45788-UlabyISMCh03

45788-UlabyISMCh03 - 107 Chapter 3: Vector Analysis Lesson...

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Unformatted text preview: 107 Chapter 3: Vector Analysis Lesson #14 Chapter -- Section: 3-1 Topics: Basic laws of vector algebra Highlights: Vector magnitude, direction, unit vector Position and distance vectors Vector addition and multiplication - Dot product - Vector product - Triple product Special Illustrations: CD-ROM Module 3.2 108 Lessons #15 and 16 Chapter -- Section: 3-2 Topics: Coordinate systems Highlights: Commonly used coordinate systems: Cartesian, cylindrical, spherical Choice is based on which one best suits problem geometry Differential surface vectors and differential volumes Special Illustrations: Examples 3-3 to 3-5 Technology Brief on "GPS" (CD-ROM) Global Positioning System The Global Positioning System (GPS), initially developed in the 1980s by the U.S. Department of Defense as a navigation tool for military use, has evolved into a system with numerous civilian applications including vehicle tracking, aircraft navigation, map displays in automobiles, and topographic mapping. The overall GPS is composed of 3 segments. The space segment consists of 24 satellites (A), each circling Earth every 12 hours at an orbital altitude of about 12,000 miles and transmitting continuous coded time signals. The user segment consists of hand-held or vehicle-mounted receivers that determine their own locations by receiving and processing multiple satellite signals. The third segment is a network of five ground stations, distributed around the world, that monitor the satellites and provide them with updates on their precise orbital information. GPS provides a location inaccuracy of about 30 m, both horizontally and vertically, but it can be improved to within 1 m by differential GPS (see illustration). 109 Lesson #17 Chapter -- Section: 3-3 Topics: Coordinate transformations Highlights: Basic logic for decomposing a vector in one coordinate system into the coordinate variables of another system Transformation relations (Table 3-2) Special Illustrations: Example 3-8 110 Lesson #18 Chapter -- Section: 3-4 Topics: Gradient operator Highlights: Derivation of T in Cartesian coordinates Directional derivative T in cylindrical and spherical coordinates Special Illustrations: Example 3-10(b) CD-ROM Modules 3.5 or 3.6 CD-ROM Demos 3.1-3.9 (any 2) 111 Lesson #19 Chapter -- Section: 3-5 Topics: Divergence operator Highlights: Concept of "flux" Derivation of .E Divergence theorem Special Illustrations: CD-ROM Modules 3.7-3.11 (any 2) CD-ROM Demos 3.10-3.15 (any 1 or 2) 112 Lesson #20 Chapter -- Section: 3-6 Topics: Curl operator Highlights: Concept of "circulation" Derivation of x B Stokes's theorem Special Illustrations: Example 3-12 113 Lesson #21 Chapter -- Section: 3-7 Topics: Laplacian operator Highlights: Definition of 2 V Definition of 2 E Special Illustrations: Technology Brief on "X-Ray Computed Tomography" X-Ray Computed Tomography Tomography is derived from the Greek words tome, meaning section or slice, and graphia, meaning writing. Computed tomography, also known as CT scan or CAT scan (for computed axial tomography), refers to a technique capable of generating 3-D images of the x-ray attenuation (absorption) properties of an object. This is in contrast with the traditional x-ray technique which produces only a 2-D profile of the object. CT was invented in 1972 by British electrical engineer Godfrey Hounsfield, and independently by Allan Cormack, a South African-born American physicist. The two inventors shared the 1979 Nobel Prize for Physiology or Medicine. Among diagnostic imaging techniques, CT has the decided advantage in having the sensitivity to image body parts on a wide range of densities, from soft tissue to blood vessels and bones. 114 CHAPTER 3 Chapter 3 Section 3-1: Vector Algebra Problem 3.1 Vector A starts at point 1 a unit vector in the direction of A. Solution: 1 3 and ends at point 2 Solution: Problem 3.3 In Cartesian coordinates, the three corners of a triangle are P1 0 4 4 , P2 4 4 4 , and P3 2 2 4 . Find the area of the triangle. ^ ^ ^ ^ ^ Solution: Let B P1 P2 x4 y8 and C P1 P3 x2 y2 z8 represent two sides of the triangle. Since the magnitude of the cross product is the area of the parallelogram (see the definition of cross product in Section 3-1.4), half of this is the area of the triangle: where the cross product is evaluated with Eq. (3.27). ^ ^ ^ ^ ^ ^ Problem 3.4 Given A x2 y3 z1 and B xBx y2 zBz : (a) find Bx and Bz if A is parallel to B; (b) find a relation between Bx and Bz if A is perpendicular to B. ^ x64 ^ y32 ^ z8 1 2 642 322 82 1 2 5184 ^ x 8 8 ^ y 4 8 ^ z4 2 82 A B C ^ x4 ^ y8 ^ x2 ^ y2 ^ z8 1 2 1 2 1 2 1 2 36 B C 8 2 6 0 A C ^ x2 ^ x2 ^ ^ y3 z ^ ^ y z3 ^ x4 ^ x4 ^ y2 ^ y2 ^ z2 ^ z2 8 6 2 0 ^ ^ ^ Problem 3.2 Given vectors A x2 y3 z, B show that C is perpendicular to both A and B. ^ ^ ^ x2 y z3, and C ^ a A 1 A A 9 3 16 ^ ^ x z3 ^ x 0 32 3 16 ^ z 0 95 A ^ x2 1 ^ y 1 1 ^ z0 3 ^ x ^ z3 ^ ^ ^ x4 y2 z2, 1 0 . Find CHAPTER 3 Solution: ^ (a) If A is parallel to B, then their directions are equal or opposite: aA ^ x2 From the y-component, 115 ^ aB , or 4 which can only be solved for the minus sign (which means that A and B must point in opposite directions for them to be parallel). Solving for B 2 B2 , x z From the x-component, and, from the z-component, This is consistent with our result for B 2 B2 . x z These results could also have been obtained by assuming AB was 0 or 180 and solving A B A B, or by solving A B 0. (b) If A is perpendicular to B, then their dot product is zero (see Section 3-1.4). Using Eq. (3.17), or There are an infinite number of vectors which could be B and be perpendicular to A, but their x- and z-components must satisfy this relation. This result could have also been obtained by assuming AB 90 and calculating A B A B. Problem 3.5 Given vectors A ^ x ^ y2 ^ z3, B ^ x2 ^ y4, and C Bz 6 2Bx 0 A B 2Bx 6 Bz ^ y2 ^ z4, find Bz 2 3 2 14 Bx 56 9 Bx 2 56 3 14 B2 x B2 z 2 14 3 2 4 20 9 4 3 3 14 2 B2 x B2 z A A ^ ^ y3 z 14 B B ^ ^ ^ xBx y2 zBz 2 4 Bx B2 z 116 (a) (b) (c) (d) (e) (f) (g) (h) ^ A and a, the component of B along C, AC , A C, A B C, A B C, ^ x B, and ^ ^ A y z. CHAPTER 3 Solution: (a) From Eq. (3.4), and, from Eq. (3.5), (b) The component of B along C (see Section 3-1.4) is given by (c) From Eq. (3.21), (d) From Eq. (3.27), (e) From Eq. (3.27) and Eq. (3.17), Eq. (3.30) could also have been used in the solution. Also, Eq. (3.29) could be used in conjunction with the result of part (d). (f) By repeated application of Eq. (3.27), Eq. (3.33) could also have been used. A B C A ^ x16 ^ y8 ^ z4 ^ x32 ^ y52 ^ z24 A B C ^ A x16 ^ y8 ^ z4 1 16 28 34 20 A C ^ x2 4 32 ^ y 30 1 4 ^ z12 20 AC cos 1 A C AC cos 1 4 12 14 20 cos 1 B cos BC B C C 8 20 ^ aA ^ x ^ ^ y2 z3 14 18 A 3 12 22 2 14 16 280 17 0 ^ x2 ^ y4 ^ z2 CHAPTER 3 (g) From Eq. (3.27), 117 (h) From Eq. (3.27) and Eq. (3.17), Eq. (3.29) and Eq. (3.25) could also have been used in the solution. ^ ^ ^ ^ ^ Problem 3.6 Given vectors A x2 y z3 and B x3 z2, find a vector C whose magnitude is 9 and whose direction is perpendicular to both A and B. Solution: The cross product of two vectors produces a new vector which is perpendicular to both of the original vectors. Two vectors exist which have a magnitude of 9 and are orthogonal to both A and B: one which is 9 units long in the direction of the unit vector parallel to A B, and one in the opposite direction. 1 5 42 Problem 3.8 By expansion in Cartesian coordinates, prove: (a) the relation for the scalar triple product given by (3.29), and (b) the relation for the vector triple product given by (3.33). Solution: (a) Proof of the scalar triple product given by Eq. (3.29): From Eq. (3.27), A B ^ x Ay Bz Az By ^ y Az Bx Ax Bz ^ z Ax By Ay Bx 2 2 A1 A1 12 12 ^ x ^ y5 ^ z4 ^ x ^ ^ y5 z4 42 ^ x0 15 ^ y0 77 Solution: The unit vector parallel to A point P 1 1 2 is ^ xx 2y ^ yy 3z ^ z 3x ^ Problem 3.7 Given A x x 2y parallel to A at point P 1 1 2 . ^ y y 3z ^ z 3x y , determine a unit vector y at the C 9 9 A A B B ^ ^ ^ ^ ^ x2 y z3 x3 z2 ^ ^ ^ ^ ^ x2 y z3 x3 z2 ^ ^ ^ x2 y13 z3 ^ 9 x1 34 22 132 32 A ^ ^ y z ^ x3 ^ ^ z z 1 ^ y8 67 ^ z2 0 ^ z0 62 ^ x B ^ z4 118 CHAPTER 3 Employing Eq. (3.17), it is easily shown that which are all the same. (b) Proof of the vector triple product given by Eq. (3.33): The evaluation of the left hand side employs the expression above for B C with Eq. (3.27): while the right hand side, evaluated with the aid of Eq. (3.17), is By rearranging the expressions for the components, the left hand side is equal to the right hand side. Problem 3.9 Find an expression for the unit vector directed toward the origin from an arbitrary point on the line described by x 1 and z 2. Solution: An arbitrary point on the given line is 1 y 2 . The vector from this point to 0 0 0 is: 5 y2 ^ a ^ x ^ yy A 1 A A 4 y2 5 y2 ^ z2 A ^ x0 1 ^ y0 y ^ z0 2 ^ x ^ yy ^ z Bz AxCx AyCy Cz Ax Bx Ay By 2^ z ^ y By AxCx AzCz Cy Ax Bx Az Bz ^ x Bx AyCy AzCz Cx Ay By Az Bz BA C CA B B AxCx AyCy AzCz C Ax Bx ^ z Ax BzCx BxCz Ay ByCz BzCy ^ y Az ByCz BzCy Ax BxCy ByCx ^ x Ay BxCy ByCx Az BzCx BxCz Ay By Az Bz A B C A ^ x ByCz BzCy ^ y BzCx BxCz ^ z BxCy C A B Cx Ay Bz Az By Cy Az Bx Ax Bz Cz Ax By Ay Bx ByCx B C A Bx Cy Az Cz Ay By Cz Ax Cx Az Bz Cx Ay Cy Ax A B C Ax ByCz BzCy Ay BzCx BxCz Az BxCy ByCx C A ^ x Cy Az Cz Ay ^ y Cz Ax Cx Az ^ z Cx Ay CyAx B C ^ x ByCz BzCy ^ y BzCx BxCz ^ z BxCy ByCx CHAPTER 3 119 Problem 3.10 Find an expression for the unit vector directed toward the point P located on the z-axis at a height h above the xy plane from an arbitrary point Q x y 3 in the plane z 3. Problem 3.11 Find a unit vector parallel to either direction of the line described by Solution: First, we find any two points on the given line. Since the line equation is not a function of y, the given line is in a plane parallel to the xz plane. For convenience, we choose the xz plane with y 0. For x 0, z 4. Hence, point P is at 0 0 4 . For z 0, x 2. Hence, point Q is at 2 0 0 . Vector A from P to Q is: Problem 3.12 Two lines in the xy plane are described by the expressions: Use vector algebra to find the smaller angle between the lines at their intersection point. Solution: Intersection point is found by solving the two equations simultaneously: 3x 4y 8 2x 4y 12 Line 1 Line 2 ^ a x 2y 3x 4y 6 8 A ^ x2 A A ^ 0 y0 ^ ^ x2 z4 20 0 ^ z0 2x z x2 y2 2 4 ^ a A A ^ xx ^ zh 3 h 321 A x 2 y 2 h 3 ^ yy 2 1 2 4 ^ x2 ^ z4 A ^ x0 x ^ y0 y ^ zh 3 ^ xx ^ yy ^ zh Solution: Point P is at 0 0 h . Vector A from Q x y 3 to P 0 0 h is: 3 120 CHAPTER 3 30 25 20 15 10 (0, 2) -35 -30 -25 -20 -15 -10 (0, -3) -10 -15 -20 -25 -30 10 15 20 25 30 35 B A (20, -13) AB Figure P3.12: Lines 1 and 2. The sum gives x 20, which, when used in the first equation, gives y 13. Hence, intersection point is 20 13 . Another point on line 1 is x 0, y 3. Vector A from 0 3 to 20 13 is Angle between A and B is Problem 3.13 A given line is described by Vector A starts at the origin and ends at point P on the line such that A is orthogonal to the line. Find an expression for A. x 2y 4 AB cos 1 A B A B cos 1 400 150 500 625 B 202 152 625 B ^ x 20 ^ y 13 2 ^ x20 ^ y15 10 3 A point on line 2 is x 0, y 2. Vector B from 0 2 to 20 A 202 102 500 A ^ x 20 ^ y 13 3 ^ x20 ^ y10 13 is CHAPTER 3 121 Solution: We first plot the given line. Next we find vector B which connects point P1 0 2 to P2 4 0 , both of which are on the line: Vector A starts at the origin and ends on the line at P. If the x-coordinate of P is x, y P1 (0,2) A (0,0) B P2 (4,0) Figure P3.13: Given line and vector A. 2 But A is perpendicular to the line. Hence, Problem 3.14 Show that, given two vectors A and B, A ^ x0 8 ^ y Hence, 4 08 2 x 08 4x 4 x 4 5 0 2 or ^ x0 8 ^ y1 6 ^ xx ^ y 4 x ^ x4 A B 0 A ^ xx ^ y then its y-coordinate has to be 4 x 4 x 2 . Vector A is x 2 in order to be on the line. Hence P is at 4 x ^ y2 0 x B ^ x4 0 ^ y0 2 ^ x4 ^ y2 122 CHAPTER 3 (a) the vector C defined as the vector component of B in the direction of A is given by AB A ^ ^ C aB a A2 ^ where a is the unit vector of A, and (b) the vector D defined as the vector component of B perpendicular to A is given by AB A D B A2 Solution: ^ ^ (a) By definition, B a is the component of B along a. The vector component of ^ B a along A is (b) The figure shows vectors A, B, and C, where C is the projection of B along A. It is clear from the triangle that B C D D B C B C D B Figure P3.14: Relationships between vectors A, B, C, and D. AB A A2 or A C ^ ^ aB a B A A A A AB A A2 CHAPTER 3 123 Problem 3.15 A certain plane is described by Find the unit vector normal to the surface in the direction away from the origin. Solution: Procedure: 1. Use the equation for the given plane to find three points, P1 , P2 and P3 on the plane. 2. Find vector A from P1 to P2 and vector B from P1 to P3 . 3. Cross product of A and B gives a vector C orthogonal to A and B, and hence to the plane. ^ 4. Check direction of c. Steps: 1. Choose the following three points: P2 at 8 0 0 P3 at 0 2. Vector A from P1 to P2 16 3 Vector B from P1 to P3 3. ^ x 64 3 ^ y 32 ^ x 0 4 ^ y 4 0 8 4 ^ z 8 0 0 ^ x Ay Bz Az By ^ y Az Bx 16 4 3 128 ^ z 3 Ax Bz ^ z Ax By C A B Ay Bx 16 3 B ^ x 0 0 ^ y 0 ^ z 0 16 3 4 ^ y 16 3 A ^ x 8 0 ^ y 0 P1 at 0 0 4 0 0 ^ z 0 4 ^ x8 ^ z4 ^ z4 2x 3y 4z 16 124 Verify that C is orthogonal to A and B CHAPTER 3 ^ c points away from the origin as desired. Problem 3.17 When sketching or demonstrating the spatial variation of a vector field, we often use arrows, as in Fig. 3-25 (P3.17), wherein the length of the arrow is made to be proportional 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. P3.17, which represents ^ the vector field E rr, consists of arrows pointing radially away from the origin and their lengths increase linearly in proportion to their distance away from the origin. Using this arrow representation, sketch each of the following vector fields: (f) E6 (e) E5 ^ r sin . (d) E4 (c) E3 (b) E2 ^ yx, ^ ^ xx yy, ^ ^ xx y2y, ^ r, (a) E1 ^ xy, ^ b B ^ x 1 B B ^ ^ y2 3 z1 ^ ^ ^ ^ x y5 z x 1 25 1 Solution: At P 1 0 1, ^ ^ x y5 ^ ^ y5 z 27 Problem 3.16 Given B to B at point P 1 0 1 . ^ x z 3y ^ y 2x 3z ^ z x y , find a unit vector parallel 322 ^ z 64 2 3 128 2 3 ^ c C C 4. C ^ 3 x 64 ^ y 32 ^ 3 z 128 ^ 3 x 64 ^ y 32 ^ 3 z 128 ^ x 0 37 ^ y 0 56 ^ z 0 74 B C 32 16 3 A C 8 32 0 64 3 64 0 3 128 4 3 128 4 3 512 512 0 3 3 512 512 0 3 3 CHAPTER 3 y 125 E E x E E Solution: (a) y E E E P2.13a: E 1 = - ^ y x Figure P3.17: Arrow representation for vector field E ^ rr (Problem 3.17). x E 126 (b) y E CHAPTER 3 E x E E (c) y E E E P2.13c: E 3 = ^ x + ^ y y x P3.17b: E2 ^ yx E x E CHAPTER 3 (d) y E E 127 x E E P2.13d: E 4 = ^ x + ^ 2y y x (e) y E E x E E P2.13e: E 5 = ^ r 128 (f) y E E CHAPTER 3 x E E P2.13f: E 6 = ^ sin r Problem 3.18 Use arrows to sketch each of the following vector fields: Solution: (d) E4 (c) E3 ^x y1, ^ r cos . (b) E2 (a) E1 ^ ^ xx yy, ^, CHAPTER 3 (a) y 129 E E E E x P2.14a: E 1 = ^ x - ^ y y x (b) y E E x E E P2.14b: E 2 = - ^ 130 (c) y CHAPTER 3 E x E Indicates |E| is infinite P2.14c: E 3 = ^ (1/x) y (d) y E E E E E x E E E E P2.14d: E 4 = ^ cos r CHAPTER 3 131 Sections 3-2 and 3-3: Coordinate Systems Problem 3.19 Convert the coordinates of the following points from Cartesian to cylindrical and spherical coordinates: (a) P1 1 2 0 , (b) P2 0 0 2 , (c) P3 1 1 3 , (d) P4 2 2 2 . Solution: Use the "coordinate variables" column in Table 3-2. (a) In the cylindrical coordinate system, In the spherical coordinate system, 5 2 rad 1 107 rad Note that in both the cylindrical and spherical coordinates, is in Quadrant I. (b) In the cylindrical coordinate system, In the spherical coordinate system, 2 0 rad 0 rad 20 0 Note that in both the cylindrical and spherical coordinates, is arbitrary and may take any value. (c) In the cylindrical coordinate system, In the spherical coordinate system, Note that in both the cylindrical and spherical coordinates, is in Quadrant I. 11 0 44 rad 4 rad P3 12 12 32 tan 1 12 12 3 tan 1 1 1 3 32 25 2 45 0 P3 12 12 tan 1 1 1 3 2 4 rad 3 1 41 45 0 3 P2 02 02 22 tan 1 02 02 2 tan 1 0 0 P2 02 02 tan 1 0 0 2 0 0 rad 2 P1 12 22 02 tan 1 12 22 0 tan 1 2 1 2 24 90 0 63 4 00 2 P1 12 22 tan 1 2 1 0 5 1 107 rad 0 2 24 63 4 0 132 (d) In the cylindrical coordinate system, CHAPTER 3 In the spherical coordinate system, 2 3 2 187 rad 3 4 rad 3 46 125 3 135 0 Note that in both the cylindrical and spherical coordinates, is in Quadrant II. Problem 3.20 Convert the coordinates of the following points from cylindrical to Cartesian coordinates: (a) P1 2 4 2 , (b) P2 3 0 2 , (c) P3 4 3 . Solution: (a) Problem 3.21 Convert the coordinates of the following points from spherical to cylindrical coordinates: (a) P1 5 0 0 , (b) P2 5 0 , (c) P3 3 2 0 . Solution: (a) P1 0 0 5 (b) P2 r z P2 5 sin 0 5 cos 0 P2 0 5 . P1 r z P1 R sin R cos P1 5 sin 0 0 5 cos 0 (b) P2 x y z (c) P3 x y z P2 3 cos 0 3 sin 0 2 P2 3 0 2 . P3 4 cos 4 sin 3 P3 4 0 3 . P1 x y z P1 r cos r sin z P1 2 cos 2 P1 1 41 1 41 2 sin 4 4 P4 2 22 2 tan 2 2 1 2 2 22 2 tan 1 2 2 2 2 2 3 4 rad 2 2 83 135 0 P4 2 2 22 tan 1 2 2 2 2 CHAPTER 3 133 Problem 3.22 Use the appropriate expression for the differential surface area ds to determine the area of each of the following surfaces: (a) r 3; 0 3; 2 z 2, (b) 2 r 5; 2 ; z 0, (c) 2 r 5; 4; 2 z 2, (d) R 2; 0 3; 0 , (e) 0 R 5; 3; 0 2. Also sketch the outlines of each of the surfaces. Solution: = /3 y 3 2 2 5 5 2 (a) (b) (d) Figure P3.22: Surfaces described by Problem 3.22. (a) Using Eq. (3.43a), z 2 0 z 2 A r r 3 d dz 3z 2 3 (c) P3 r z P3 3 sin 0 3 cos 2 2 P3 3 0 0 . x (c) (e) 3 0 2 4 134 (b) Using Eq. (3.43c), r 2 2 CHAPTER 3 2 Problem 3.23 Find the volumes described by (a) 2 r 5; 2 ; 0 z 2, (b) 0 R 5; 0 3; 0 2. Also sketch the outline of each volume. Solution: z z 5 2 2 x 5 y x (a) (b) Figure P3.23: Volumes described by Problem 3.23 . (a) From Eq. (3.44), 2 z 0 2 r 2 z 0 V 2 5 r dr d dz 1 2 2 r z 5 r 2 R 0 0 0 R 0 y 2 21 2 A R sin 3 d dR 1 2 2 R sin 2 25 3 2 5 2 3 5 (e) Using Eq. (3.50c), 0 0 R 2 0 A R2 sin d d 4 cos 3 (d) Using Eq. (3.50b), 3 0 z 2 r 2 2 r 2 A 1 4 dr dz rz 2 5 2 z (c) Using Eq. (3.43b), 5 12 2 A r 5 z 0 d dr 1 2 2r 5 r 2 21 4 CHAPTER 3 (b) From Eq. (3.50e), 135 (a) the surface area of the spherical section, (b) the enclosed volume. Also sketch the outline of the section. Solution: z y x =30o Figure P3.24: Outline of section. Problem 3.24 A section of a sphere is described by 0 30 90 . Find: R 0 0 0 R 2, 0 R3 cos 3 5 V 0 0 2 3 5 R 0 R2 sin dR d d 3 2 125 3 90 and 136 S CHAPTER 3 0 Problem 3.25 A vector field is given in cylindrical coordinates by (a) the vector component of E perpendicular to the cylinder, (b) the vector component of E tangential to the cylinder. Solution: (a) En At P 2 (b) Et At P 2 Problem 3.26 At a given point in space, vectors A and B are given in spherical coordinates by Find: (a) the scalar component, or projection, of B in the direction of A, (b) the vector component of B in the direction of A, (c) the vector component of B perpendicular to A. Solution: B A ^ ^ ^ R4 2 ^ ^ R2 3 ^ ^ ^ ^^ ^ rr E r r rr cos r sin ^ ^ 3 , En r2 cos r2. ^ ^ E En r sin zz2 . ^ ^ ^ 3 , Et 2 sin z32 z9. ^ zz2 ^ rr cos . Point P 2 3 is located on the surface of the cylinder described by r find: E ^ rr cos ^ r sin ^ zz2 cos 0 R 3 2 6 V R 0 6 0 3 2 2 2 2 R2 sin dR d d 2 8 33 8 9 (m3 ) 4 cos 0 2 6 2 4 6 0 2 2 R2 sin d d R 2 3 4 3 (m2 ) 2. At point P, CHAPTER 3 (a) Scalar component of B in direction of A: 137 8 21 3 (b) Vector component of B in direction of A: (c) Vector component of B perpendicular to A: Problem 3.27 Given vectors ^ r sin ^ z cos find (a) AB at 2 2 0 , (b) a unit vector perpendicular to both A and B at 2 3 1 . Solution: It doesn't matter whether the vectors are evaluated before vector products are calculated, or if the vector products are directly calculated and the general results are evaluated at the specific point in question. ^ ^ ^ 8 z2 and B r. From Eq. (3.21), (a) At 2 2 0 , A ^2 ^ ^2 ^2 r 1 3 z 1 . Since A B is (b) At 2 3 1 , A r 7 4 1 1 3 and B 2 perpendicular to both A and B, a unit vector perpendicular to both A and B is given by ^ r0 487 ^ 0 228 ^ z 0 843 21 3 3 2 3 1 2 7 2 4 2 A A B B ^ r 41 3 ^ z41 3 1 2 1 2 ^ 7 2 2 1 2 1 2 1 2 3 AB cos 1 A B AB cos 1 0 AB B 90 A ^ r cos 3z ^ 2r 4 sin ^ zr 2z D B C ^ ^ ^ 3 R 2 09 1 05 ^ ^ ^ R 0 09 1 05 2 48 ^ R2 ^ 0 52 ^ R 2 09 C ^ aC A ^ R4 ^ 2 ^ C A 24 21 ^ 1 05 0 52 ^ 11 21 24 C ^ B a B ^ R2 ^ 3 A A ^ R4 ^ ^ 2 16 4 1 138 CHAPTER 3 Problem 3.28 Find the distance between the following pairs of points: (b) P3 1 4 3 and P4 3 4 4 in cylindrical coordinates, (c) P5 4 2 0 and P6 3 0 in spherical coordinates. Solution: (a) (b) (c) Solution: (a) From Eq. (3.66), (b) From Eq. (3.67), d 2 2 4 cos 3 1 24 8 3 22 42 2 3 2 d 0 1 2 1 3 2 2 2 Problem 3.29 (a) P1 1 1 2 (b) P3 2 3 (c) P5 3 Determine the distance between the following pairs of points: and P2 0 2 3 , 1 and P4 4 2 3 , 2 and P6 4 2 . 9 16 0 1 2 25 5 2 3 9 16 2 3 4 cos cos 2 sin sin cos 0 2 0 d R2 2 R2 1 2R1 R2 cos 2 cos 1 sin 1 sin 2 cos 2 10 6 1 1 2 9 1 2 3 4 3 2 d 2 r2 2 r1 2r1 r2 cos 2 1 z2 1 cos 4 4 1 2 5 2 24 z1 2 1 2 1 2 1 1 2 1 2 3 18 d 2 1 2 3 2 2 2 3 2 1 2 9 25 25 1 (a) P1 1 2 3 and P2 2 3 2 in Cartesian coordinates, 2 59 7 68 CHAPTER 3 (c) From Eq. (3.68), 139 Problem 3.30 Transform the following vectors into cylindrical coordinates and then evaluate them at the indicated points: ^ (a) A x x y at P1 1 2 3 , ^ ^ (b) B x y x y x y at P2 1 0 2 , 2 x2 ^ ^ ^ (c) C xy y2 yx2 x2 y2 z4 at P3 1 1 2 , ^ cos cos2 at P4 2 2 4 , ^ ^ (d) D R sin ^ ^ ^ (e) E R cos sin sin2 at P5 3 2 . Solution: From Table 3-2: (a) (b) (c) (d) D ^ r sin ^ z cos sin C P3 ^ r0 707 ^ z4 ^ r cos ^ z sin cos ^ cos2 P3 12 1 2 tan 1 1 1 2 2 45 2 ^ sin3 ^ r C ^ r cos r2 sin2 ^ sin r2 ^ r sin cos sin cos B P2 ^ r ^ P2 12 02 tan 1 0 1 2 10 2 ^ cos ^ r sin r2 cos2 r2 3 ^ cos z4 ^ z4 ^ cos2 B ^ ^ ^ sin r sin r cos cos r sin r cos r sin 2 2 ^ ^ ^ ^ rr 2 sin cos 1 r cos sin rr sin 2 1 r cos 2 ^ r cos A P1 894 P1 22 tan 1 2 1 3 ^ ^ r 0 447 0 894 5 447 12 5 63 4 3 ^ r1 34 ^ rr cos cos sin ^ r sin cos A ^ r cos ^ sin r cos r sin sin ^ 2 68 d 32 42 23 4 cos cos 2 sin sin cos 2 2 5 140 CHAPTER 3 2 (e) Problem 3.31 Transform the following vectors into spherical coordinates and then evaluate them at the indicated points: ^ ^ ^ (a) A xy2 yxz z4 at P1 1 1 2 , 2 2 ^ ^ (b) B y x y z2 z x2 y2 at P2 1 0 2 , ^ ^ ^ (c) C r cos sin z cos sin at P3 2 4 2 , and 2 2 x2 ^ ^ ^ (d) D xy y yx2 x2 y2 z4 at P4 1 1 2 . Solution: From Table 3-2: (a) (b) 5 26 6 180 P2 1 02 22 tan 2 1 1 2 02 2 tan 1 0 1 B ^ ^ ^ ^ ^ R sin sin cos sin cos R2 R cos sin R2 sin2 ^ ^ ^ RR2 sin sin sin cos R2 cos sin sin3 R2 cos A P1 ^ R2 856 ^ 2 888 6 35 3 45 ^ 2 123 P1 12 1 22 tan 2 1 12 1 2 2 tan 1 ^ R R2 sin2 sin cos sin sin cos 4 cos ^ R2 sin cos sin cos sin sin cos 4 sin ^ R2 sin cos cos2 sin sin3 1 1 A ^ ^ ^ R sin cos cos cos sin R sin sin 2 ^ ^ ^ R sin sin cos sin cos R sin cos R cos ^ ^ R cos sin 4 E P5 P5 E ^ ^ ^ ^ ^ r sin z cos cos r cos z sin sin sin2 3 2 ^ ^ ^ ^ ^ z cos z sin r sin r cos cos sin sin2 2 2 2 2 2 D P4 ^ r ^ P4 2 sin 2 4 2 cos 2 ^ ^ r 1 2 45 0 CHAPTER 3 141 (c) (d) Sections 3-4 to 3-7: Gradient, Divergence, and Curl Operators Problem 3.32 Find the gradient of the following scalar functions: (a) T 3 x2 z2 , (b) V xy2 z4 , ^ cos 35 26 cos 45 sin2 45 ^ cos3 45 sin3 45 ^ ^ ^ R 3 67 1 73 0 707 cos 35 26 sin 45 cos2 45 4 sin 35 26 D P4 ^ R sin 35 26 cos 45 sin2 45 sin 35 26 sin P4 6 35 26 45 45 cos2 45 P4 1 12 P4 1 1 4 tan 1 1 1 2 tan 1 1 1 4 cos 35 26 ^ cos cos sin ^ cos3 sin3 2 cos sin cos 2 4 sin ^ R sin cos sin2 ^ R cos ^ sin 4 sin sin cos2 4 cos ^ R sin sin ^ cos sin D ^ R sin cos ^ cos cos ^ sin R2 sin2 sin2 R2 sin2 sin2 R2 sin2 cos2 R2 sin2 cos2 ^ cos 2 2 R sin sin2 R2 sin2 cos2 C P3 ^ R0 854 ^ 0 146 ^ 0 707 P3 2 2 45 45 22 22 tan 1 2 2 4 C ^ ^ ^ ^ ^ R sin cos cos sin R cos sin cos sin ^ ^ ^ R cos sin cos sin cos cos sin sin sin B P2 ^ R0 896 ^ 0 449 ^ 5 142 CHAPTER 3 U z cos 1 r2 , W e R sin , S 4x2 e z y3 , N r2 cos2 , M R cos sin . Solution: (a) From Eq. (3.72), z2 2 2 (b) From Eq. (3.72), (c) From Eq. (3.82), (d) From Eq. (3.83), (e) From Eq. (3.72), (f) From Eq. (3.82), (g) From Eq. (3.83), M M R cos sin M ^ 1 M ^ R R R ^ 1 M R sin ^ R cos sin ^ sin sin N N r2 cos2 N ^ 1 N ^ r r r ^ z N z ^ r2r cos2 ^ 2r sin cos S S 4x2 e S ^ x x z y3 S ^ y y ^ z S z ^ x8xe z ^ y3y2 ^ z4x2 e W ^ Re R sin ^ e R R cos 1 r2 2 U ^ r 2rz cos ^ z sin r 1 r2 ^ z cos 1 r2 V ^ xy2 z4 ^ y2xyz4 ^ z4xy2 z3 x2 x2 z2 T (c) (d) (e) (f) (g) ^ x 6x ^ z 6z z ^ cos tan CHAPTER 3 Problem 3.33 The gradient of a scalar function T is given by 143 Solution: 0 0 z 0 0 Hence, Problem 3.34 Follow a procedure similar to that leading to Eq. (3.82) to derive the expression given by Eq. (3.83) for in spherical coordinates. Solution: From the chain rule and Table 3-2, T T T ^ ^ y z x y z T R T T ^ x R x x x T R T T ^ y R y y y T R T T ^ z R z z z T T T ^ x x2 y2 z2 x2 y2 z tan 1 tan 1 y x R x x x T T T ^ tan 1 tan 1 y x y x2 y2 z2 x2 y2 z R y y y T T T ^ tan 1 tan 1 y x z x2 y2 z2 x2 y2 z R z z z ^ x T T z T 0 e 3z 10 1 1 3 1 1 3 e 3z e 3z dz e 3z 3 z 1 1 3 T z T 0 ^ ze T dl 3z By choosing P1 at z 0 and P2 at any point z, (3.76) becomes z z ^ x dx If T 10 at z 0, find T z . T ^ ze 3z ^ y dy ^ z dz e 3z T ^ ze 3z 144 CHAPTER 3 which is Eq. (3.83). 1 z2 ^ al ^ x ^ yz Solution: The directional derivative is given by Eq. (3.75) as dV dl the unit vector in the direction of A is given by Eq. (3.2): Problem 3.35 For the scalar function V xy2 ^ x derivative along the direction of vector A P1 14. z2 , determine its directional ^ yz and then evaluate it at ^ V al , where T R sin cos T R cos R sin cos T R R R2 R sin T R sin sin T R cos R sin sin T ^ y R R R2 R sin T R cos T R sin ^ z R R R2 T T cos cos T sin ^ sin cos x R R R sin T T cos sin T cos ^ sin sin y R R R sin T T sin ^ cos z R R T ^ ^ ^ x sin cos y sin sin z cos R 1 T ^ ^ ^ x cos cos y cos sin z sin R 1 T ^ ^ x sin y cos R sin T ^ 1 T ^ 1 T ^ R R R R sin ^ x R sin sin R2 sin2 R sin cos R2 sin2 x2 y2 z2 ^ z T R z T x2 1 y2 z2 x2 y2 T 0 ^ y T R y x2 y2 z2 T x2 z y2 y y2 z2 x2 x T x2 y2 x2 y2 z2 z2 x2 y2 ^ x T R x T x2 z y2 x T y 2 x y2 CHAPTER 3 and the gradient of V in Cartesian coordinates is given by Eq. (3.72): 145 Therefore, by Eq. (3.75), 1 r 5 cos , determine its directional Problem 3.36 For the scalar function T 2e ^ derivative along the radial direction r and then evaluate it at P 2 4 3 . Solution: 2 1 Problem 3.37 For the scalar function U R sin , determine its directional ^ derivative along the range direction R and then evaluate it at P 5 4 2 . Solution: 5 4 2 dU dl dU dl U U 1 2 sin R U ^ 1 U ^ 1 U ^ R R R R sin 2 sin ^ U R R2 2 sin 4 0 02 25 ^ R sin2 R2 2 43 dT dl dT dl 10 2 10 2r ^ 2 sin cos R T ^ r e r 5 cos ^e r 5 sin T 1 r5 e cos 2 T ^ 1 T T ^ ^ r z r r z r 5 cos e ^ T r 10 e 2 5 cos 4 4 74 10 1 14 At P 1 14, dV dl 9 17 dV dl y2 2xyz 1 z2 2 18 V ^ xy2 ^ y2xy ^ z2z 146 CHAPTER 3 Problem 3.38 Vector field E is characterized by the following properties: (a) E ^ points along R, (b) the magnitude of E is a function of only the distance from the origin, (c) E vanishes at the origin, and (d) E 12, everywhere. Find an expression for E that satisfies these properties. Solution: According to properties (a) and (b), E must have the form where ER is a function of R only. 0 0 0 Hence, and ^ ^ ^ Problem 3.39 For the vector field E xxz yyz2 zxy, verify the divergence theorem by computing: (a) the total outward flux flowing through the surface of a cube centered at the origin and with sides equal to 2 units each and parallel to the Cartesian axes, and (b) the integral of E over the cube's volume. Solution: (a) For a cube, the closed surface integral has 6 sides: E ds Ftop Fbottom Fright Fleft Ffront E ^ R4R ER 4R R ER 2 4R 3 R2 ER R 0 12R3 3 R R 2 R ER dR R R 12R2 dR R2 ER R 12R2 Fback E E ^ RER 1 R2 ER R2 R 12 CHAPTER 3 Ftop ^ xxz ^ zxy 147 E ds 0 0 0 0 4 3 4 3 8 3 z 1 y 1 z 1 y 1 z dz dy 1 1 yz2 2 1 y 1 z 1 x 1 1 0 Fback ^ xxz ^ zxy ^ yyz2 1 1 ^ x dz dy z 1 y 1 z 1 y 1 z dz dy 1 1 yz2 2 1 y 1 z 1 x 1 1 Ffront ^ xxz ^ zxy ^ yyz2 ^ x dz dy 1 1 z 1 x 1 z 1 x 1 0 z dz dx 2 1 1 xz3 3 x 1 z 1 1 y 1 1 Fleft ^ xxz ^ zxy ^ yyz2 ^ y dz dx 1 1 z 1 x 1 z 1 x 1 4 3 z dz dx 2 1 1 xz3 3 x 1 z 1 y 1 1 Fright ^ xxz ^ zxy ^ yyz2 ^ y dz dx 1 1 y 1 x 1 y 1 x 1 1 xy dy dx 1 1 x2 y2 4 x 1 y 1 1 z 1 1 Fbottom ^ xxz ^ zxy ^ yyz2 ^ z dy dx 1 1 y 1 x 1 y 1 x 1 0 xy dy dx 1 1 x2 y2 4 x 1 y 1 z 1 1 1 ^ yyz2 1 1 ^ z dy dx 0 4 3 148 (b) CHAPTER 3 ^ ^ Problem 3.40 For the vector field E r10e r z3z, verify the divergence theorem for the cylindrical region enclosed by r 2, z 0, and z 4. Solution: r 0 0 2 4 r 0 ^ Problem 3.41 A vector field D rr3 exists in the region between two concentric cylindrical surfaces defined by r 1 and r 2, with both cylinders extending between z 0 and z 5. Verify the divergence theorem by evaluating: S (a) D ds, 160e 2 48 82 77 8 10e r 10e r 1 r 8 10e 1 r 3r dr 3r2 2 2 r 0 z 0 r 0 0 2 r r E dV 4 2 2 10e r 1 160e 48 82 77 r 3 r d dr dz 0 z 0 2 0 r 0 0 10e 2 2 d dz 2 2 12r dr d r 0 0 2 4 ^ r10e ^ z3z ^ zr dr d r z 4 0 z 0 2 2 ^ r10e ^ z3z ^ rr d dz r r 2 E ds ^ r10e ^ z3z r 2 2 ^ zr dr d y 1 z 1 x 1 z 0 z2 xy 2 z3 3 1 1 x 1 y 1 z 1 1 1 1 x 1 y 1 z 1 z z2 dz dy dx 1 E dv ^ xxz 1 1 1 ^ yyz2 ^ zxy dz dy dx 8 3 CHAPTER 3 149 V Solution: (a) ^ Problem 3.42 For the vector field D R3R2 , evaluate both sides of the divergence theorem for the region enclosed between the spherical shells defined by R 1 and R 2. Solution: The divergence theorem is given by Eq. (3.98). Evaluating the left hand side: V 0 0 R 1 2 cos 0 3R4 2 R 1 180 D dV 2 2 1 2 R 3R2 R2 R R2 sin dR d d z 0 0 r 1 0 z 0 D dV 5 2 2 4r2 r dr d dz 1 r r rr 3 Therefore, D ds 150. (b) From the back cover, D 4r2 . Therefore, r4 2 r 1 2 5 r 1 0 z 5 Ftop ^ zr d dr ^ rr 3 r 1 0 2 2 z 0 0 Fbottom ^ rr 3 ^ zr d dr 0 z 0 2 2 r 2 160 r4 dz d 0 z 0 2 5 Fouter ^ rr dz d ^ rr 3 0 z 0 2 5 r 1 r 2 r4 dz d 10 0 z 0 2 5 Finner ^ rr 3 2 5 ^ rr dz d D ds (b) D dV . Finner Fouter Fbottom Ftop r 1 0 150 150 The right hand side evaluates to CHAPTER 3 C S Solution: In addition to the independent condition that z 0, the three lines of the triangle are represented by the equations y 0, x 1, and y x, respectively. y 1 y L3 L2 x 1 L3 L1 1 (b) L2 2 x 0 L1 1 (a) 0 Figure P3.43: Contours for (a) Problem 3.43 and (b) Problem 3.44. (a) x 0 y 0 z 0 z 0 xy y 0 z 0 dx dy 0 x2 2y2 1 0 L1 ^ xxy ^ y x2 2y2 E dl L1 L2 L3 ^ x dx ^ y dy ^ z dz (b) (a) E dl around the triangular contour shown in Fig. P3.43(a), and E ds over the area of the triangle. Problem 3.43 For the vector field E ^ xxy ^ y x2 0 0 2y2 , calculate 0 y 0 dz 2 2 3 sin d 48 sin d 180 0 0 ^ R3R2 ^ RR2 sin d d R 2 S 0 0 2 D ds ^ R3R2 2 ^ RR2 sin d d R 1 0 CHAPTER 3 151 y 0 x 1 ^ z3x ^ z dy dx Problem 3.44 Repeat Problem 3.43 for the contour shown in Fig. P3.43(b). Solution: In addition to the independent condition that z 0, the three lines of the triangle are represented by the equations y 0, y 2 x, and y x, respectively. (a) x 2 x2 4y 2y2 y3 1 y 0 x3 3 1 0 11 3 x 2 y 0 z 0 xy z 0 y 2 x dx x 2 yz dy 0 0 x2 2y2 1 1 L2 ^ xxy ^ y x2 2y2 ^ x dx ^ y dy ^ z dz x 0 y 0 z 0 z 0 y 2 x dz xy y 0 z 0 dx dy 0 0 x2 2y2 2 0 L1 ^ xxy ^ y x2 2y2 E dl L1 L2 L3 ^ x dx ^ y dy ^ z dz 0 y 0 dz x 0 y 0 x 0 0 3x dy dx 3x x 1 x 0 y 0 1 x E ds 1 x (b) From Eq. (3.105), E z 0 E dl Therefore, 5 2 3 3 ^ z3x so that 0 y3 0 y 1 0 x3 3 0 2 3 1 0 dx x3 x 1 y 1 z 0 1 0 xy y x z 0 dx x y z dy 0 0 x2 2y2 0 0 L3 ^ xxy ^ y x2 2y2 ^ x dx 0 y 0 ^ y dy 2y3 3 1 5 3 ^ z dz x 1 y 0 z 0 0 z xy 0 dx x 1z dy 0 0 x2 2y2 1 1 L2 ^ xxy ^ y x2 2y2 ^ x dx ^ y dy ^ z dz 0 x 1 dz y x dz 1 152 CHAPTER 3 x 1 Therefore, ^ z3x ^ z dy dx ^ z3x ^ z dy dx 3x dy dx C S Solution: (a) 0 2 1 2 2r 2 r 0 L1 r 0 0 z 0 0 B dl r cos dr 2 0 r sin d z 0 B dl ^ rr cos ^ sin L1 L2 L3 ^ r dr ^ r d B dl B dl B dl B dl (b) (a) B dl over the semicircular contour shown in Fig. P3.46(a), and B ds over the surface of the semicircle. ^ z dz r cos dr r sin d Problem 3.45 by evaluating: Verify Stokes's theorem for the vector field B 3x2 x3 x 0 1 x3 0 x 1 2 x 1 3 ^ rr cos 3x x 0 dx 3x 2 x 0 y 0 1 x 1 y 0 2 x 1 y 0 1 x 2 2 x 3x dy dx x 0 dx x 0 y 0 2 2 x E ds 1 x (b) From Eq. (3.105), E ^ z3x so that E dl 0 11 3 y3 0 y 1 0 2 3 x3 3 0 2 3 3 z 0 z 0 x 1 y 1 z 0 y xz xy 0 dx x yz dy 0 0 x2 2y2 0 0 L3 ^ xxy ^ y x2 2y2 ^ x dx ^ y dy ^ z dz 0 y x dz ^ sin CHAPTER 3 y 2 y 153 L2 2 1 L3 L4 1 L2 L1 2 (b) x -2 L3 0 L1 (a) 2 x 0 Figure P3.46: Contour paths for (a) Problem 3.45 and (b) Problem 3.46. (b) Problem 3.46 Solution: Repeat Problem 3.45 for the contour shown in Fig. P3.46(b). 0 r 0 0 sin r 1 dr d cos 1 r2 2 2 B ds 1 ^ r cos sin 0 0 r z z r 1 ^ z r sin r cos r r 1 1 ^ ^ ^ ^ r0 0 z sin r sin z sin 1 r r 2 1 ^ ^ z sin 1 zr dr d r 0 r 0 ^ r B ^ rr cos B dl 2 4 2 8 ^ sin 0 2 1 2 2r 0 r 2 L3 r 2 z 0 B dl r cos dr 0 r sin d 0 2 cos 4 0 z 0 L2 r 2 z 0 0 B dl r cos dr 2 r sin d r 2 z 0 r 2 r 0 8 154 (a) CHAPTER 3 L1 L2 L3 L4 (b) r 1 0 cos 1 r2 2 r 2 r 1 0 sin r 2 2 1 dr d 2 5 2 B ds 1 ^ r cos 0 0 sin r z z r 1 ^ z r sin r cos r r 1 1 ^ ^ ^ ^ r0 0 z sin r sin z sin 1 r r 2 2 1 ^ ^ z sin 1 zr dr d r 0 r 1 ^ r B ^ rr cos B dl 3 2 2 0 1 5 2 ^ sin 0 cos 0 2 1 L4 r 1 z 0 2 r 1 z 0 B dl r cos dr r sin d 1 0 L3 r 2 2 z 0 2 z 0 B dl r cos dr 1 2 r sin d 0 2 cos 2 0 2 L2 r 2 z 0 0 B dl r cos dr 2 r sin d r 2 z 0 1 2 2r 0 2 r 1 3 2 L1 r 1 0 z 0 0 B dl r cos dr 2 0 r sin d z 0 0 B dl ^ rr cos ^ sin ^ r dr ^ r d ^ z dz r cos dr r sin d B dl B dl B dl B dl B dl CHAPTER 3 155 ^ R cos ^ sin by Solution: The contour C is the circle in the xy plane bounding the hemispherical surface. Problem 3.48 Determine if each of the following vector fields is solenoidal, conservative, or both: ^ ^ (a) A xx2 yy2xy, ^ ^ ^ (b) B xx2 yy2 z2z, ^ ^ (c) C r sin r2 cos r2 , ^ R, (d) D R ^ ^ zz, (e) E r 3 1 r r 2 ^ ^ xy yx x (f) F y2 , 2 2 2 ^ ^ ^ y y x2 z y2 z2 , (g) G x x z ^ Re R . (h) H R C 0 0 A dl ^ R cos ^ ^ sin R d 2 R 1 R sin 2 0 R 1 4R 2 2 d 2 R 1 sin 2 2 2 0 0 R 0 0 2 2 A ds ^ R2 cos 2 2 For the hemispherical surface, ds ^ RR2 sin d d. ^ sin R ^ sin R ^ RR2 sin d d R 1 A 1 ^ 1 RA 1 AR ^ A sin R sin R R R 1 ^ 1 R sin 1 cos ^ ^ R sin2 R sin R R R 2 cos ^ sin ^ sin ^ R R R R ^ R Hence, AR cos , A 0, A sin . 2 A ^ R cos ^ sin ^ RAR ^ A ^ A Problem 3.47 Verify Stokes's Theorem for the vector field A evaluating it on the hemisphere of unit radius. 156 Solution: (a) CHAPTER 3 The field A is solenoidal but not conservative. (b) The field B is conservative but not solenoidal. (c) The field C is neither solenoidal nor conservative. C C sin ^ cos 2 r2 r 1 sin 1 cos r 0 2 2 r r r r r z sin sin 2 sin 0 3 3 r r r3 sin ^ cos ^ r 2 2 r r 1 cos ^ sin ^ r 0 0 r z r2 z r2 r 1 cos sin ^ z r 2 r r r r2 cos cos 2 cos 1 ^ ^ ^ ^ r0 0 z z 2 2 r r r r3 ^ r ^ x0 ^ y0 ^ z0 ^ x y2 ^ y ^ z y2 2z y z 2 x z 2z x B ^ xx2 ^ yy2 ^ z2z x B ^ xx2 ^ yy2 ^ z2z ^ x0 ^ y0 ^ z 2y 0 2 x x 2 y y 2z z 2x 2y 2 0 2 x y ^ x 2xy ^ y ^ z 2xy 0 y z 2 x z 0 x A ^ xx2 ^ y2xy x 2 x y A ^ xx2 ^ y2xy 2 x x 2xy y 2x 2x 0 CHAPTER 3 (d) 157 The field D is conservative but not solenoidal. (e) 1 r Hence, E is conservative, but not solenoidal. (f) x2 y2 x2 y2 x2 2xy y2 y2 x2 2 F y x 2 y x y2 2xy 0 x2 y2 2 F ^ xy x2 x ^ yx y2 ^ x y ^ y x E 1 Er r E 1 E Ez rEr r r z 2 r 3r 1 r 1 r 2r r2 3 1 1 r 1 r 2 3 3r2 6r 2r 2r2 r2 1 1 r 2 1 Ez E ^ Er Ez ^ r r z z r 1 r 1 r 1 r 1 r E ^ r 3 r ^ zz 2r2 4r 3 r 1 r 2 1 ^ z rE r r ^ r0 ^ 0 ^ 0 1 0 ^ R 1 0 sin 0 R sin 1 ^1 R 0 R R R ^1 R 1 sin D ^ R R 1 R R0 R 0 D R2 ^ R R 1 R2 R 1 R 1 0 sin R sin 1 0 R sin 1 R2 158 F ^ x0 ^ z 0 ^ y0 0 ^ z x x x2 y2 y 1 2y2 x2 y2 x2 y2 CHAPTER 3 x2 2 y2 Hence, F is neither solenoidal nor conservative. (g) Hence, G is neither solenoidal nor conservative. (h) Hence, H is conservative, but not solenoidal. Problem 3.49 Find the Laplacian of the following scalar functions: (a) V 4xy2 z3 , (b) V xy yz zx, (c) V 3 x2 y2 , (d) V 5e r cos , (e) V 10e R sin . Solution: (a) From Eq. (3.110), 2 4xy2 z3 H 8xz3 24xy2 z H R R R3 e R e R 1 R3 e R2 R 0 H ^ R Re R 1 3R2 e R2 ^ x 2y ^ y 2z ^ z 2x 0 3 R 0 G ^ x ^ y z2 2 2 y z2 y x2 y z 2 2 2 ^ y x2 x z z x y G G ^ ^ ^ x x2 z2 y y2 x2 z y2 z2 2 2 2 x z2 y x2 y x y z 2x 2y 2z 0 0 z2 2 x z x2 y2 2 y2 x2 ^ z 2 x y2 2 2 2 y x z2 1 2x2 x2 y2 y CHAPTER 3 (b) 2 xy yz zx 0 (c) From the inside back cover of the book, x2 y2 2 159 (d) (e) Problem 3.50 Find a vector G whose magnitude is 4 and whose direction is ^ ^ ^ ^ ^ perpendicular to both vectors E and F, where E x y 2 z 2 and F y 3 z 6. Solution: The cross product of two vectors produces a third vector which is perpendicular to both of the original vectors. Two vectors exist that satisfy the stated conditions, one along E F and another along the opposite direction. Hence, Problem 3.51 A given line is described by the equation: Vector A starts at point P1 0 2 and ends at point P2 on the line such that A is orthogonal to the line. Find an expression for A. Solution: We first plot the given line. y x 1 4 9 ^ x6 ^ y6 ^ z3 4 G 4 4 E E F F ^ x ^ x ^ ^ ^ y2 z2 y3 ^ ^ ^ y2 z2 y3 ^ ^ ^ x6 y6 z3 36 36 9 ^ z6 ^ z6 ^ x 8 3 ^ y 8 3 ^ z 4 3 2 10e R sin 10e R sin 1 2 R cos2 sin2 R2 sin 2 5e r cos 5e r cos 1 1 r 1 r2 x2 y2 2 2 3r 3 2 12r 4 12 160 y P1 (0, 2) CHAPTER 3 A B P2 (x, x-1) x P4 (1, 0) P3 (0, -1) Next we find a vector B which connects point P3 0 1 to point P4 1 0 , both of which are on the line. Hence, Vector A starts at P1 0 2 and ends on the line at P2 . If the x-coordinate of P2 is x, then its y-coordinate has to be y x 1, per the equation for the line. Thus, P2 is at x x 1 , and vector A is Since A is orthogonal to B, Finally, ^ y 3 2 A ^ xx ^ y x 3 3 2 3 ^ x 2 ^ x ^ y 3 2 x x x 3 ^ xx ^ y x 3 ^ x A B 0 ^ y 0 0 3 2 3 A ^ x x 0 ^ y x 1 2 ^ xx ^ y x 3 B ^ x 1 0 ^ y 0 1 ^ x ^ y CHAPTER 3 Problem 3.52 Vector field E is given by 161 Solution: At P, E is given by ^ The R component is normal to the spherical surface while the other two are tangential. Hence, ^ ^ Et 1 5 2 6 Problem 3.53 Transform the vector into cylindrical coordinates and then evaluate it at P 2 2 2 . Solution: From Table 3-2, ^ r sin3 cos cos cos2 ^ sin ^ r sin ^ z cos sin2 cos ^ r cos ^ z sin cos2 At P 2 2 2 , y P1 (0, 3) P2 (-3, 0) Problem 3.54 Evaluate the line integral of E of the circular path shown in the figure. A ^ ^ xx ^ y y along the segment P1 to P2 x ^ z cos sin2 cos A A ^ R sin2 cos ^ cos2 ^ R 8 67 ^ 1 5 ^ sin ^ sin sin cos2 E ^ R5 2 cos 30 ^ 12 sin 30 cos 60 2 ^ 2 6 ^ 3 sin 60 ^ 12 ^ sin cos 3 sin R Determine the component of E tangential to the spherical surface R P 2 30 60 . E ^ R 5R cos 2 at point 162 Solution: We need to calculate: P2 P1 CHAPTER 3 Since the path is along the perimeter of a circle, it is best to use cylindrical coordinates, which requires expressing both E and d in cylindrical coordinates. Using Table 3-2, Hence, P1 90 r 3 figure, and S C (b) (a) B d over the path comprising a quarter section of a circle, as shown in the B ds over the surface of the quarter section. y (0, 3) L2 L1 x (-3, 0) L3 Problem 3.55 evaluating: Verify Stokes's theorem for the vector field B 2r2 9 sin2 2 180 90 r 3 ^ ^ r cos sin by 2r2 sin cos d 90 180 E d ^ r r cos2 sin2 ^ ^ 2r sin cos r d P2 180 The designated path is along the -direction at a constant r applicable component of d is: ^ r d d 3. From Table 3-1, the E ^ xx ^ yy ^ ^ ^ sin r cos r sin cos r sin ^ ^ r r cos2 sin2 2r sin cos ^ r cos E d r 3 CHAPTER 3 Solution: (a) C L1 L2 L3 163 Given the shape of the path, it is best to use cylindrical coordinates. B is already expressed in cylindrical coordinates, and we need to choose d in cylindrical coordinates: ^ ^ ^ r dr r d z dz d Hence, C (b) 90 2r 3 r 0 cos S 180 6 1 Br rB r r 1 ^ z r sin cos r r 1 2 ^ ^ z sin sin z sin r r 3 180 2 ^ ^ B ds z sin z r dr d r r 0 90 B ^ z B d 0 3 3 r 3 6 180 r cos dr 0 3 3 L3 r 3 0 B d 0 ^ r cos Along L3 , dz 0 and d 0. Hence, d ^ r dr and ^ ^ sin r dr 180 3 3 cos 180 90 L2 B d 90 180 ^ r cos ^ ^ sin r d Along L2 , dr dz 0. Hence, d ^ r d and r 3 r 0 90 0 90 cos dr r cos 3 r 3 L1 r 0 B d r 3 ^ r cos ^ ^ sin r dr Along path L1 , d 0 and dz 0. Hence, d ^ r dr and 90 0 B d B d B d B d 164 Hence, Stokes's theorem is verified. Problem 3.56 Find the Laplacian of the following scalar functions: CHAPTER 3 Solution: (a) (b) 2V2 1 1 1 V2 2V2 V2 R2 sin R2 R R R2 sin R2 sin2 2 1 2 cos sin R2 2 R R R R2 1 2 sin cos sin 2 sin R R2 1 2 2 cos sin 2 sin2 2 R2 R 4 4 2 cos cos sin cos sin sin 4 4 R R R4 sin2 2 cos sin R4 sin2 2V1 V1 1 1 2V1 2V r r r r r2 2 z2 1 1 2 r 10r3 sin 2 10r3 sin 2 r r r r2 2 1 1 30r3 sin 2 10r3 4 sin 2 r r r2 90r sin 2 40r sin 2 50r sin 2 (b) V2 (a) V1 10r3 sin 2 2 R2 cos sin 0 ...
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This note was uploaded on 02/09/2011 for the course EE 172 taught by Professor . during the Spring '10 term at UCLA.

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