Calculate Riemann Sums to Evaluate a Definite Integral (TI-85)Select PRGM to get the program menu, then select EDIT to enter a program. When you enter a program, you must rst give it a name (for example, RSUMS). To nish editing, hit
Southeastern Bible - MT - 101
281 Prgm: INTEGRAL :Disp LOWER LIMIT :Input A :Disp UPPER LIMIT :Input B :Disp DIVNS :Input N :B , A=N :A ! X :0 ! L Where to Find The Commands Disp and Input are accessed via PRGM, I/O Enter lower limit of integration. Enter upper limit of integrati
BC - MT - 101
281 Prgm: INTEGRAL :Disp LOWER LIMIT :Input A :Disp UPPER LIMIT :Input B :Disp DIVNS :Input N :B , A=N :A ! X :0 ! L Where to Find The Commands Disp and Input are accessed via PRGM, I/O Enter lower limit of integration. Enter upper limit of integrati
Southeastern Bible - MT - 101
290 Name= INTEGRAL :Disp LOWER LIMIT :Input A :Disp UPPER LIMIT :Input B :Disp DIVNS :Input N :B , A=N :A ! X :0 ! L :0 ! M :1 ! I :Lbl P Where to Find The Commands Disp and Input are accessed via PRGM, I/O Enter lower limit of integration. Enter upp
BC - MT - 101
290 Name= INTEGRAL :Disp LOWER LIMIT :Input A :Disp UPPER LIMIT :Input B :Disp DIVNS :Input N :B , A=N :A ! X :0 ! L :0 ! M :1 ! I :Lbl P Where to Find The Commands Disp and Input are accessed via PRGM, I/O Enter lower limit of integration. Enter upp
Southeastern Bible - MT - 101
299 Name=INTEG :Disp LOWER LIMIT :Input A :Disp UPPER LIMIT :Input B :Disp DIVNS :Input N :B , A=N :A ! x :0 ! L Where to Find The Commands Disp, Input, and are accessed via PRGM, I/O. Enter lower limit of integration. Enter upper limit of integrat
BC - MT - 101
299 Name=INTEG :Disp LOWER LIMIT :Input A :Disp UPPER LIMIT :Input B :Disp DIVNS :Input N :B , A=N :A ! x :0 ! L Where to Find The Commands Disp, Input, and are accessed via PRGM, I/O. Enter lower limit of integration. Enter upper limit of integrat
Southeastern Bible - MT - 101
323Program Goto start Label eqn f=: : : Return Label start Print l-limit Input a Print u-limit Input b Print divns Input n x=a s=0 m=0 i=1 h=(b,a)/n Label 1 Gosub eqn s=s+f*h x=x+ .5h Gosub eqn m=m+f*h x=x+ .5h i=i+1 If i &lt;= n Goto 1 Print left/rig
BC - MT - 101
323Program Goto start Label eqn f=: : : Return Label start Print l-limit Input a Print u-limit Input b Print divns Input n x=a s=0 m=0 i=1 h=(b,a)/n Label 1 Gosub eqn s=s+f*h x=x+ .5h Gosub eqn m=m+f*h x=x+ .5h i=i+1 If i &lt;= n Goto 1 Print left/rig
Southeastern Bible - MT - 101
Program EULER for TI-81 Prgm:EULER :All-Off :Disp &quot;INITIAL X&quot; : Input X :Disp &quot;INITIAL Y&quot; : Input Y :Disp &quot;FINAL X&quot; : Input B : Input N :1-&gt; I : (B-X)/N-&gt;H :Lbl P : X+H-&gt;U :Y+H*Y1-&gt;V :Line(X,Y,U,V) :U-&gt;X :V-&gt;Y :Is&gt;(I,N) :Goto P :Disp &quot;X:&quot;, X :Disp &quot;Y
BC - MT - 101
Program EULER for TI-81 Prgm:EULER :All-Off :Disp &quot;INITIAL X&quot; : Input X :Disp &quot;INITIAL Y&quot; : Input Y :Disp &quot;FINAL X&quot; : Input B : Input N :1-&gt; I : (B-X)/N-&gt;H :Lbl P : X+H-&gt;U :Y+H*Y1-&gt;V :Line(X,Y,U,V) :U-&gt;X :V-&gt;Y :Is&gt;(I,N) :Goto P :Disp &quot;X:&quot;, X :Disp &quot;Y