7 Chapter : Vector Analysis Lesson #4 Chapter Section: - 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.
8 Lessons #5 and 6 Chapter Section: - 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: xamples - to -5 Technology Brief on GPS (CD-ROM) Global Positioning System The Global Positioning System (GPS), initially developed in the 98s 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 segments. The space segment consists of 4 satellites (A), each circling arth every hours at an orbital altitude of about, 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 m, both horizontally and vertically, but it can be improved to within m by differential GPS (see illustration).
9 Lesson #7 Chapter Section: - Topics: Coordinate transformations Highlights: Basic logic for decomposing a vector in one coordinate system into the coordinate variables of another system Transformation relations (Table -) Special Illustrations: xample -8
Lesson #8 Chapter Section: -4 Topics: Gradient operator Highlights: Derivation of T in Cartesian coordinates Directional derivative T in cylindrical and spherical coordinates Special Illustrations: xample -(b) CD-ROM Modules.5 or.6 CD-ROM Demos.-.9 (any )
Lesson #9 Chapter Section: -5 Topics: Divergence operator Highlights: Concept of flux Derivation of. Divergence theorem Special Illustrations: CD-ROM Modules.7-. (any ) CD-ROM Demos.-.5 (any or )
Lesson # Chapter Section: -6 Topics: Curl operator Highlights: Concept of circulation Derivation of x B Stokes s theorem Special Illustrations: xample -
Lesson # Chapter Section: -7 Topics: Laplacian operator Highlights: Definition of Definition of V 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 -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 -D profile of the object. CT was invented in 97 by British electrical engineer Godfrey Hounsfield, and independently by Allan Cormack, a South African-born American physicist. The two inventors shared the 979 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.
4 CHAPTR Chapter Section -: Vector Algebra Problem. a unit vector in the direction of A. Vector A starts at point and ends at point. Find A ˆx ŷ ẑ ˆx ẑ A 9 6 A ˆx â ẑ A ˆx 6 ẑ 95 Problem. Given vectors A ˆx ŷ ẑ, B ˆx ŷ ẑ, and C ˆx4 ŷ ẑ, show that C is perpendicular to both A and B. A C ˆx ŷ ẑ ˆx4 ŷ ẑ 8 6 B C ˆx ŷ ẑ ˆx4 ŷ ẑ 8 6 Problem. In Cartesian coordinates, the three corners of a triangle are P 4 4, P 4 4 4, and P 4. Find the area of the triangle. Let B P P ŷ8 and ˆx4 C P P ẑ8 represent two sides of ˆx ŷ the triangle. Since the magnitude of the cross product is the area of the parallelogram (see the definition of cross product in Section -.4), half of this is the area of the triangle: A B C ˆx4 ŷ8 ˆx ŷ ẑ8 ˆx 8 8 ŷ 4 8 ẑ 4 8 ˆx64 ŷ ẑ8 64 8 where the cross product is evaluated with q. (.7). Problem.4 Given A ˆx ŷ ẑ and B ˆxB x ŷ ẑb z : (a) find B x and B z if A is parallel to B; (b) find a relation between B x and B z if A is perpendicular to B. 584 6
CHAPTR 5 (a) If A is parallel to B, then their directions are equal or opposite: â A â B, or From the y-component, A A B B ˆxB x ŷ ẑb z 4 Bx B z ˆx ŷ ẑ 4 4 4 B x B z 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 Bx B z, Bx B z 4 4 9 From the x-component, and, from the z-component, B x 4 56 9 B 56 x 4 4 B z This is consistent with our result for B x B z. These results could also have been obtained by assuming θ AB was or 8 and solving A B A B, or by solving A B. (b) If A is perpendicular to B, then their dot product is zero (see Section -.4). Using q. (.7), A B B x 6 B z or B z 6 B x 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 θ AB 9 and calculating A B A B. Problem.5 Given vectors A ˆx ŷ ẑ, B ˆx ŷ4, and C ŷ ẑ4, find
6 CHAPTR (a) A and â, (b) the component 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. (a) From q. (.4), and, from q. (.5), A 4 â A ˆx ŷ ẑ 4 (b) The component of B along C (see Section -.4) is given by (c) From q. (.), Bcosθ BC B C C 8 8 θ AC cos A C AC cos 4 4 cos 6 8 7 (d) From q. (.7), A C ˆx 4 ŷ 4 ẑ ˆx ŷ4 ẑ (e) From q. (.7) and q. (.7), A B C A ˆx6 ŷ8 ẑ4 6 8 4 q. (.) could also have been used in the solution. Also, q. (.9) could be used in conjunction with the result of part (d). (f) By repeated application of q. (.7), A B C A ˆx6 ŷ8 ẑ4 ˆx ŷ5 ẑ4 q. (.) could also have been used.
CHAPTR 7 (g) From q. (.7), ˆx B ẑ4 (h) From q. (.7) and q. (.7), A ŷ ẑ ˆx ẑ ẑ q. (.9) and q. (.5) could also have been used in the solution. Problem.6 magnitude is 9 and whose direction is perpendicular to both A and B. Given vectors A ˆx ŷ ẑ and B ˆx ẑ, find a vector C whose 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. C A 9 A B ˆx ŷ ẑ 9 ˆx ẑ B ŷ ˆx ẑ ˆx ẑ 9 ˆx ŷ ẑ ˆx 4 ŷ8 67 ẑ Problem.7 Given A ˆx x y ŷ y z ẑ x y, determine a unit vector parallel to A at point P. The unit vector parallel to A ˆx x y ŷ y z ẑ x y at the point P is A ŷ5 A ˆx ẑ4 ŷ5 ˆx 5 4 ẑ4 4 ˆx 5 ŷ 77 ẑ 6 Problem.8 By expansion in Cartesian coordinates, prove: (a) the relation for the scalar triple product given by (.9), and (b) the relation for the vector triple product given by (.). (a) Proof of the scalar triple product given by q. (.9): From q. (.7), 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
8 CHAPTR 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 mploying q. (.7), 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 triple product given by q. (.): The evaluation of the left hand side employs the expression above for B C with q. (.7): 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. (.7), 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 expressions for the components, the left hand side is equal to the right hand side. Problem.9 Find an expression for the unit vector directed toward the origin from an arbitrary point on the line described by x and z. An arbitrary point on the given line is y. The vector from this point to is: A ˆx ŷ y ẑ ˆx ŷy ẑ A y 4 5 y â A ẑ ˆx ŷy A 5 y
CHAPTR 9 Problem. Find an expression for the unit vector directed toward the point P located on the z-axis at a height h above the x y plane from an arbitrary point Q x y in the plane z. Point P is at h. Vector A from Q x y to P h is: A ˆx x ŷ y ẑ h ŷy ˆxx ẑ h A x y h A ŷy â ˆxx ẑ h A x y h Problem. Find a unit vector parallel to either direction of the line described by x z 4 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 x z plane. For convenience, we choose the x z plane with y. For x, z 4. Hence, point P is at 4. For z, x. Hence, point Q is at. Vector A from P to Q is: A ˆx ŷ ẑ 4 ˆx ẑ4 A ẑ4 â ˆx A Problem. Two lines in the x y plane are described by the expressions: Line x y 6 Line x 4y 8 Use vector algebra to find the smaller angle between the lines at their intersection point. Intersection point is found by solving the two equations simultaneously: x 4y x 4y 8
CHAPTR 5 5 (, ) -5 - -5 - -5 - (, -) - -5 A 5 5 5 B (, -) - -5 - θ AB Figure P.: Lines and. The sum gives x, which, when used in the first equation, gives y. Hence, intersection point is. Another point on line is x, y. Vector A from to is A ˆx ŷ ˆx ŷ A 5 A point on line is x, y. Vector B from to is B ˆx ŷ ˆx ŷ5 B 5 65 Angle between A and B is Problem. θ AB cos A B A B cos A given line is described by x y 4 4 5 5 65 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.
CHAPTR We first plot the given line. Next we find vector B which connects point P to P 4, both of which are on the line: B ˆx 4 ŷ ˆx4 ŷ Vector A starts at the origin and ends on the line at P. If the x-coordinate of P is x, y P (,) (,) A B P (4,) x Figure P.: Given line and vector A. then its y-coordinate has to be 4 x in order to be on the line. Hence P is at x 4 x. Vector A is A ˆxx x ŷ 4 But A is perpendicular to the line. Hence, Hence, A B ˆxx x ŷ 4 ˆx4 ŷ 4x 4 x or 4 x 8 5 A ˆx 8 ŷ 4 8 ˆx 8 ŷ 6 Problem.4 Show that, given two vectors A and B,
CHAPTR (a) the vector C defined as the vector component of B in the direction of A is given by C â B â A B A A where â is the unit vector of A, and (b) the vector D defined as the vector component of B perpendicular to A is given by D B A B A A (a) By definition, B â is the component of B along â. The vector component of B â along A is C â B â A A B A A B A A A (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 or B C D D B C B A B A A A C B D Figure P.4: Relationships between vectors A, B, C, and D.
CHAPTR Problem.5 A certain plane is described by x y 4z 6 Find the unit vector normal to the surface in the direction away from the origin. Procedure:. Use the equation for the given plane to find three points, P, P and P on the plane.. Find vector A from P to P and vector B from P to P.. Cross product of A and B gives a vector C orthogonal to A and B, and hence to the plane. 4. Check direction of ĉ. Steps:. Choose the following three points:. Vector A from P to P P at 4 P at 8 6 P at A ˆx 8 ŷ ẑ 4 ẑ4 ˆx8 Vector B from P to P 6 B ˆx ŷ ẑ 4 ŷ 6 ẑ4. C 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 6 ˆx 4 4 ŷ 4 8 4 ẑ ˆx 64 ŷ ẑ 8 8 6
4 CHAPTR Verify that C is orthogonal to A and B 64 8 A C 8 64 6 8 B C 4. C ˆx 64 ŷ ẑ 8 C ĉ C ˆx 64 5 4 5 4 5 5 ŷ ẑ 8 ˆx 8 7 ŷ 56 ẑ 74 64 ĉ points away from the origin as desired. Problem.6 Given B ˆx z y ŷ x z ẑ x y, find a unit vector parallel to B at point P. At P, B ˆx ŷ ẑ ˆx ŷ5 ẑ B ˆx ˆb ẑ ŷ5 B 5 ˆx ẑ ŷ5 7 Problem.7 When sketching or demonstrating the spatial variation of a vector field, we often use arrows, as in Fig. -5 (P.7), 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. P.7, which represents the vector field ˆrr, 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: (a) ˆxy, (b) ŷx, (c) ˆxx ŷy, (d) 4 ˆxx ŷy, (e) 5 ˆφr, (f) 6 ˆrsinφ.
CHAPTR 5 y x Figure P.7: Arrow representation for vector field ˆrr (Problem.7). (a) y x P.a: = - x^ y
6 CHAPTR (b) y x (c) P.7b: ŷx y x P.c: = x^ x + y^ y
CHAPTR 7 (d) y x (e) P.d: 4 = x^ x + y^ y y x P.e: 5 = φ^ r
8 CHAPTR (f) y x P.f: 6 = r^ sinφ Problem.8 Use arrows to sketch each of the following vector fields: (a) ŷy, ˆxx (b) ˆφ, (c) ŷ x, (d) 4 ˆrcosφ.
CHAPTR 9 (a) y x (b) P.4a: = x^ x - y^ y y x P.4b: = - φ^
CHAPTR (c) y x Indicates is infinite (d) P.4c: = y^ (/x) y x P.4d: 4 = r^ cosφ
CHAPTR Sections - and -: Coordinate Systems Problem.9 Convert the coordinates of the following points from Cartesian to cylindrical and spherical coordinates: (a) P, (b) P, (c) P, (d) P 4. Use the coordinate variables column in Table -. (a) In the cylindrical coordinate system, P tan 5 7 rad 4 6 4 In the spherical coordinate system, P tan tan 5 π rad 7 rad 4 9 6 4 Note that in both the cylindrical and spherical coordinates, φ is in Quadrant I. (b) In the cylindrical coordinate system, P In the spherical coordinate system, tan rad P tan tan rad rad Note that in both the cylindrical and spherical coordinates, φ is arbitrary and may take any value. (c) In the cylindrical coordinate system, P tan π 4 rad 4 45 In the spherical coordinate system, P tan tan 44 rad π 4 rad 5 45 Note that in both the cylindrical and spherical coordinates, φ is in Quadrant I.
CHAPTR (d) In the cylindrical coordinate system, P 4 In the spherical coordinate system, P 4 tan π 4 rad 8 5 tan 87 rad π 4 rad 46 5 5 tan Note that in both the cylindrical and spherical coordinates, φ is in Quadrant II. Problem. Convert the coordinates of the following points from cylindrical to Cartesian coordinates: (a) P π 4, (b) P, (c) P 4 π. (a) P x y z P r cos φ r sin φ z P cos π 4 sin π 4 P 4 4 (b) P x y z P cos sin P. (c) P x y z P 4cos π 4sin π P 4. Problem. Convert the coordinates of the following points from spherical to cylindrical coordinates: (a) P 5, (b) P 5 π, (c) P π. (a) P r φ z P Rsinθ φ Rcos θ P 5sin 5cos P 5 (b) P r φ z P 5sin π 5cos P π 5.
CHAPTR (c) P r φ z P sin π cos π P. Problem. Use the appropriate expression for the differential surface area ds to determine the area of each of the following surfaces: (a) r ; φ π ; z, (b) r 5; π φ π; z, (c) r 5; φ π 4; z, (d) R ; θ π ; φ π, (e) R 5; θ π ; φ π. Also sketch the outlines of each of the surfaces. Φ = π/ y 5 5 x (a) (b) (c) (d) (e) Figure P.: Surfaces described by Problem.. (a) Using q. (.4a), A z π φ r r dφ dz φz π φ z 4π
4 CHAPTR (b) Using q. (.4c), A 5 r (c) Using q. (.4b), A π z (d) Using q. (.5b), A π θ π φ φ π 5 r r z dφ dr r φ 5 r π φ π π 4 φ π 4 dr dz rz z 5 r R sinθ dφ dθ R (e) Using q. (.5c), 5 π Rsinθ dr θ π dφ A R Problem. φ Find the volumes described by (a) r 5; π φ π; z, (b) R 5; θ π ; φ π. Also sketch the outline of each volume. 4φcos θ π θ π φ π R φsin π π φ 5 R 5 π z z 5 5 y y x (a) x (b) Figure P.: Volumes described by Problem.. (a) From q. (.44), π V z φ π 5 r r dr dφ dz r φz 5 r π φ π z π
CHAPTR 5 (b) From q. (.5e), V π φ π θ 5 R cosθ R φ R sin θ dr dθ dφ 5 R π π θ φ 5π Problem.4 A section of a sphere is described by R, θ 9 and φ 9. Find: (a) the surface area of the spherical section, (b) the enclosed volume. Also sketch the outline of the section. z y x φ= o Figure P.4: Outline of section.
6 CHAPTR Problem.5 π S π 4 φ π 6 π θ V R R π R sinθ dθ dφ R π π cosθ π 6 4 4π π φ π 6 π θ R sin θ dr dθ dφ π π 8 π cosθ 6 8π 9 (m ) (m ) A vector field is given in cylindrical coordinates by ˆrr cos φ ˆφr sinφ ẑz Point P π is located on the surface of the cylinder described by r. At point P, find: (a) the vector component of perpendicular to the cylinder, (b) the vector component of tangential to the cylinder. (a) n ˆr ˆr ˆr ˆr ˆrr cosφ ˆφr sin φ ẑz ˆrr cos φ. At P π, n ˆrcos π ˆr. (b) t n ˆφr sinφ ẑz. At P π, t ˆφsin π ẑ ẑ9. Problem.6 coordinates by Find: At a given point in space, vectors A and B are given in spherical A ˆR4 ˆθ ˆφ B ˆR ˆφ (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.
CHAPTR 7 (a) Scalar component of B in direction of A: A C B â B A (b) Vector component of B in direction of A: ˆR ˆφ 6 4 8 4 ˆR4 ˆθ ˆφ C âc A C ˆR4 A ˆθ ˆφ 4 ˆR 9 ˆθ 5 ˆφ 5 (c) Vector component of B perpendicular to A: Problem.7 D B C ˆR ˆφ ˆR 9 ˆθ 5 ˆφ 5 ˆR 9 ˆθ 5 ˆφ 48 Given vectors A ˆr cos φ z ˆφ r 4sin φ ẑ r z B ˆrsinφ ẑcosφ find (a) θ AB at π, (b) a unit vector perpendicular to both A and B at π. 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. (a) At π, A ˆφ8 ẑ and B ˆr. From q. (.), θ AB cos A B AB cos AB 9 (b) At π, A ˆr 7 ˆφ4 B and ˆr ẑ. Since A B is perpendicular to both A and B, a unit vector perpendicular to both A and B is given by A B A B ˆr 4 7 ˆφ ẑ 4 7 4 ˆr 487 ˆφ 8 ẑ 84
8 CHAPTR Problem.8 Find the distance between the following pairs of points: (a) P and P in Cartesian coordinates, (b) P π 4 and P 4 π 4 4 in cylindrical coordinates, (c) P 5 4 π and P 6 π in spherical coordinates. (a) d 9 5 5 59 7 68 (b) (c) d r r r r cos φ φ z z 9 π π cos 4 4 4 6 5 4 d R R R R cosθ cosθ sinθ sinθ cos φ φ 9 6 4 cos πcos π sin π sinπcos 9 6 5 5 Problem.9 Determine the distance between the following pairs of points: (a) P and P, (b) P π and P 4 4 π, (c) P 5 π π and P 6 4 π π. (a) From q. (.66), (b) From q. (.67), d d 4 4 cos π π 8 4 8
CHAPTR 9 (c) From q. (.68), d 4 4 cos π cos π sinπsin π cos π π 5 Problem. Transform the following vectors into cylindrical coordinates and then evaluate them at the indicated points: (a) A ˆx x y at P, (b) B ˆx y x ŷ x y at P, (c) C ˆxy x y ŷx x y ẑ4 at P, (d) D ˆRsinθ ˆθcosθ ˆφcos φ at P 4 π π 4, (e) ˆRcosφ ˆθsinφ ˆφsin θ at P 5 π π. From Table -: (a) A ˆrcosφ ˆφsinφ r cos φ r sinφ ˆrr cos φ cos φ sinφ ˆφr sin φ cos φ sinφ P tan 5 6 4 A P ˆr 447 ˆφ 894 5 447 894 ˆr 4 ˆφ 68 (b) B ˆrcosφ ˆφsinφ r sin φ r cosφ ˆφcosφ ˆrsinφ r cos φ r sinφ ˆrr sin φcos φ ˆφr cos φ sin φ ˆrr sin φ ˆφr cos φ P tan B P ˆr ˆφ (c) (d) r C ˆrcosφ ˆφsinφ sin φ ˆφcos r φ r ˆrsinφ cos φ r ˆrsinφcos φ sin φ cosφ ˆφ sin φ cos φ ẑ4 P tan 45 C P ˆr 77 ẑ4 ẑ4 D ˆrsinθ ẑcosθ sinθ ˆrcos θ ẑsin θ cos θ ˆφcos φ ˆr ˆφcos φ
4 CHAPTR P 4 sin π π 4 cos π 45 D P 4 ˆr ˆφ (e) ˆrsin θ ẑcos θ cos φ ˆrcosθ ẑsinθ sin φ ˆφsin θ P 5 π π P 5 ˆrsin π ẑcos π cos π ˆrcos π ẑsin π sinπ ˆφsin π ˆr ˆφ Problem. Transform the following vectors into spherical coordinates and then evaluate them at the indicated points: (a) A ˆxy ŷxz ẑ4 at P, (b) B ŷ x y z ẑ x y at P, (c) C ˆrcos ˆφsin φ φ ẑcos φsinφ at P π 4, and (d) D ˆxy x y ŷx x y ẑ4 at P 4. From Table -: (a) A ˆRsin θcosφ ˆθcosθcos φ ˆφsinφ Rsinθsinφ ˆRsinθsin φ ˆθcosθsin φ ˆφcos φ Rsinθcos φ Rcosθ ˆRcosθ ˆθsinθ 4 ˆR R sin θsin φcosφ sin θsin φ cosθ 4cosθ ˆθ R sinθcos θsinφcos φ sin θsin φ cosθ 4sinθ ˆφR sinθ cos θcos φ sinθsin φ P tan tan 6 5 45 (b) A P ˆR 856 ˆθ 888 ˆφ B ˆRsinθsinφ ˆθcosθsin φ ˆφcosφ R ˆRcosθ ˆRR sinθ sin φ sinθcosθ ˆθR cos θsinφ sin θ ˆφR cosφ tan tan 5 6 6 8 P ˆθsinθ R sin θ
CHAPTR 4 (c) B P ˆR 896 ˆθ 449 ˆφ5 C ˆRsin θ ˆθcos θ ˆφsin φ φ ˆRcosθ ˆθsinθ cosφsin φ ˆRcosφ sin θ cosθsinφ ˆθcos φ cos sinθsin θ φ ˆφsinφ P tan π 4 45 45 C P ˆR 854 ˆθ 46 ˆφ 77 (d) R D ˆRsinθcos φ ˆθcosθcos ˆφsin φ φ sin θsin φ R sin θsin φ R sin θcos φ ˆRsinθsin φ ˆθcos θsinφ R ˆφcosφ sin θcos φ ˆRcosθ ˆθsinθ 4 ˆR sinθcos φsin φ sinθsin φcos φ 4cosθ ˆθ cosθcos φsin φ cosθsinφcos φ 4sinθ ˆφ cos φ sin φ R sin θsin φ R sin θcos φ P 4 P 4 4 tan tan P 4 6 5 6 45 D P 4 ˆR sin5 6 45 sin 45 sin5 6 sin 45 cos 45 4cos5 6 ˆθ cos5 6 cos45 sin 45 cos5 6 sin 45 cos 45 4sin5 6 ˆφ cos 45 sin 45 ˆR 67 ˆθ 7 ˆφ 77 Sections -4 to -7: Gradient, Divergence, and Curl Operators Problem. (a) T x z, (b) V xy z 4, Find the gradient of the following scalar functions:
4 CHAPTR (c) U zcos φ r, (d) W e R sin θ, (e) S 4x e z y, (f) N r cos φ, (g) M Rcosθsinφ. (a) From q. (.7), (b) From q. (.7), (c) From q. (.8), (d) From q. (.8), (e) From q. (.7), T ˆx 6x x z ẑ 6z x z V ˆxy z 4 ŷxyz 4 ẑ4xy z U ˆr rzcos φ r ˆφ zsinφ r r ẑ cosφ r W ˆRe R sinθ ˆθ e R R cos θ S 4x e z y S ˆx S ŷ x S ẑ y S ˆx8xe z ŷy z ẑ4x e z (f) From q. (.8), N r cos φ N ˆr N ˆφ r N ẑ r φ N z φ ˆrr cos ˆφr sin φcos φ (g) From q. (.8), M Rcosθsin φ M M ˆR R ˆθ M ˆφ R θ Rsinθ M ˆRcosθsin φ φ ˆθsinθsin φ ˆφ cosφ tanθ
CHAPTR 4 Problem. The gradient of a scalar function T is given by T ẑe z If T at z, find T z. T ẑe z By choosing P at z and P at any point z, (.76) becomes z z T z T T dl Hence, T z T z ẑe ˆx dx ŷ dy ẑ dz z dz e z z z e e z e z e z Problem.4 Follow a procedure similar to that leading to q. (.8) to derive the expression given by q. (.8) for in spherical coordinates. From the chain rule and Table -, T ˆx T ŷ x T ẑ y T z T R T θ ˆx R x θ x T R T θ ŷ R y θ y T R T θ ẑ R z θ z T ˆx R T ŷ R T ẑ R T φ φ x x x y z y x y z z x y z T φ φ y T φ φ z T θ tan x x y z tan x y y z tan x z y z T θ T θ T φ tan y x x tan y x y T φ T φ z tan y x
44 CHAPTR ˆx T R x x y z T θ z x y z x x y T φ y x y T y T z y T x ŷ R x y z θ x y z x y φ x y T z T ẑ R x y z θ x y z x y T φ T Rsinθcosφ T Rcosθ Rsinθcosφ T ˆx R R θ R Rsinθ Rsinθsin φ φ R sin θ T Rsinθsin φ T Rcosθ Rsinθsin φ T Rsinθcos φ ŷ R R θ R Rsinθ φ R sin θ T Rcosθ T ẑ R R Rsinθ θ R T ˆx R sinθcos φ T cosθcos φ T θ R sinφ φ Rsinθ T ŷ R sinθsin φ T cosθsin φ T cosφ θ R φ Rsinθ T ẑ R cosθ T sinθ θ R ˆxsinθcos φ ŷsinθsin φ T ẑcosθ R ˆxcosθcos φ ŷcosθsin φ ẑsinθ ˆxsinφ ŷcosφ ˆR T ˆθ R R which is q. (.8). T Rsinθ φ T T ˆφ θ Rsinθ φ Problem.5 For the scalar function V xy z, determine its directional derivative along the direction of vector A ˆx ŷz and then evaluate it at P 4. The directional derivative is given by q. (.75) as dv dl V â l, where the unit vector in the direction of A is given by q. (.): R â l ˆx ŷz T θ z
CHAPTR 45 and the gradient of V in Cartesian coordinates is given by q. (.7): V ˆxy ŷxy ẑz Therefore, by q. (.75), At P 4, dv dv y xyz dl z dl 4 9 8 7 Problem.6 For the scalar function T e r 5 cosφ, determine its directional derivative along the radial direction ˆr and then evaluate it at P π 4. dt dl π 4 T e r cosφ 5 T ˆr T ˆφ r T ẑ r φ T ˆr z e r 5 cosφ ˆφ e r 5 sinφ r dt r T ˆr e dl 5 cosφ e 5 cos π 4 4 74 Problem.7 For the scalar function U R sin θ, determine its directional derivative along the range direction ˆR and then evaluate it at P 5 π 4 π. du dl 5 π 4 π U R sin θ U U ˆR ˆθ R R du U ˆR dl U ˆφ θ Rsinθ sin θ R sin π 4 5 U sin ˆR φ θ sin θcos θ ˆθ R R
46 CHAPTR Problem.8 Vector field is characterized by the following properties: (a) points along ˆR, (b) the magnitude of is a function of only the distance from the origin, (c) vanishes at the origin, and (d), everywhere. Find an expression for that satisfies these properties. According to properties (a) and (b), must have the form ˆRR where R is a function of R only. Hence, and R R R R R R R R R dr R R R R R R R 4R R 4R ˆR4R R R R R dr R Problem.9 For the vector field ŷyz ˆxxz ẑxy, 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 units each and parallel to the Cartesian axes, and (b) the integral of over the cube s volume. (a) For a cube, the closed surface integral has 6 sides: ds F top F bottom F right F left F front F back
CHAPTR 47 F top F bottom F right F left F front F back x x x x x x x x y y y y y y y y z ˆxxz ŷyz ẑxy z ẑ dy dx xy dy dx x y 4 ˆxxz ŷyz ẑxy z xy dy dx x y 4 y ẑ dy dx y ˆxxz ŷyz ẑxy y ŷ dz dx z z dz dx z xz ˆxxz ŷyz ẑxy y z z dz dx z z z z xz x x z ŷ dz dx z ˆxxz ŷyz ẑxy x ˆx dz dy z dz dy yz z ˆxxz ŷyz ẑxy x z dz dy yz z ds 4 4 8 y x x ˆx dz dy y 4 4
48 CHAPTR (b) dv x x y y xy z z z ˆxxz ŷyz ẑxy dz dy dx z z z dz dy dx z y x 8 Problem.4 For the vector field ˆre ẑz, verify the divergence theorem r for the cylindrical region enclosed by r, z, and z 4. ds π r φ π 4 φ z π r φ π 4 6πe 4 dv z 8π r ˆre ẑz r ˆre ẑr dr dφ z ẑz ˆrr dφ dz r ẑz ẑr dr dφ z 4 r ˆre e dφ dz φ z 48π 8 77 π r r r 8π e 6πe 48π e r r φ r r e r r dr e r r 8 77 r π φ r dr dφ r dφ dr dz r r Problem.4 A vector field D ˆrr exists in the region between two concentric cylindrical surfaces defined by r and r, with both cylinders extending between z and z 5. Verify the divergence theorem by evaluating: (a) D ds, S
CHAPTR 49 (b) D V dv. (a) D ds F inner F outer F bottom F top F inner F outer π φ π φ π φ π φ F bottom F top r r 5 z 5 z 5 z 5 z π φ π φ ˆrr ˆrr dz dφ r r 4 dz dφ r π ˆrr ˆrr dz dφ r r 4 dz dφ 6π r ˆrr ẑr dφ dr z ˆrr ẑr dφ dr z 5 Therefore, D ds 5π. (b) From the back cover, D r r rr 4r. Therefore, 5 π D dv 4r r dr dφ dz r 4 π 5 r z φ r φ z 5π Problem.4 For the vector field D ˆRR, evaluate both sides of the divergence theorem for the region enclosed between the spherical shells defined by R and R. The divergence theorem is given by q. (.98). valuating the left hand side: π π D dv V φ θ R R R R R R sinθ dr dθ dφ π π cosθ θ R 4 8π R
5 CHAPTR The right hand side evaluates to π D ds S π φ θ π π π π φ θ θ ˆRR sinθ dθ dφ R ˆRR ˆRR sinθ dθ dφ ˆRR sin θ dθ π π R 48sin θ dθ 8π θ Problem.4 (a) (b) For the vector field ˆxxy ŷ x y, calculate dl around the triangular contour shown in Fig. P.4(a), and C ds over the area of the triangle. S In addition to the independent condition that z, the three lines of the triangle are represented by the equations y, x, and y x, respectively. y y L L L x (a) L L L (b) x Figure P.4: Contours for (a) Problem.4 and (b) Problem.44. (a) dl L L L L ˆxxy ŷ x y ˆx dx ŷ dy ẑ dz x xy y z dx y x y z dy z y dz
CHAPTR 5 Therefore, L ˆxxy ŷ x y ˆx dx ŷ dy ẑ dz xy z dx x y x y x z dy y y 5 y L ŷ x ˆxxy y ˆx dx ŷ dy ẑ dz x x xy y x z dx y x y y dl z x y x y z dy 5 (b) From q. (.5), ẑx so that ds Problem.44 x x x y ẑx ẑ dy dx z x x dy dx y x x dz z y x dz x x dx x Repeat Problem.4 for the contour shown in Fig. P.4(b). In addition to the independent condition that z, the three lines of the triangle are represented by the equations y, y x, and y x, respectively. (a) dl L L L L ŷ x ˆxxy y ˆx dx ŷ dy ẑ dz xy dx y z x y x y z dy y dz z L ŷ x ˆxxy y ˆx dx ŷ dy ẑ dz xy z y x x dx x y y x y z dy y x z dz x x x 4y y y y
5 CHAPTR Therefore, L ˆxxy ŷ x y ˆx dx ŷ dy ẑ dz x x xy y x z dx y x y y dl x y x y z dy (b) From q. (.5), ẑx so that Problem.45 by evaluating: (a) (b) ds x z y x dz x y ẑx ẑ dy dx z x x y ẑx ẑ dy dx z x x x dy x dy dx x y dx x y x x x x dx x x dx x x x x Verify Stokes s theorem for the vector field B ˆrr cos φ ˆφsinφ B dl over the semicircular contour shown in Fig. P.46(a), and C B ds over the surface of the semicircle. S (a) B dl B dl B dl B L dl L L B dl ˆrr cos φ ˆφsin φ ˆr dr ˆφr dφ ẑ dz r cos φ dr r sinφ dφ L B dl r r r cos φ dr r φ z φ r sin φ dφ z
CHAPTR 5 y y L - L L (a) x L L L 4 L (b) x Figure P.46: Contour paths for (a) Problem.45 and (b) Problem.46. (b) L B dl L B dl π cosφ φ 4 r r cos φ dr r r cos φ dr z r r B dl 4 8 B ds φ π r φ π z π φ r sinφ dφ π φ π r sin φ dφ B ˆrr cos φ ˆφsinφ sinφ ˆr ˆφ r φ z z φ r cos ẑ r r r sin φ φ φ r cos ˆr ˆφ ẑ r sinφ r sin φ ẑsinφ π ẑsinφ r ẑr dr dφ φ r r z z r r sinφ r dr dφ cosφ r r π r 8 φ Problem.46 Repeat Problem.45 for the contour shown in Fig. P.46(b).
54 CHAPTR (a) (b) B dl B dl B dl B dl B L dl L L L 4 B dl ˆrr cos φ ˆφsin φ ˆr dr ˆφr dφ ẑ dz r cos φ dr r sinφ dφ L B dl L B dl L B dl L 4 B dl B dl r r r cos φ dr r r r cos φ dr φ z π z π φ π z z π cosφ φ r r cos φ dr cosφ φ π 5 B ds r r cos φ dr φ r sin φ dφ φ r sinφ dφ z r z r sin φ dφ φ π r sin φ dφ φ π r z B ˆrr cos φ ˆφsin φ sinφ ˆr ˆφ r φ z z φ r cos ẑ r r r sin φ φ φ r cos ˆr ˆφ ẑ r sinφ r sin φ ẑsinφ π ẑsin φ φ r r ẑr dr dφ π φ r cosφ sinφ r dr dφ r r π r 5 φ z r r
CHAPTR 55 Problem.47 evaluating it on the hemisphere of unit radius. Verify Stokes s Theorem for the vector field A ˆRcosθ ˆφsinθ by A ˆRcos θ ˆφsinθ ˆRAR Hence, A R cosθ, A θ, A φ sinθ. A ˆR Rsinθ θ A φ sinθ ˆR θ Rsinθ θ sin ˆθ R ˆR cos θ R ˆθ sin θ R ˆφ sinθ R For the hemispherical surface, ds ˆRR sinθ dθ dφ. π φ π θ π φ A ds π θ 4πR sin θ ˆRcos θ π R R ˆθ sinθ R π ˆθA θ ˆφA φ ˆθ R Rsinθ ˆφ R R R RA φ ˆφ R θ A R θ cos θ ˆφ sin θ R ˆRR sinθ dθ dφ R The contour C is the circle in the x y plane bounding the hemispherical surface. π A dl ˆRcosθ π ˆφsinθ ˆφR dφ θ π Rsinθ dφ θ π π C φ R R Problem.48 Determine if each of the following vector fields is solenoidal, conservative, or both: (a) A ˆxx ŷyxy, (b) B ˆxx ŷy ẑz, (c) C ˆr sin φ r ˆφ cosφ r, (d) D ˆR R, r (e) ˆr r ẑz, (f) F ˆxy ŷx x y, (g) G ˆx x z ŷ y x ẑ y z, (h) H ˆR Re R.
56 CHAPTR (a) A ˆxx ŷxy x x A ˆxx ŷxy ˆx y z xy ŷ ˆx ẑ ŷ y xy x x y z x The field A is solenoidal but not conservative. (b) B ˆxx ŷy ẑz x x B ˆxx ŷy ẑz y z ŷ z ˆx y ˆx ŷ ẑ y y z x The field B is conservative but not solenoidal. (c) C ˆr sinφ ˆφ r cosφ r sinφ r r r sinφ r C ˆr r ẑ r x ẑ x xy z z x y x z ẑ cosφ r r φ r sin φ r sinφ r ˆr sinφ r φ r ˆr ˆφ ẑ r ˆφ cos φ r cosφ z r cosφ r r cosφ ˆφ φ r The field C is neither solenoidal nor conservative. z sinφ z r sinφ r cosφ r x y r ẑ cosφ r y x y x
CHAPTR 57 (d) D D ˆR R R R ˆR R ˆR Rsinθ ˆφ R θ θ sin R R R R θ φ Rsinθ ˆθ R The field D is conservative but not solenoidal. (e) r ˆr r r r r ˆr r ẑz r r r φ r φ r r r r θ θ sin sinθ φ R ˆr ˆθ ˆφ z z R Rsinθ R R r r r r r r 6r r r r r 4r r r z φ r φ r z ˆφ z z ẑ r r Hence, is conservative, but not solenoidal. (f) ˆxy F ŷx y x x y ˆx x y ŷ x y y x x y y x y xy xy x y x y F x φ r r φ R r r φ
58 CHAPTR x y F ˆx ŷ ẑ x x y y x y x y ẑ x y x y ẑ y x x y Hence, F is neither solenoidal nor conservative. (g) x y G ˆx x z ŷ y x ẑ y z G x x z y y x z y z z x y G ˆx y y z z y x ŷ ẑ x y x y x z ˆxy ẑx ŷz Hence, G is neither solenoidal nor conservative. (h) x y z x z H ˆR Re R H e R e R R R R R R R e R e R R H Hence, H is conservative, but not solenoidal. x y z Problem.49 (a) V 4xy z, (b) V xy yz zx, (c) V x y, (d) V 5e r cosφ, (e) V e R sinθ. Find the Laplacian of the following scalar functions: (a) From q. (.), 4xy z 8xz 4xy z
CHAPTR 59 (b) xy yz zx (c) From the inside back cover of the book, x y r r 4 x y (d) (e) Problem.5 5e r cosφ 5e r cosφ e R sinθ e R sinθ R r r cos θ sin θ R sinθ Find a vector G whose magnitude is 4 and whose direction is perpendicular to both vectors and F, where ˆx ŷ ẑ and F ŷ ẑ6. 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 F and another along the opposite direction. Hence, Problem.5 G 4 F 4 ˆx ŷ ẑ ŷ ẑ6 F ˆx ŷ ẑ ŷ ẑ6 ˆx6 4 ŷ6 ẑ 6 6 9 4 ˆx6 9 ŷ6 ẑ ˆx 8 ŷ 8 ẑ 4 A given line is described by the equation: y x Vector A starts at point P and ends at point P on the line such that A is orthogonal to the line. Find an expression for A. We first plot the given line.
6 CHAPTR y P (, ) A B P 4 (, ) P (x, x-) x P (, -) Next we find a vector B which connects point P to point P 4, both of which are on the line. Hence, B ˆx ŷ ˆx ŷ Vector A starts at P and ends on the line at P. If the x-coordinate of P is x, then its y-coordinate has to be y x, per the equation for the line. Thus, P is at x x, and vector A is A ˆx x ŷ x ˆxx ŷ x Since A is orthogonal to B, Finally, A B ˆxx ŷ x ˆx ŷ x x x A ˆxx ŷ x ˆx ŷ ˆx ŷ
CHAPTR 6 Problem.5 Vector field is given by ˆR 5Rcos θ ˆθ R sin θcosφ ˆφsin φ Determine the component of tangential to the spherical surface R at point P 6. At P, is given by ˆR 5 cos ˆR 8 67 ˆθ 5 ˆφ 6 ˆθ sin cos6 ˆφsin 6 The ˆR component is normal to the spherical surface while the other two are tangential. Hence, t ˆθ 5 ˆφ 6 Problem.5 Transform the vector A ˆRsin θcos φ ˆθcos φ ˆφsinφ into cylindrical coordinates and then evaluate it at P π π. From Table -, A ˆr sinθ ẑcosθ sin θcosφ ˆr cos ẑsin θ θ cos φ ˆφsinφ ˆr sin θcosφ cosθcos φ ˆφsinφ ẑ cosθsin θcos sinθcos φ φ At P π π, A ˆφ Problem.54 valuate the line integral of ˆxx ŷy along the segment P to P of the circular path shown in the figure. y P (, ) P (-, ) x
6 CHAPTR We need to calculate: P P d Since the path is along the perimeter of a circle, it is best to use cylindrical coordinates, which requires expressing both and d in cylindrical coordinates. Using Table -, ˆxx ŷy ˆr cos φ ˆφsin φ r cos φ ˆr sinφ ˆφcosφ r sin φ ˆr r cos φ sin φ ˆφr sin φcos φ The designated path is along the φ-direction at a constant r. From Table -, the applicable component of d is: d ˆφ r dφ Hence, P φ 8 d P ˆrr cos sin φ 9 φ φ ˆφ r sin φcosφ ˆφ r dφ r 8 9 r sinφcos φ dφ r r sin φ 8 φ 9 r 9 Problem.55 evaluating: Verify Stokes s theorem for the vector field B ˆr cosφ ˆφsinφ by (a) B d over the path comprising a quarter section of a circle, as shown in the C figure, and (b) S B ds over the surface of the quarter section. y L (-, ) L (, ) L x
CHAPTR 6 (a) B d B d B d B d C L L L 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: d ˆr dr ˆφr dφ ẑ dz Along path L, dφ and dz. Hence, d ˆr dr and r B d L ˆr cosφ ˆφsinφ ˆr dr r φ 9 cosφ dr r cos φ r r φ 9 φ 9 Along L, dr dz. Hence, d ˆφr dφ and 8 B d L ˆr cosφ ˆφsinφ ˆφr dφ φ 9 r 8 cos φ 9 Along L, dz and dφ. Hence, d ˆr dr and B d L ˆr cosφ ˆφsinφ ˆr dr r φ 8 Hence, (b) S r B d C r ẑ r cosφ dr φ 8 r 6 B ẑ B r rb r φ φ φ r r sin φ φ cos ẑ r sinφ sinφ ẑ sin φ r B ds 8 r φ 9 r r cosφ 8 ẑ r sinφ ẑr dr dφ φ 9 6
64 CHAPTR Hence, Stokes s theorem is verified. Problem.56 Find the Laplacian of the following scalar functions: (a) V r sinφ (b) V R cos θsin φ (a) (b) V r V R R 4 R r r r V r V r φ r sinφ r r V z r sinφ r r r r r 4sin φ 9r sin 4r sin φ φ 5r sin φ R R R V R R R R sinθ θ R sinθ θ R cosθsin φ sinθ θ R sin θ φ R cos θsinφ 4 cos cos 4 θsinφ R4 θsinφ cosθsin φ R 4 sin θ sinφ r φ r sinθ V θ R cosθsinφ cosθ R 4 sin θ sin φ R sin θ V φ