INSTRUCTOR S GUIDE ELECTROMAGNETICS FOR E NGINEERS FAWWAZ T. ULABY. Upper Saddle River, New Jersey 07458

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1 INSTRUCTOR S GUID LCTROMAGNTICS FOR NGINRS FAWWAZ T. ULABY Uer Saddle River, New Jersey 07458

2 Associate ditor: Alice Dworkin xecutive Managing ditor: Vince O'Brien Managing ditor: David A. George Production ditor: Craig Little Sulement Cover Manager: Daniel Sandin Manufacturing Buyer: Ilene Kahn 2005 by Pearson ducation, Inc. Pearson Prentice Hall Pearson ducation, Inc. Uer Saddle River, NJ All rights reserved. No art of this book may be reroduced in any form or by any means, without ermission in writing from the ublisher. The author and ublisher of this book have used their best efforts in rearing this book. These efforts include the develoment, research, and testing of the theories and rograms to determine their effectiveness. The author and ublisher make no warranty of any kind, exressed or imlied, with regard to these rograms or the documentation contained in this book. The author and ublisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, erformance, or use of these rograms. Pearson Prentice Hall is a trademark of Pearson ducation, Inc. This work is rotected by United States coyright laws and is rovided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any art of this work (including on the World Wide Web) will destroy the integrity of the work and is not ermitted. The work and materials from it should never be made available to students excet by instructors using the accomanying text in their classes. All reciients of this work are exected to abide by these restrictions and to honor the intended edagogical uroses and the needs of other instructors who rely on these materials. Printed in the United States of America ISBN Pearson ducation Ltd., London Pearson ducation Australia Pty. Ltd., Sydney Pearson ducation Singaore, Pte. Ltd. Pearson ducation North Asia Ltd., Hong Kong Pearson ducation Canada, Inc., Toronto Pearson ducación de Mexico, S.A. de C.V. Pearson ducation Jaan, Tokyo Pearson ducation Malaysia, Pte. Ltd. Pearson ducation, Inc., Uer Saddle River, New Jersey

3 Table of Contents Chater 1 Lesson Plan 1 1 Chater 2 2 Lesson Plans Problem Solutions Chater 3 Lesson Plans Problem Solutions 39 Chater 4 Lesson Plans Problem Solutions 68 Chater 5 Lesson Plans Problem Solutions 135 Chater 6 Lesson Plans Problem Solutions 186 Chater 7 Lesson Plans Problem Solutions 211 Chater 8 Lesson Plans Problem Solutions 268 Chater 9 Lesson Plans Problem Solutions 343 Chater 10 Lesson Plans Problem Solutions 387

4 Dear Instructor: This Instructor s Guide and Solutions Manual is intended for use by the course instructor only. It was develoed to rovide: (1) Individual lesson lans that secify the book section(s) to be covered, toic highlights, and secial materials that can be used from the text or the CD-ROM to illustrate key oints of interest. (2) Solutions to all end-of-chater roblems. As the course instructor, you are authorized to share the solutions of selected roblems with students, but lease limit access to the Manual to yourself and to your teaching assistants. Your feedback regarding the textbook to ulaby@umich.edu is always welcomed. Sincerely, Fawwaz Ulaby University of Michigan

5 Chater 1: Introduction Lesson #1 Chater Section: Chater 1 Toics: M history and how it relates to other fields Highlights: M in Classical era: 1000 BC to 1900 (Table 1-1) xamles of Modern ra Technology timelines (Tables 1-2 and 1-3) Concet of fields (gravitational, electric, magnetic) Static vs. dynamic fields (Table 1-6) The M Sectrum Secial Illustrations: Timelines (Tables 1-1 to 1-3) from CD-ROM Timeline for lectromagnetics in the Classical ra ca. 900 BC ca. 600 BC Legend has it that while walking across a field in northern Greece, a sheherd named Magnus exeriences a ull on the iron nails in his sandals by the black rock he was standing on. The region was later named Magnesia and the rock became known as magnetite [a form of iron with ermanent magnetism]. Greek hilosoher Thales describes how amber, after being rubbed with cat fur, can ick u feathers [static electricity]. ca Magnetic comass used as a navigational device Benjamin Franklin (American) invents the lightning rod and demonstrates that lightning is electricity Charles-Augustin de Coulomb (French) demonstrates that the electrical force between charges is roortional to the inverse of the square of the distance between them Alessandro Volta (Italian) develos the first electric battery Hans Christian Oersted (Danish) demonstrates the interconnection between electricity and magnetism through his discovery that an electric current in a wire causes a comass needle to orient itself erendicular to the wire.

6 2 Chater 2: Vector Algebra Lesson #2 Chater Section: 2-1 Toics: Basic laws of vector algebra Highlights: Vector magnitude, direction, unit vector Position and distance vectors Vector addition and multilication - Dot roduct - Vector roduct - Trile roduct Secial Illustrations: CD-ROM Module 2.2

7 3 Lesson #3 Chater Section: 2-2 Toics: Coordinate systems Highlights: Commonly used coordinate systems: Cartesian, cylindrical, sherical Choice is based on which one best suits roblem geometry Differential surface vectors and differential volumes Secial Illustrations: xamle 2-5 Technology Brief on GPS (CD-ROM) Global Positioning System The Global Positioning System (GPS), initially develoed in the 1980s by the U.S. Deartment of Defense as a navigation tool for military use, has evolved into a system with numerous civilian alications including vehicle tracking, aircraft navigation, ma dislays in automobiles, and toograhic maing. The overall GPS is comrised of 3 segments. The sace segment consists of 24 satellites (A), each circling arth every 12 hours at an orbital altitude of about 12,000 miles and transmitting continuous coded time signals. The user segment consists of handheld or vehicle-mounted receivers that determine their own locations by receiving and rocessing multile satellite signals. The third segment is a network of five ground stations, distributed around the world, that monitor the satellites and rovide them with udates on their recise orbital information. GPS rovides a location inaccuracy of about 30 m, both horizontally and vertically, but it can be imroved to within 1 m by differential GPS (see right).

8 4 Lesson #4 Chater Section: 2-3 Toics: Coordinate transformations Highlights: Basic logic for decomosing a vector in one coordinate system into the coordinate variables of another system Transformation relations (Table 2-2) Secial Illustrations: xamle 2-8

9 CHAPTR 2 5 Chater 2 Section 2-1: Vector Algebra Problem 2.1 Vector A starts at oint (1; 1; 2) and ends at oint (2; 1;0). Find a unit vector in the direction of A. Solution: A = ˆx(2 1)+ŷ( 1 ( 1)) + ẑ(0 ( 2)) = ˆx + ẑ2; jaj = = 2:24; â A ˆx + ẑ2 = = = ˆx0:45 +ẑ0:89: jaj 2:24 Problem 2.2 Given vectors A = ˆx2 ŷ3+ẑ, B = ˆx2 ŷ +ẑ3, and C = ˆx4+ŷ2 ẑ2, show that C is erendicular to both A and B. Solution: A C =(ˆx2 ŷ3 + ẑ) (ˆx4 + ŷ2 ẑ2) =8 6 2 = 0; B C =(ˆx2 ŷ + ẑ3) (ˆx4 + ŷ2 ẑ2) =8 2 6 = 0: Problem 2.3 In Cartesian coordinates, the three corners of a triangle are P 1 (0;2;2), P 2 (2; 2;2), and P 3 (1;1; 2). Find the area of the triangle.! Solution: Let B = P 1 P 2 = ˆx2 ŷ4 and C = P 1 P 3 = ˆx ŷ ẑ4 reresent two sides of the triangle. Since the magnitude of the cross roduct is the area of the arallelogram (see the definition of cross roduct in Section 2-1.4), half of this is the area of the triangle: A = 1 2 jb Cj = 1 2j(ˆx2 ŷ4) (ˆx ŷ ẑ4)j = 1 2jˆx( 4)( 4)+ŷ( (2)( 4)) + ẑ(2( 1) ( 4)1)j = 1 2 jˆx16 + ŷ8 + ẑ2j = = = 9;! where the cross roduct is evaluated with q. (2.27). Problem 2.4 Given A = ˆx2 ŷ3 + ẑ1 and B = ˆxB x + ŷ2 + ẑb z : (a) find B x and B z if A is arallel to B; (b) find a relation between B x and B z if A is erendicular to B.

10 6 CHAPTR 2 Solution: (a) If A is arallel to B, then their directions are equal or oosite: â A = ±â B,or A=jAj = ±B=jBj; ˆx2 ŷ3 + ẑ 14 = ± ˆxB x + ŷ2 + ẑb z 4 + B 2 x + B 2 z : From the y-comonent, 3 ±2 = B 2 x + B 2 z which can only be solved for the minus sign (which means that A and B must oint in oosite directions for them to be arallel). Solving for B 2 x + B2 z, From the x-comonent, and, from the z-comonent, 2 B 2 x + B2 z 14 2 = 4 20 = 3 9 : 2 B x = ; B 2 56 x = 14 56= = 3 B z = 2 3 : This is consistent with our result for B 2 x + B2 z. These results could also have been obtained by assuming θ AB was 0 ffi or 180 ffi and solving jajjbj = ±A B, or by solving A B = 0. (b) If A is erendicular to B, then their dot roduct is zero (see Section 2-1.4). Using q. (2.17), or 0 = A B = 2B x 6 + B z ; B z = 6 2B x : There are an infinite number of vectors which could be B and be erendicular to A, but their x- and z-comonents must satisfy this relation. This result could have also been obtained by assuming θ AB = 90 ffi and calculating jajjbj = ja Bj. Problem 2.5 Given vectors A = ˆx + ŷ2 ẑ3, B = ˆx3 ŷ4, and C = ŷ3 ẑ4, find

11 CHAPTR 2 7 (a) A and â, (b) the comonent of B along C, (c) θ AC, (d) A C, (e) A (B C), (f) A (B C), (g) ˆx B, and (h) (A ŷ) ẑ. Solution: (a) From q. (2.4), and, from q. (2.5), A = q ( 3) 2 = 14; â A = ˆx + ŷ2 ẑ3 14 : (b) The comonent of B along C (see Section 2-1.4) is given by (c) From q. (2.21), Bcosθ BC = B C C = 12 5 : θ AC = cos 1 A C AC = cos = cos = 15:8ffi : (d) From q. (2.27), A C = ˆx(2( 4) ( 3)3)+ŷ(( 3)0 1( 4)) +ẑ(1(3) 0( 3)) = ˆx + ŷ4 + ẑ3: (e) From q. (2.27) and q. (2.17), A (B C) =A (ˆx16 + ŷ12 + ẑ9) =1(16)+2(12)+( 3)9 = 13: q. (2.30) could also have been used in the solution. Also, q. (2.29) could be used in conjunction with the result of art (d). (f) By reeated alication of q. (2.27), A (B C)=A (ˆx16 + ŷ12 + ẑ9) =ˆx54 ŷ57 ẑ20: q. (2.33) could also have been used.

12 8 CHAPTR 2 (g) From q. (2.27), ˆx B = ẑ4: (h) From q. (2.27) and q. (2.17), (A ŷ) ẑ =(ˆx3 + ẑ) ẑ = 1: q. (2.29) and q. (2.25) could also have been used in the solution. Problem 2.6 Given vectors A = ˆx2 ŷ +ẑ3 and B = ˆx3 ẑ2, find a vector C whose magnitude is 6 and whose direction is erendicular to both A and B. Solution: The cross roduct of two vectors roduces a new vector which is erendicular to both of the original vectors. Two vectors exist which have a magnitude of 6 and are orthogonal to both A and B: one which is 6 units long in the direction of the unit vector arallel to A B, and one in the oosite direction. C = ±6 A B (ˆx2 ŷ + ẑ3) (ˆx3 ẑ2) = ±6 ja Bj j(ˆx2 ŷ + ẑ3) (ˆx3 ẑ2)j ˆx2 + ŷ13 + ẑ3 = ±6 ß±(ˆx0:89 +ŷ5:78 +ẑ1:33): Problem 2.7 Given A = ˆx(2x +3y) ŷ(2y+3z) +ẑ(3x y), determine a unit vector arallel to A at oint P(1; 1;2). Solution: The unit vector arallel to A = ˆx(2x + 3y) ŷ(2y + 3z)+ẑ(3x y) at the oint P(1; 1;2) is A(1; 1;2) ja(1; 1;2)j = q ˆx ŷ4 + ẑ4 ( 1) 2 +( 4) = ˆx ŷ4 + ẑ4 33 ß ˆx0:17 ŷ0:70 +ẑ0:70: Problem 2.8 By exansion in Cartesian coordinates, rove: (a) the relation for the scalar trile roduct given by (2.29), and (b) the relation for the vector trile roduct given by (2.33). Solution: (a) Proof of the scalar trile roduct given by q. (2.29): From q. (2.27), A B = ˆx(A y B z A z B y )+ŷ(a z B x A x B z )+ẑ(a x B y A y B x );

13 CHAPTR 2 9 B C = ˆx(B y C z B z C y )+ŷ(b z C x B x C z )+ẑ(b x C y B y C x ); C A = ˆx(C y A z C z A y )+ŷ(c z A x C x A z )+ẑ(c x A y C y A x ): mloying q. (2.17), it is easily shown that A (B C) =A x (B y C z B z C y )+A y (B z C x B x C z )+A z (B x C y B y C x ); B (C A) =B x (C y A z C z A y )+B y (C z A x C x A z )+B z (C x A y C y A x ); C (A B) =C x (A y B z A z B y )+C y (A z B x A x B z )+C z (A x B y A y B x ); which are all the same. (b) Proof of the vector trile roduct given by q. (2.33): The evaluation of the left hand side emloys the exression above for B C with q. (2.27): A (B C)=A (ˆx(B y C z B z C y )+ŷ(b z C x B x C z )+ẑ(b x C y B y C x )) = ˆx(A y (B x C y B y C x ) A z (B z C x B x C z )) + ŷ(a z (B y C z B z C y ) A x (B x C y B y C x )) + ẑ(a x (B z C x B x C z ) A y (B y C z B z C y )); while the right hand side, evaluated with the aid of q. (2.17), is B(A C) C(A B) =B(A x C x + A y C y + A z C z ) C(A x B x + A y B y + A z B z ) = ˆx(B x (A y C y + A z C z ) C x (A y B y + A z B z )) + ŷ(b y (A x C x + A z C z ) C y (A x B x + A z B z )) + ẑ(b z (A x C x + A y C y ) C z (A x B x + A y B y )): By rearranging the exressions for the comonents, the left hand side is equal to the right hand side. Problem 2.9 Find an exression for the unit vector directed toward the origin from an arbitrary oint on the line described by x = 1 and z = 2. Solution: An arbitrary oint on the given line is (1;y;2). The vector from this oint to (0;0;0) is: A = ˆx(0 1)+ŷ(0 y)+ẑ(0 2) = ˆx ŷy 2ẑ; jaj = 1 + y = 5 + y 2 ; â = A jaj ˆx ŷy ẑ2 = : 5 + y 2

14 10 CHAPTR 2 Problem 2.10 Find an exression for the unit vector directed toward the oint P located on the z-axis at a height h above the x y lane from an arbitrary oint Q(x;y;2) in the lane z = 2. Solution: Point P is at (0;0;h). Vector A from Q(x;y;2) to P(0;0;h) is: A = ˆx(0 x)+ŷ(0 y)+ẑ(h 2) = ˆxx ŷy + ẑ(h 2); jaj =[x 2 + y 2 +(h 2) 2 ] 1=2 ; â = A ˆxx ŷy + ẑ(h 2) = jaj [x 2 + y 2 +(h 2) 2 ] 1=2 : Problem 2.11 Find a unit vector arallel to either direction of the line described by 2x z = 4: Solution: First, we find any two oints on the given line. Since the line equation is not a function of y, the given line is in a lane arallel to the x z lane. For convenience, we choose the x z lane with y = 0. For x = 0, z = 4. Hence, oint P is at (0;0; 4). For z = 0, x = 2. Hence, oint Q is at (2;0;0). Vector A from P to Q is: A = ˆx(2 0)+ŷ(0 0)+ẑ(0 + 4) =ˆx2 + ẑ4; â A ˆx2 + ẑ4 = = : jaj 20 Problem 2.12 Two lines in the x y lane are described by the exressions: Line 1 x + 2y = 6; Line 2 3x + 4y = 8: Use vector algebra to find the smaller angle between the lines at their intersection oint. Solution: Intersection oint is found by solving the two equations simultaneously: 2x 4y = 12; 3x + 4y = 8:

15 CHAPTR (0, 2) (0, -3) A B (20, -13) θ AB Figure P2.12: Lines 1 and 2. The sum gives x = 20, which, when used in the first equation, gives y = 13. Hence, intersection oint is (20; 13). Another oint on line 1 is x = 0, y = 3. Vector A from (0; 3) to (20; 13) is A = ˆx(20)+ŷ( ) =ˆx20 ŷ10; jaj = = 500: A oint on line 2 is x = 0, y = 2. Vector B from (0;2) to (20; 13) is Angle between A and B is A B θ AB = cos 1 jajjbj B = ˆx(20)+ŷ( 13 2) =ˆx20 ŷ15; jbj = = 625: = cos = 10:3 ffi : Problem 2.13 A given line is described by x + 2y = 4: Vector A starts at the origin and ends at oint P on the line such that A is orthogonal to the line. Find an exression for A.

16 12 CHAPTR 2 Solution: We first lot the given line. Next we find vector B which connects oint P 1 (0;2) to P 2 (4;0), both of which are on the line: B = ˆx(4 0)+ŷ(0 2) =ˆx4 ŷ2: Vector A starts at the origin and ends on the line at P. If the x-coordinate of P is x, y P (0,2) 1 (0,0) A B P (4,0) 2 x Figure P2.13: Given line and vector A. then its y-coordinate has to be (4 x)=2 in order to be on the line. Hence P is at (x; (4 x)=2). Vector A is 4 x A = ˆxx + ŷ : 2 But A is erendicular to the line. Hence,» ˆxx + ŷ A B = 0; 4 x 2 4x (4 x) =0; (ˆx4 ŷ2) =0; x = 4 5 = 0:8: or Hence, 4 0:8 A = ˆx0:8 + ŷ = ˆx0:8 + ŷ1:6: 2 Problem 2.14 Show that, given two vectors A and B,

17 CHAPTR 2 13 (a) the vector C defined as the vector comonent of B in the direction of A is given by A(B A) C = â(b â) = jaj 2 ; where â is the unit vector of A, and (b) the vector D defined as the vector comonent of B erendicular to A is given by A(B A) D = B jaj 2 : Solution: (a) By definition, B â is the comonent of B along â. The vector comonent of (B â) along A is C = â(b â) = A jaj B A jaj = A(B A) jaj 2 : (b) The figure shows vectors A, B, and C, where C is the rojection of B along A. It is clear from the triangle that B = C + D; or D = B C = B A(B A) jaj 2 : A C D B Figure P2.14: Relationshis between vectors A, B, C, and D.

18 14 CHAPTR 2 Problem 2.15 A certain lane is described by 2x + 3y + 4z = 16: Find the unit vector normal to the surface in the direction away from the origin. Solution: Procedure: 1. Use the equation for the given lane to find three oints, P 1, P 2 and P 3 on the lane. 2. Find vector A from P 1 to P 2 and vector B from P 1 to P Cross roduct of A and B gives a vector C orthogonal to A and B, and hence to the lane. 4. Check direction of ĉ. Stes: 1. Choose the following three oints: 2. Vector A from P 1 to P 2 3. Vector B from P 1 to P 3 C = A B P 1 at (0;0;4); P 2 at (8;0;0); P 3 at (0; 16 3 ;0): A = ˆx (8 0)+ŷ (0 0)+ẑ (0 4) =ˆx8 ẑ4 B = ˆx (0 0)+ŷ ẑ (0 4) =ŷ 16 3 ẑ4 = ˆx (A y B z A z B y )+ŷ(a z B x A x B z )+ẑ(a x B y A y B x ) = ˆx 0 ( 4) ( 4) 16 + ŷ (( 4) 0 8 ( 4)) +ẑ 3 = ˆx ŷ32 + ẑ

19 CHAPTR 2 15 Verify that C is orthogonal to A and B A C = (32 0) ( 4) = = 0 B C = ( 4) = = 0 4. C = ˆx ŷ32 + ẑ ĉ = C ˆx jcj = 3 + ŷ32 + ẑ q 3 = 64 2 ˆx0:37 +ŷ0:56 +ẑ0:74: ĉ oints away from the origin as desired. Problem 2.16 Given B = ˆx(2z 3y) +ŷ(2x 3z) ẑ(x + y), find a unit vector arallel to B at oint P(1;0; 1). Solution: At P(1; 0; 1), B = ˆx( 2)+ŷ(2 + 3) ẑ(1) = ˆx2 + ŷ5 ẑ; ˆb B ˆx2 + ŷ5 ẑ ˆx2 + ŷ5 ẑ = = = : jbj Problem 2.17 When sketching or demonstrating the satial variation of a vector field, we often use arrows, as in Fig (P2.17), wherein the length of the arrow is made to be roortional to the strength of the field and the direction of the arrow is the same as that of the field s. The sketch shown in Fig. P2.17, which reresents the vector field = ˆrr, consists of arrows ointing radially away from the origin and their lengths increase linearly in roortion to their distance away from the origin. Using this arrow reresentation, sketch each of the following vector fields: (a) 1 = ˆxy, (b) 2 = ŷx, (c) 3 = ˆxx + ŷy, (d) 4 = ˆxx + ŷ2y, (e) 5 = ˆφr, (f) 6 = ˆrsinφ.

20 16 CHAPTR 2 y x Figure P2.17: Arrow reresentation for vector field = ˆrr (Problem 2.17). Solution: (a) y x P2.17a: = - x^ y 1

21 CHAPTR 2 17 (b) y x (c) P2.17b: 2 = ŷx y x P2.17c: 3 = x^ x + y^ y

22 18 CHAPTR 2 (d) y x (e) P2.17d: 4 = x^ x + y^ 2y y x P2.17e: 5 = φ^ r

23 CHAPTR 2 19 (f) y x P2.17f: 6 = r^ sinφ Problem 2.18 Use arrows to sketch each of the following vector fields: (a) 1 = ˆxx ŷy, (b) 2 = ˆφ, (c) 3 = ŷ 1 x, (d) 4 = ˆrcosφ. Solution:

24 20 CHAPTR 2 (a) y x (b) P2.18a: 1 = x^ x - y^ y y x P2.18b: = - φ^ 2

25 CHAPTR 2 21 (c) y x Indicates is infinite (d) P2.18c: 3 = y^ (1/x) y x P2.18d: 4 = r^ cosφ

26 22 CHAPTR 2 Sections 2-2 and 2-3: Coordinate Systems Problem 2.19 Convert the coordinates of the following oints from Cartesian to cylindrical and sherical coordinates: (a) P 1 (1;2;0), (b) P 2 (0;0;3), (c) P 3 (1;1;2), (d) P 4 ( 3;3; 3). Solution: Use the coordinate variables column in Table 2-2. (a) In the cylindrical coordinate system, P 1 =( ;tan 1 (2=1);0) =( 5;1:107 rad;0) ß (2:24;63:4 ffi ;0): In the sherical coordinate system, P 1 =( ;tan 1 ( =0);tan 1 (2=1)) =( 5;π=2 rad;1:107 rad) ß (2:24;90:0 ffi ;63:4 ffi ): Note that in both the cylindrical and sherical coordinates, φ is in Quadrant I. (b) In the cylindrical coordinate system, P 2 =( ;tan 1 (0=0);3) =(0;0 rad;3) =(0;0 ffi ;3): In the sherical coordinate system, P 2 =( ;tan 1 ( =3);tan 1 (0=0)) =(3;0 rad;0 rad) =(3;0 ffi ;0 ffi ): Note that in both the cylindrical and sherical coordinates, φ is arbitrary and may take any value. (c) In the cylindrical coordinate system, P 3 =( ;tan 1 (1=1);2) =( 2;π=4 rad;2) ß (1:41;45:0 ffi ;2): In the sherical coordinate system, P 3 =( ;tan 1 ( =2);tan 1 (1=1)) =( 6;0:616 rad;π=4 rad) ß (2:45;35:3 ffi ;45:0 ffi ): Note that in both the cylindrical and sherical coordinates, φ is in Quadrant I.

27 CHAPTR 2 23 (d) In the cylindrical coordinate system, q P 4 =( ( 3) ;tan 1 (3= 3); 3) =(3 2;3π=4 rad; 3) ß (4:24;135:0 ffi ; 3): In the sherical coordinate system, q q P 4 =( ( 3) ( 3) 2 ;tan 1 ( ( 3) = 3);tan 1 (3= 3)) =(3 3;2:187 rad;3π=4 rad) ß (5:20;125:3 ffi ;135:0 ffi ): Note that in both the cylindrical and sherical coordinates, φ is in Quadrant II. Problem 2.20 Convert the coordinates of the following oints from cylindrical to Cartesian coordinates: (a) P 1 (2;π=4; 3), (b) P 2 (3;0;0), (c) P 3 (4;π;2). Solution: (a) P 1 (x;y;z) =P 1 (r cos φ;r sin φ;z) =P 1 2cos π 4 ;2sin π 4 ; 3 = P 1 (1:41;1:41; 3): (b) P 2 (x;y;z) =P 2 (3cos 0;3sin 0;0) =P 2 (3;0;0). (c) P 3 (x;y;z) =P 3 (4cos π;4sin π;2) =P 3 ( 4;0;2). Problem 2.21 Convert the coordinates of the following oints from sherical to cylindrical coordinates: (a) P 1 (5;0;0), (b) P 2 (5;0;π), (c) P 3 (3;π=2;π). Solution: (a) P 1 (r;φ;z) =P 1 (Rsinθ;φ;Rcos θ) =P 1 (5sin 0;0;5cos 0) (b) P 2 (r;φ;z) =P 2 (5sin 0;π;5cos 0) =P 2 (0;π;5). = P 1 (0;0;5):

28 24 CHAPTR 2 (c) P 3 (r;φ;z) =P 3 (3sin π 2 ;π;3cos π 2 )=P 3(3;π;0). Problem 2.22 Use the aroriate exression for the differential surface area ds to determine the area of each of the following surfaces: (a) r = 3; 0» φ» π=3; 2» z» 2, (b) 2» r» 5; π=2» φ» π; z = 0, (c) 2» r» 5; φ = π=4; 2» z» 2, (d) R = 2; 0» θ» π=3; 0» φ» π, (e) 0» R» 5; θ = π=3; 0» φ» 2π. Also sketch the outlines of each of the surfaces. Solution: Φ = π/3 3 2 y x (a) (b) (c) (d) (e) Figure P2.22: Surfaces described by Problem (a) Using q. (2.43a), Z 2 Z π=3 A = z= 2 φ=0 (r)j r=3 dφ dz = (3φz)j π=3 φ=0 fi fifi 2 z= 2 = 4π:

29 CHAPTR 2 25 (b) Using q. (2.43c), Z 5 Z π A = r=2 φ=π=2 Z 2 Z 5 A = z= 2 r=2 Z π=3z π A = θ=0 φ=0 (r)j z=0 dφ dr = φ fi fi 1 2 r2 5 fifi π 21π r=2 = φ=π=2 4 : (c) Using q. (2.43b), fi (1)j drdz φ=π=4 = (rz)j 2 fifi 5 z= 2 = 12: r=2 (d) Using q. (2.50b), R 2 sinθ fi fi R=2 dφ dθ = ( 4φcos θ)j π=3 fifi π θ=0 = 2π: φ=0 (e) Using q. (2.50c), (Rsinθ)j dφ dr θ=π=3 = 1 2 R2 φsin π fi fi fifi 2π fififi 5 3 φ=0 Z 5 Z 2π A = R=0 φ=0 R=0 = 25 3π 2 : Problem 2.23 Find the volumes described by (a) 2» r» 5; π=2» φ» π; 0» z» 2, (b) 0» R» 5; 0» θ» π=3; 0» φ» 2π. Also sketch the outline of each volume. Solution: z z y y x (a) x (b) Figure P2.23: Volumes described by Problem (a) From q. (2.44), Z 2 Z π Z 5 V = z=0 φ=π=2 r=2 rdrdφdz φz fi fi fi 1 = 2 r2 5 fifi π fififi 2 21π = r=2 φ=π=2 z=0 2 :

30 26 CHAPTR 2 (b) From q. (2.50e), Z 2π Z π=3z 5 V = φ=0 θ=0 R=0 = R 2 sin θ drdθ dφ ψ fi cosθ R3 fififi 5 3 φ R=0!fi fififi fi π=3 θ=0 1fi fi fi Afi fi fi 2π φ=0 = 125π 3 : Problem 2.24 A section of a shere is described by 0» R» 2, 0» θ» 90 ffi ; and 30 ffi» φ» 90 ffi. Find: (a) the surface area of the sherical section, (b) the enclosed volume. Also sketch the outline of the section. Solution: z y x φ=30 o Figure P2.24: Outline of section.

31 CHAPTR 2 27 Z π=2 Z π=2 S = φ=π=6 θ=0 π 2 π 6 h R 2 sinθ dθ dφj R=2 cosθj π=2 0 = 4 Z 2 Z π=2 Z π=2 V = R=0 φ=π=6 θ=0 fi = R3 2 fi fi 3 fi 0 π 2 π 6 i = 4 π 4π 3 = 3 R 2 sin θ drdθ dφ [ cosθj π=2 8 π 8π 0 ]= 3 3 = 9 (m 2 ); (m 3 ): Problem 2.25 A vector field is given in cylindrical coordinates by = ˆrr cos φ + ˆφr sinφ + ẑz 2 : Point P(4;π;2) is located on the surface of the cylinder described by r = 4. At oint P, find: (a) the vector comonent of erendicular to the cylinder, (b) the vector comonent of tangential to the cylinder. Solution: (a) n = ˆr(ˆr ) =ˆr[ˆr (ˆrr cosφ + ˆφr sin φ + ẑz 2 )] = ˆrr cos φ. At P(4;π;2), n = ˆr4cos π = ˆr4. (b) t = n = ˆφr sinφ + ẑz 2. At P(4;π;2), t = ˆφ4sin π + ẑ2 2 = ẑ4. Problem 2.26 coordinates by At a given oint in sace, vectors A and B are given in sherical A = ˆR4 + ˆθ2 ˆφ; B = ˆR2 + ˆφ3: Find: (a) the scalar comonent, or rojection, of B in the direction of A, (b) the vector comonent of B in the direction of A, (c) the vector comonent of B erendicular to A. Solution:

32 28 CHAPTR 2 (a) Scalar comonent of B in direction of A: C = B â = B A jaj =( ˆR2 + ˆφ3) ( ˆR4 + ˆθ2 ˆφ) = = 11 = 2:4: (b) Vector comonent of B in direction of A: C = âc = A C jaj =(ˆR4 ( 2:4) + ˆθ2 ˆφ) 21 (c) Vector comonent of B erendicular to A: = ( ˆR2:09 + ˆθ1:05 ˆφ0:52): D = B C =( ˆR2 + ˆφ3)+(ˆR2:09 + ˆθ1:05 ˆφ0:52) = ˆR0:09 + ˆθ1:05 + ˆφ2:48: Problem 2.27 Given vectors A = ˆr(cos φ + 3z) ˆφ(2r + 4sin φ)+ẑ(r 2z); B = ˆrsinφ + ẑcosφ; find (a) θ AB at (2;π=2;0), (b) a unit vector erendicular to both A and B at (2; π=3; 1). Solution: It doesn t matter whether the vectors are evaluated before vector roducts are calculated, or if the vector roducts are directly calculated and the general results are evaluated at the secific oint in question. (a) At (2;π=2;0), A = ˆφ8 + ẑ2 and B = ˆr. From q. (2.21), θ AB = cos 1 A B AB = cos 1 0 AB = 90 ffi : (b) At (2;π=3;1), A = ˆr 7 2 ˆφ4( ) and B = ˆr ẑ 1 2. Since A B is erendicular to both A and B, a unit vector erendicular to both A and B is given by ± A B ja Bj = ± ˆr( 4( ))( 2 ) ˆφ( 7 2 )( 1 2 ) ẑ(4( ))( 2 3) q (2( )) +( 7 4 )2 +( ) 2 ß (ˆr0:487 + ˆφ0:228 +ẑ0:843):

33 CHAPTR 2 29 Problem 2.28 Find the distance between the following airs of oints: (a) P 1 (1;2;3) and P 2 ( 2; 3;2) in Cartesian coordinates, (b) P 3 (1;π=4;2) and P 4 (3;π=4;4) in cylindrical coordinates, (c) P 5 (2;π=2;0) and P 6 (3;π;0) in sherical coordinates. Solution: (a) (b) (c) d =[( 2 1) 2 +( 3 2) 2 +(2 3) 2 ] 1=2 =[ ] 1=2 = 35 = 5:92: d =[r2 2 + r2 1 2r 1 r 2 cos(φ 2 φ 1 )+(z 2 z 1 ) 2 ] 1=2 h π = cos 4 π +(4 2) 2i 1=2 4 =( ) 1=2 = 8 1=2 = 2:83: d = fr R2 1 2R 1 R 2 [cosθ 2 cosθ 1 + sinθ 1 sinθ 2 cos(φ 2 φ 1 )]g 1=2 n = hcosπcos π 2 + sin π io 1=2 2 sin πcos(0 0) = f g 1=2 = 13 = 3:61: Problem 2.29 Determine the distance between the following airs of oints: (a) P 1 (1;1;2) and P 2 (0;2;2), (b) P 3 (2;π=3;1) and P 4 (4;π=2;0), (c) P 5 (3;π;π=2) and P 6 (4;π=2;π). Solution: (a) From q. (2.66), d = (b) From q. (2.67), r d = (2)(4)cos q (0 1) 2 +(2 1) 2 +(2 2) 2 = 2 : π 2 π q +(0 1) 2 = ß 2:67: 3

34 30 CHAPTR 2 (c) From q. (2.68), r d = (3)(4) cos π 2 cos π + sinπsin π 2 cos π π = 5: 2 Problem 2.30 Transform the following vectors into cylindrical coordinates and then evaluate them at the indicated oints: (a) A = ˆx(x + y) at P 1 (1;2;3), (b) B = ˆx(y x)+ŷ(x y) at P 2 (1;0;2), (c) C = ˆxy 2 =(x 2 + y 2 ) ŷx 2 =(x 2 + y 2 )+ẑ4 atp 3 (1; 1;2), (d) D = ˆRsinθ + ˆθcosθ + ˆφcos 2 φ at P 4 (2;π=2;π=4), (e) = ˆRcosφ + ˆθsinφ + ˆφsin 2 θ at P 5 (3;π=2;π). Solution: From Table 2-2: (a) (b) A =(ˆrcosφ ˆφsinφ)(r cos φ + r sinφ) = ˆrr cos φ(cos φ + sinφ) ˆφr sin φ(cos φ + sinφ); ;tan 1 (2=1);3) =( 5;63:4 ffi ;3); P 1 =( A(P 1 )=(ˆr0:447 ˆφ0:894) 5 (:447 + :894) =ˆr1:34 ˆφ2:68: B =(ˆrcosφ ˆφsinφ)(r sin φ r cosφ)+(ˆφcosφ + ˆrsinφ)(r cos φ r sinφ) = ˆrr(2sin φcos φ 1)+ˆφr(cos 2 φ sin 2 φ)=ˆrr(sin 2φ 1)+ˆφr cos 2φ; ;tan 1 (0=1);2) =(1;0 ffi ;2); P 2 =( B(P 2 )= ˆr + ˆφ: (c) (d) C =(ˆrcosφ ˆφsinφ) r2 sin 2 φ r 2 (ˆφcos φ + ˆrsinφ) r2 cos 2 φ r 2 = ˆrsinφcos φ(sin φ cosφ) ˆφ(sin 3 φ + cos 3 φ)+ẑ4; q 1 2 +( 1) 2 ;tan 1 ( 1=1);2) =( 2; 45 ffi ;2); P 3 =( C(P 3 )=ˆr0:707 +ẑ4: + ẑ4 D =(ˆrsinθ + ẑcosθ)sinθ +(ˆrcos θ ẑsin θ)cos θ + ˆφcos 2 φ = ˆr + ˆφcos 2 φ;

35 CHAPTR 2 31 P 4 =(2sin (π=2);π=4;2cos (π=2))=(2;45 ffi ;0); D(P 4 )=ˆr + ˆφ 1 2 : (e) =(ˆrsin θ + ẑcos θ)cos φ +(ˆrcosθ ẑsinθ)sin φ + ˆφsin 2 θ; 3; π 2 ;π ; P 5 = (P 5 )= ˆrsin π 2 + ẑcos π 2 cos π + ˆrcos π 2 ẑsin π sinπ + 2 ˆφsin 2 π = ˆr + ˆφ: 2 Problem 2.31 Transform the following vectors into sherical coordinates and then evaluate them at the indicated oints: (a) A = ˆxy 2 + ŷxz + ẑ4 at P 1 (1; 1;2), (b) B = ŷ(x 2 + y 2 + z 2 ) ẑ(x 2 + y 2 ) at P 2 ( 1;0;2), (c) C = ˆrcos φ ˆφsin φ + ẑcos φsinφ at P 3 (2;π=4;2), and (d) D = ˆxy 2 =(x 2 + y 2 ) ŷx 2 =(x 2 + y 2 )+ẑ4 at P 4 (1; 1;2). Solution: From Table 2-2: (a) (b) A =(ˆRsin θcosφ + ˆθcosθcos φ ˆφsinφ)(Rsinθsinφ) 2 P 1 = +(ˆRsinθsin φ + ˆθcosθsin φ + ˆφcos φ)(rsinθcos φ)(rcosθ) +(ˆRcosθ ˆθsinθ)4 = ˆR(R 2 sin 2 θsin φcosφ(sin θsin φ + cosθ)+4cosθ) + ˆθ(R 2 sinθcos θsinφcos φ(sin θsin φ + cosθ) 4sinθ) + ˆφR 2 sinθ(cos θcos 2 φ sinθsin 3 φ); q q1 2 +( 1) ;tan ( 1) 2 =2 =( 6;35:3 ffi ; 45 ffi ); A(P 1 ) ß ˆR2:856 ˆθ2:888 + ˆφ2:123: ;tan 1 ( 1=1) B =(ˆRsinθsinφ + ˆθcosθsin φ + ˆφcosφ)R 2 ( ˆRcosθ ˆθsinθ)R 2 sin 2 θ = ˆRR 2 sinθ(sin φ sinθcosθ)+ˆθr 2 (cos θsinφ + sin 3 θ)+ˆφr 2 cosφ; q q( 1) ;tan 1 ( 1) =2 P 2 = =( 5;26:6 ffi ;180 ffi ); ;tan 1 (0=( 1))

36 32 CHAPTR 2 (c) (d) B(P 2 ) ß ˆR0:896 + ˆθ0:449 ˆφ5: C =(ˆRsin θ + ˆθcos θ)cos φ ˆφsin φ +(ˆRcosθ ˆθsinθ)cosφsin φ = ˆRcosφ(sin θ + cosθsinφ)+ˆθcos φ(cos θ sinθsin φ) ˆφsinφ; ;tan 1 (2=2);π=4 =(2 2;45 ffi ;45 ffi ); P 3 = C(P 3 ) ß ˆR0:854 + ˆθ0:146 ˆφ0:707: D =(ˆRsinθcos φ + ˆθcosθcos φ ˆφsin R 2 sin 2 θsin 2 φ φ) R 2 sin 2 θsin 2 φ + R 2 sin 2 θcos 2 φ R ( ˆRsinθsin φ + ˆθcos 2 sin 2 θcos 2 φ θsinφ + ˆφcosφ) R 2 sin 2 θsin 2 φ + R 2 sin 2 θcos 2 φ +(ˆRcosθ ˆθsinθ)4 = ˆR(sinθcos φsin 2 φ sinθsin φcos 2 φ + 4cosθ) + ˆθ(cosθcos φsin 2 φ cosθsinφcos 2 φ 4sinθ) ˆφ(cos 3 φ + sin 3 φ); P 4 (1; 1;2) =P 4 h ;tan 1 ( 1 + 1=2);tan 1 ( 1=1) i = P 4 ( 6;35:26 ffi ; 45 ffi ); D(P 4 )= ˆR(sin35:26 ffi cos 45 ffi sin 2 45 ffi sin35:26 ffi sin( 45 ffi )cos 2 45 ffi + 4cos35:26 ffi ) + ˆθ(cos35:26 ffi cos45 ffi sin 2 45 ffi cos35:26 ffi sin( 45 ffi )cos 2 45 ffi 4sin35:26 ffi ) ˆφ(cos 3 45 ffi + sin 3 45 ffi ) = ˆR3:67 ˆθ1:73 ˆφ0:707: Problem 2.32 Find a vector G whose magnitude is 4 and whose direction is erendicular to both vectors and F, where = ˆx + ŷ2 ẑ2 and F = ŷ3 ẑ6. Solution: The cross roduct of two vectors roduces a third vector which is erendicular to both of the original vectors. Two vectors exist that satisfy the stated

37 CHAPTR 2 33 conditions, one along F and another along the oosite direction. Hence, G = ±4 F (ˆx + ŷ2 ẑ2) (ŷ3 ẑ6) = ±4 j Fj j(ˆx + ŷ2 ẑ2) (ŷ3 ẑ6)j ( ˆx6 + ŷ6 + ẑ3) = ± = ± 4 9 ( ˆx6 + ŷ6 + ẑ3) =± ˆx ŷ ẑ 4 3 : Problem 2.33 A given line is described by the equation: y = x 1: Vector A starts at oint P 1 (0;2) and ends at oint P 2 on the line such that A is orthogonal to the line. Find an exression for A. Solution: We first lot the given line. y P 1 (0, 2) A B P 4 (1, 0) P 2 (x, x-1) x P 3 (0, -1) Next we find a vector B which connects oint P 3 (0;1) to oint P 4 (1;0), both of which are on the line. Hence, B = ˆx (1 0)+ŷ (0 + 1) =ˆx + ŷ : Vector A starts at P 1 (0;2) and ends on the line at P 2. If the x-coordinate of P 2 is x, then its y-coordinate has to be y = x 1, er the equation for the line. Thus, P 2 is at (x;x 1), and vector A is A = ˆx (x 0)+ŷ (x 1 2) =ˆxx + ŷ (x 3):

38 34 CHAPTR 2 Since A is orthogonal to B, A B = 0; [ˆxx + ŷ (x 3)] (ˆx + ŷ) =0 x + x 3 = 0 x 3 = 2 : Finally, A = ˆxx + ŷ (x 3) =ˆx ŷ = ˆx 3 2 ŷ 3 2 : Problem 2.34 Vector field is given by = ˆR 5Rcos θ ˆθ 12 R sin θcosφ + ˆφ3sin φ: Determine the comonent of tangential to the sherical surface R = 2 at oint P(2;30 ffi ;60 ffi ). Solution: At P, is given by = ˆR 5 2cos 30 ffi ˆθ 12 2 sin30ffi cos60 ffi + ˆφ3sin 60 ffi = ˆR 8:67 ˆθ1:5 + ˆφ2:6: The ˆR comonent is normal to the sherical surface while the other two are tangential. Hence, t = ˆθ1:5 + ˆφ2:6: Problem 2.35 Transform the vector A = ˆRsin 2 θcos φ + ˆθcos 2 φ ˆφsinφ into cylindrical coordinates and then evaluate it at P(2; π=2; π=2). Solution: From Table 2-2, A =(ˆr sinθ + ẑcosθ)sin 2 θcosφ +(ˆr cos θ ẑsin θ)cos 2 φ ˆφsinφ = ˆr (sin 3 θcosφ + cosθcos 2 φ) ˆφsinφ + ẑ (cosθsin 2 θcos φ sinθcos 2 φ) At P(2; π=2; π=2), A = ˆφ:

39 35 Chater 3: Vector Calculus Lesson #5 Chater Section: 3-1 Toics: Gradient oerator Highlights: Derivation of T in Cartesian coordinates Directional derivative T in cylindrical and sherical coordinates Secial Illustrations: xamle 3-2(b) CD-ROM Modules 3.1 or 3.2 CD-ROM Demos (any 2)

40 36 Lesson #6 Chater Section: 3-2 Toics: Divergence oerator Highlights: Concet of flux Derivation of. Divergence theorem Secial Illustrations: CD-ROM Modules (any 2) CD-ROM Demos (any 1 or 2)