Introduction to Vector Calculus (9) SOLVED EXAMPLES Q. If vector A i ˆ ˆj k, ˆ B i ˆ ˆj, C i ˆ 3j ˆ kˆ (a) A B (e) A B C (g) Solution: (b) A B (c) A. B C (d) B. C A then find (f) a unit vector perpendicular to both B and C Component of A along B. (a) A B (b) A B i ˆ ˆ j k ˆ i ˆ ˆ j 4iˆ j ˆ kˆ i ˆ ˆ j b ˆ i ˆ ˆ j ˆ k (c) A. B C (d) B. C A (e) A B C 0 3 + 8 8 0 3 0 + 8 B A.C C A.B i ˆ ˆj i ˆ ˆj k ˆ. i ˆ 3j ˆ kˆ i ˆ 3j ˆ k ˆ i ˆ ˆ j k ˆ. i ˆ ˆ j iˆ 3kˆ
(30) Introduction to Vector Calculus (f) ˆn i ˆ ˆ j ˆ ˆ ˆ i 3j k B C B C 4 ˆi j ˆ 4bˆ (g) A B A cos Bˆ A.B Bˆ ˆ 0i ˆ 5j ˆ 5 i ˆ ˆj Q.: Find the angle between A i ˆ ˆj kˆ Solution : as A.B B B as i ˆ ˆ j k. i ˆ ˆ j i ˆ ˆ j A.B A B cos cos B ˆB B and B ˆi ˆj 3kˆ. i ˆ ˆ j k ˆ ˆ ˆ ˆ. i j 3k A.B A B 3 cos 4 6 60.5 approx. Q.3: If A i ˆ 3j ˆ 7kˆ and B i ˆ ˆj kˆ perpendicular to each other. Solution : so A.B i ˆ 3j ˆ 7k ˆ. i ˆ ˆ j k ˆ 4 3 + 7 0 A.B A B cos 0, then show that A and B are cos 0
Introduction to Vector Calculus (3) 90 so A is perpendicular to B. Q.4: Find the unit vector perpendicular to both A i ˆ 3j ˆ 5kˆ and B i ˆ 3j ˆ kˆ. Also find the angle between them. Solution: As A B ˆ ˆ ˆ i j k 3 5 3 iˆ 8j ˆ unit vector perpendicular to both A and B A B ˆn A B Again sin i ˆ 8j ˆ i ˆ 8j ˆ 8 08 A B A B 08 08 4 9 5 4 9 38 4 08 sin 38 4 38.7 Approx. Q.5: A particle is acted upon by two constant forces F ˆ ˆ ˆ i 4j 3k and F ˆ ˆ ˆ 3i j k due to which particle is displaced from ˆi j ˆ 3kˆ to 4i ˆ 5j ˆ kˆ. Calculate the total work done. Solution: Displacement of the particle r 4i ˆ 5j ˆ k ˆ ˆ i j ˆ 3k ˆ
(3) Introduction to Vector Calculus Hence total work done 3i ˆ 3j ˆ kˆ Total force. displacement F F.r ˆi 4ˆj 3kˆ 3i ˆ ˆj k ˆ. 3i ˆ 3j ˆ kˆ ˆ ˆ ˆ ˆ ˆ ˆ 4i 5j 4k. 3i 3jk 5 8 35 units. Q.6: A rigid body is rotating with angular velocity of 5 rad/s about an axis parallel to 3j ˆ kˆ and passing through the point ˆi ˆj 3kˆ. Find the velocity vector of the particle, when it is at the point i ˆ 4 ˆj kˆ. Solution : Suppose r is the position vector then r i ˆ 4j ˆ k ˆ ˆ i ˆ j 3k ˆ angular velocity linear velocity ˆi 5j ˆ 4kˆ 3j ˆ kˆ 5 5 3j ˆ kˆ 3j ˆ kˆ 0 v 5 ˆ ˆ r 3j k ˆ i 5j ˆ 4k ˆ 0 ˆi ˆj kˆ 5 0 3 0 5 4 5 7i ˆ ˆj 3kˆ 0 units. Q.7 : Calculate the torque of a force i ˆ j ˆ 5kˆ about the point 8j ˆ acting through the point 6i ˆ 4j ˆ kˆ. Solution : Here
Introduction to Vector Calculus (33) r 8j ˆ 6i ˆ 4j ˆ k ˆ 6i ˆ 4j ˆ kˆ r F 6i ˆ 4j ˆ k ˆ i ˆ j ˆ 5k ˆ torque ˆ ˆ ˆ i j k 6 4 5 4i ˆ 34ˆj 4kˆ Q.8: A force vector 0ˆi 5ˆj 35kˆ passes through a point (, 5, 7). Prove that force is also passing through the origin. Solution: The position vector r i ˆ 5j ˆ 7k ˆ and moment of the form about this point i.e. torque r F i ˆ 5j ˆ 7k ˆ 0i ˆ 5j ˆ 35k ˆ ˆ ˆ ˆ i j k 5 7 0 5 35 0 As the moment is zero, which shows that forces is passing through the origin. Q.9: A force 4i ˆ 3j ˆ kˆ passes through the point ( 9,, ). Find the component of moment of the force about the axis of reference. Sol.: Here so moment of force i.e. torque r F r 9i ˆ j ˆ k ˆ 9i ˆ ˆ j k ˆ 4i ˆ 3j ˆ k ˆ
(34) Introduction to Vector Calculus ˆ ˆ ˆ i j k 9 4 3 7iˆ ˆj 9kˆ Hence components of moment of force are 7 unit, units and 9 units in x, y and z direction respectively. Q.0: A proton is moving with velocity 0 8 cm/s along z-axis through an electric field of intensity 3 0 4 volt/cm along x-axis and magnetic field of intensity 000 gauss along y-axis. Calculate the magnitude and direction of total force. Solution: Intensity of electric field E 4 30 iˆ volt / cm 00 iˆ esu/cm Proton charge 9 0.6 0 C 4.80 esu Magnetic field B velocity v ˆ 000 j gauss 8 ˆ 0 k cm / s so total force acting on the proton F v B q E C 0 8 4.8 0 00ˆi 0 kˆ 000ˆj 0 30 8 4.47 0 ˆi dyne Hence total force acting on the proton has magnitude +4.47 0 8 dyne along the +ve x- direction. Q.: Find the value of the constant p so that A i ˆ ˆj 3k, ˆ B i ˆ 3j ˆ kˆ and C 3i ˆ pj ˆ kˆ are coplanar. A. B C 0 Solution: We know that three vectors are said to be coplanar if 3 3 3 p 0
Introduction to Vector Calculus (35) 4p 38 0 or 4p 38 Q.: Evaluate A B C Solution: p 38 9.5 4 where A i ˆ ˆj, B ˆi ˆj kˆ A B C and C 5i ˆ 3j ˆ kˆ. B A.C C A.B ˆi ˆj kˆ i ˆ ˆj 0k ˆ. 5i ˆ 3j ˆ kˆ 5i ˆ 3j ˆ kˆ i ˆ ˆj 0k ˆ. ˆi ˆj kˆ 7 ˆ i ˆ j k ˆ 5i ˆ 3j ˆ k ˆ i ˆ 4j ˆ 8kˆ Q.3: If r si the position vector of any point (x, y, z) and A then show that r.a. A 0 is the equation of a constant plane. (i) (ii) r A.r is the equation of a sphere. is a constant vector Also show that result of (i) is of the form Ax By Cz D 0 where D A B C and that of (ii) is of the from x y z r. [RU 005] Solution: (i) Suppose A A, B, C and r x, y, z r A. A x AA y BB z CC xa A yb B zc C xa yb zc A B C Ax By Cz D
(36) Introduction to Vector Calculus where D A B C r A.A so which is an equation of a plane. r A.A (ii) r A.A if x y z 0 Ax By Cz D 0 x Ax y B y z Cz 0 then x y z A B C 0 Which is the equation of sphere whose surface touches the origin. Q.4: A particle moves on the curve x t, y t 4t, z 3t 5 where t is the time. Find the components of velocity and acceleration at time t in the direction ˆi 3j ˆ kˆ. Solution: Position vector so velocity vector acceleration r dr v dt t ˆi t 4t ˆj 3t 5 kˆ d ˆ d ˆ d t i t 4t j 3t 5 kˆ dt dt dt 4t ˆ i t 4 ˆ j 3k ˆ a dv dt d ˆ d ˆ d 4t i t 4 j 3 kˆ dt dt dt 4iˆ j ˆ 0 at` t, velocity v 4iˆ j ˆ 3kˆ acceleration a 4i ˆ j ˆ and the component of v along ˆi 3j ˆ kˆ
Introduction to Vector Calculus (37) is 4i ˆ ˆ j 3k ˆ ˆ ˆ ˆ. i 3j k 3 6 8 4 4 7 and component of a along i 3j ˆ kˆ is ˆ ˆ ˆ ˆ ˆ 4i j. i 3j k 3 4 4 7 Q.5: Calculate the unit vector, which is normal to the surface Solution : Here At (,, ), x y xy 3xyz at the point (,, ). ˆ ˆ ˆ i j k x y xy 3xyz x y z ˆ x x y xy 3xyz i x y xy 3xyz ˆj y x y xy 3xyz kˆ z ˆ xy y 3yz i x xy 3xz j 3xy kˆ 3 ˆ i 3 ˆ j 3k ˆ 3k ˆ so the unit vector normal to the surface at (,, ) is ˆ 3k 3 ˆk ˆ 3k 3 Q.6: Find the direction derivative of x, y,z x y xy at the point (,, 4) along the direction of the vector (,, ). Solution: as
(38) Introduction to Vector Calculus,, 4 x y xy ˆ ˆ ˆ x y z i j k x, y,z x y xy iˆ x y xy ˆj x y xy kˆ x y z ˆ xy y i x xy ˆj ˆ 3i Position vector ˆr ˆi j ˆ kˆ and unit vector along this position vector and direction derivative ˆn ˆn ˆi j ˆ kˆ iˆ j ˆ kˆ 4 6 ˆi j ˆ kˆ 3i ˆ. 3 6 6 Q.7: Find the equation of the tangent plane and normal line to the surface x y z 3 at the point (,, 3). Solution : Here x,y,z x y z 3 x y z 6 x x y z 4x y x y z y z x y z so the components, and at the point (,, 3) will be x y z
Introduction to Vector Calculus (39) x 4 8,, y z Hence the equation of the tangent plane to the surface at the point (,, 3) is X 8 Y Z 3 0 or 4X + y + Z 6 so the equation of normal to the surface at (,, 3) is or X 8 X 4 Y Z 3 Y Z + 3 Q.8: Find the angle between the surfaces x y z 9 and x y z 3 at the (,, ) Solution: Suppose x y z and x y z so x ˆi yˆj z kˆ and x iˆ yˆj kˆ and,,,, iˆ 4j ˆ 4kˆ iˆ 4j ˆ kˆ since and are normal to and then. cos surfaces and. so. cos where is the angle between the cos 4 6 4 6 cos 36 6 54.4 approx. Q.9 (i) Provle that P cos ˆ ˆ ˆ ˆ i sin j and cosi sin j are unit vectors in
(40) Introduction to Vector Calculus the xy-plane respectively making and with the x-axis. (ii) By means of dot product, obtain the formula for cos. by similarly formulating P and Q, obtain the formula for cos. (iii) If is the angle between P and Q find P Q in terms of. Solution: (i) Given P cos ˆ ˆ i sin j Q cos ˆ ˆ i sin j y Q P (ii) hence P and Q But P cos sin Q cos sin are unit vectors. P.Q P Q cos. cos...() P.Q cos ˆ i sin ˆ ˆ ˆ j. cosi sin j so cos cos sin sin...() cos cos cos sin sin let P P cos ˆ ˆ i sin j and Q cos ˆ ˆ i sin j then P.Q. cos
Introduction to Vector Calculus (4) (iii) P and Q are unit vectors. so P Q cos cos sin sin Q PQ cos cos cos sin W 4x y ˆi 7x z ˆj 4xy z kˆ Q.0: A vector field is given as (i) What is the magnitude of the field at point (. 3, 4). (ii) At what point on z-axis is the magnitude of W equal to unity? [RU 00] Solution: (i) 4x y iˆ 7x z ˆj 4xy z kˆ W 4 3 iˆ 7 4 ˆj 4 3 4 kˆ at P(, 3, 4), W 48iˆ ˆj 8kˆ W 48 8 53.4 (ii) As the required point is on z-axis so x 0, y 0 W ˆ z j z k for that point. W so 4 4z 4z 0 taking z as positive z 0.07 4 z z 4z 4z 4 6 6 z 8 z.07 and 0.07 z ± 0.455 Q.: Calculate the differential volume to obtain the expression for volume of the (i) sphere of radius 'b'
(4) Introduction to Vector Calculus (ii) Semispherical shell of inner radius 'a' and outer radius 'b'. (iii) Cylinder of radius 'b' and height 'h'. Solution: (i) Differential volume in spherical coordinates dv r sin dr d d here r 0 to b, 0 to, 0 to. So volume of sphere V b dv r sin dr d d V 0 0 0 b r dr sin d d 0 0 0 b r dr cos 0 0 r dr b 0 4 3 b 3 so V sphere (ii) For semispherical shell r a, r b 4 b 3 3 so dv r sin dr d d here r a to b, 0 to, 0 to so V b a 0 0 r sin dr d d
Introduction to Vector Calculus (43) b a 0 0 r dr sin d d b r drcos a 0 r 3 b a (iii) 3 3 b a Differential volume for a cylinder dv r dr d dz here r 0 to b, z 0 to h, and 0 to. 3 so V dv V V b h a 0 0 r dr d dz b h r dr dz d 0 0 0 r.h. b 0 so V cyl. b h Q. : For positive x, y, z let 40 xyz c/m 3. Find the total charge within the region bounded by x 0, y 0, 0 x 3y 0 and 0 z. Solution : Here Q 5 y z 0 0 0 40 xyz dx dydz
(44) Introduction to Vector Calculus 0x 5 3 y z 40x dx 0 0 0 5 40 3 9 0 00x 40x 4x dx 40 00x 40x 4x 9 3 4 Q 95.96 C 3 4 Q.3: Given point P in Cartesian coordinate system as P(,, 3). Calculate its coordinates in cylindrical system. Solution: As given x, y, z 3 5 0 y x y tan, z z x so 5.36 z 3 tan 63.43 so P cyl. (.36, 63.43, 3) Q.4: The cooridnate of a point P in cylindrical system is P(, 45, ). find its equivalent in cartesion system. Solution: Here, 45, z and x cos, y sin, z z x. cos 45 0.707 y 0.707 z so P cart. (0.707, 0.707,) Q.5: Find the constant m such that the vector
Introduction to Vector Calculus (45) x 3y ˆi y z ˆj x mz kˆ is solenoidal. Solution: The vector will be solenoidal if ˆ ˆ i j kˆ x 3y ˆi y z ˆj x mz kˆ 0 x y z so x 3y y z x mz 0 x y z i.e. or + + m 0 m Q.6: Find div F and curl F if F grad x 3 y 3 z 3 3xyz Solution: Here F grad x 3 y 3 z 3 3xyz. [WBUT 00] 3 3 3 ˆ 3 3 3 x x y z 3xyz i x y z 3xyz ˆj y 3 3 3 x y z 3xyz kˆ z 3x 3yz ˆi 3y 3xz ˆj 3z 3xy kˆ div F 3x 3yz 3y 3xz 3z 3xy x y z 6x y z ˆ ˆ ˆ i j k F x y z 3x 3yz 3y 3xz 3z 3xy 3x 3x ˆi 3y 3y ˆj 3z 3z kˆ 0 Q.7 Show that curl grad f 0 where f x y + xy + z.
(46) Introduction to Vector Calculus Solution: grad f f ˆi f ˆj f kˆ x y z xy y ˆi x x ˆj z kˆ curl grad f ˆ ˆ ˆ i j k x y z xy y x x z 0 0 x x k ˆ 0 Q.8: If the scalar function x,y,z xy z,. is its corresponding scalar field is solenoidal or irrotational? Solution : Let so.f So field is not solenoidal. F yˆi x ˆj zkˆ x y z y x z 0 +0 + 0 Now F ˆ ˆ ˆ i j k x y z y x z so field is irrotational. 0 Q.9: Verify the divergence theorem for the vector function F ˆ 4xz i y ˆ j yz k ˆ taken over the cube bounded by x 0, y 0,, z 0,. [WBUT (math) 00] Solution.:
Introduction to Vector Calculus (47) z 3 4 5 0 y for face 4567; 7 x ˆn ˆi and x 6 F. n ˆ ds 0 0 F. n ˆ ds 4z dydz î dy dz 0 for 30 for and for 3074 for face 345, and total Again 3 56 n ˆ ĵ, y F. n ˆ ds F. n ˆ ds 0 4 F. n ˆ ds 5 F. n ˆ ds 0 for 067 6 F. n ˆ ds S F 3 0 0 0 x y z 4xz y yz 4z y
(48) Introduction to Vector Calculus.F dv 4z ydx dy dz 0 0 0 0 0 0 4z yz 0 0 0 dx dy 3 F. n ˆ ds F dv V s Hence divergence theorem is verified. Q.30: Calculate the line integral of A cos ˆ z sin zˆ wedge defined by 0 4, 0 30, z 0. Solution: Given A cos ˆ z sin zˆ differential length dl y d ˆ d ˆ dzzˆ around the edge L of the (3) () 0 () x Circulation of A around path is A.dl A.dl A.dl A.dl L 3 for () 0, d 0, z 0, dz 0 A.dl cosˆ z sin d ˆ d ˆ dz zˆ
Introduction to Vector Calculus (49) 4 4 cos d d 0 0 for () for (3) 4, d 0 6 8 z 0, dz 0 A. dl cos d zsin dz 0 / 6, d 0, z 0, dz 0 A. dl cos d 3 3 0 cos d 6 4 So total L 0 0 3 d 0.866 4 4 6.93 A. dl 8 + 0 6.93.07 Q.3 Given A x xy, calculate A. ds Solution: So y x, ds dx dy over the region y x, 0 < x <.
(50) Introduction to Vector Calculus y 0 A. ds x xy dx dy x x x x dx dy 0 y 0 0 y 0 xy dx dy 5 4 x x dx x 0 x 0 dx 3 64 5.7 x y z Q.3 For a scalar function sin sin e 3. Calculate the magnitude direction of maximum rate of increase of at the point (,, ). Solution: As gradient of a scalar function gives the magnitude and direction of max. rate of change of that So ˆi ˆj kˆ x y z x y z ˆ x y z sin sin e i sin sin e ˆ j x 3 y 3 at (,, ) x y z sin sin e kˆ z 3 z x ˆ x y z e sin j sin sin e kˆ 6 3
Introduction to Vector Calculus (5),, and,, ˆ e sin j sin sin e kˆ 6 3 6 0.367 j 0.866 0.367 kˆ ˆ 0.9 ˆj 0.38 kˆ 0.9 0.38 0.37 / and ĵ 0.9j ˆ 0.38kˆ 0.37 0.5j ˆ 0.86 kˆ Q.33 : Determine the divergence of the following vector fields at given points A yzi ˆ 4 x y ˆ j xyz k ˆ at,, (i) (ii) (iii) B z sin ˆ 5 z cos ˆ z zˆ at 5,, C r sin cos rˆ cos ˆ r ˆ at,, 3 3 Solution: In cartesion (a).a x y z yz 4x 4y xyz.a 0 + 4 + xy 4 + xy at,,,.a,, 4 + ( ) (b) In cylindrical.b B B Bz z
(5) Introduction to Vector Calculus given B z sin B 5z cos, Bz z at 5,,.B z sin 5. zsin 3z sin (c) In spherical system.b (3 ).C C r r r r sin rsin r C sin C cos sin cos r sin r r r sin rsin 3 at,, 3 3 cos cot 6sin cos r.c cos cot 6sin cos 3 3 3 3.6 + 0.88.88 Q.34: Find the nature of the vector ˆ ˆ F 30i xy j 5xz kˆ. [RU 003] Solution :.F 30 xy 5xz x y y x + 0xz 0
Introduction to Vector Calculus (53) Curl F ˆ ˆ ˆ i j k F x y z 30 xy 5xz 5z ˆj y kˆ 0 as.f 0 fields is not solenoidal and F 0 field is rotational. G 6xy 3 ˆi 8x ˆj xkˆ. Q.35: Given the vector field (i) Is G irrotational (or conservative)? (ii) Find the net flux of G over the cube 0<x, y, z <. (iii) Determine the circulation of G around the edge of the square z 0, 0 < x, y <. Assume anticlockwise direction. [RU 003] Solution: (i) So G is irrotational. G ˆ ˆ ˆ i j k x y z 6xy z 8x x (ii) Net flux of G over the cube. G dv so V. G. G dv 0ˆi ˆj 6x 6x kˆ 0 V x y z 6xy z 8x x 6y 0 0 6y 6y dx dydz 6 dx dz y dy 0 0 0
(54) Introduction to Vector Calculus (iii) G. dl 0 y 6.. 8 0 y 0 x 6xy z dx 8x dy x 0 z 0 y 0 z 0 y x 0 0 0 6xy z dx 8x dy x z 0 y z 0 0 x 0 8 y 6 0 0 8 8 0. SUMMARY A vector is with magnitude and direction. In space a quantity is specified by a function. When the result of multiplication of two vectors is a scalar then it is called scalar product or dot product. When product is a vector then it is called vector product. A. B C Multiplication of three vectors can give scalar or a vector A B C Vector differentiation is done using dal () operator the gradient of a scalar field is, divergence as. A and curl by A and laplacian by A. In Cartesion coordinate system dl dx ˆi dy ˆj dz k, ˆ dv dx dy dz. Gradient ˆi ˆj kˆ x y z Divergences. A A A x y A x y z z
Introduction to Vector Calculus (55) Curl A ˆ ˆ ˆ i j k x x z A A A x y z Laplacian In cylindrical system dl x y z d ˆ d ˆ dz z, ˆ dv d d dz gradient T T T ˆ T ˆ ẑ z divergence. A A A A z z Curl A ˆ ˆ ẑ z A A A z Laplacian In spherical system z dl dr rˆ rd ˆ r sin d ˆ dv r sin d dr d gradient T T ˆr T ˆ T ˆ r r r sin divergence A r r r sin rsin r Ar A sin
(56) Introduction to Vector Calculus Curl A r sin ˆr r ˆ rsin ˆ r A ra rsin A Laplacian r sin r r r sin Gauss Divergence theorem A. ds Stoke's theorem s sin L A. dl V. A dv V A. ds A vector field is solenoidal if Irrotational or conservative if Triple products A. B C B C A C A B A B C B A. C C A. B Second derivatives. A 0 f 0 A. A A EXERCISE
Introduction to Vector Calculus (57). Give the basic concepts of transformation of one coordinate system to another. Derive necessary relations for rectangular, cylindrical and spherical systems. [RU 003]. Write short note on "Physical significance of curl, divergence and gradient". 3. State-Gauss divergence theorem. Write its applications, advantages and limitations. [RU 00] 4. State and prove stoke's theorem. [RU 000] 5. Explain how stoke's theorem enables us to obtain the integral form of ampere circuital law. 6. Explain various types of vector fields. (i) (ii) (iii) Solenoidal and irrotational fields. Irrotational but not solenoidal fields Solenoidal but not irrotaitonal fields (iv) Neither irrotational nor solenoidal fields. 7. For the vectors A iˆ 3kˆ and B 5i ˆ j ˆ 6kˆ calculate (i) A B (ii) A B (iii) A. B (iv) A B (v) Angle between A and B (vi) A unit vector parallel to 3A B. (vii) Length of the projection of A on B. i 6i ˆ j ˆ 3k ii 4i ˆ j ˆ 9k iii 3 iv 6i ˆ j ˆ k Ans.: ˆ ˆ 8i ˆ j ˆ 3kˆ v 60 vi vii.6m 77 8. Use the differential volume dv to find volume of region. (i) 0 x, y, 3 z 3 [Ans.: 6] (ii) 5,, z 4 [Ans.: 0] 3 9. Find area of the region 0 on the spherical shell of radius 'b'. [Ans. b ] 0. Evaluate the gradient of the following scalar fields
(58) Introduction to Vector Calculus (a) z [Ans.: z z P e sin x P cos xe ˆi sin xe kˆ ] (b) q zcos ˆ Ans.: q zcos ˆ zsin cosz (c) s 0r sin cos. If f xy + yz + xz then (i) (ii) Ans.: r 0sin cos rˆ 0r cos cos ˆ 0r sin sin ˆ Find the magnitude and direction of the maximum rate of change of the function at point (,, 3) Find the rate of change of the function at the same point in the direction of the vector. and V xyz evaluate. VT. If ˆ ˆ ˆ T x i 3y j 4z k Ans.: i f 4 f i ˆ 3j ˆ kˆ ˆ f f 4 (ii) df f.dl. [Ans.: xyz] 3. If U xz x y y z find div (grad U). [Ans.: ( y + z + y )] 4. Given ˆ ˆ ˆ 3 D 6xyz i 3x z j 6x y k C/m Find the total charge lying within the region bounded by 0 < x <, < y < and z by separately evaluating each side of divergence theorem. A x y ˆ i ˆ j x y k ˆ 5. If A. 0 then prove that A x y z ˆ i y zx ˆ j z xy k ˆ 6. Prove that vector 7. If x y z find. 8. Show that is not solenoidal. [Ans.: 6C] [WBUT 005]. [WBUT 005] ˆ ˆ B xyz i x z y j x ykˆ is irrotational. [WBUT 005] 9. find a unit vector perpendicular to x y z 00 at (,, 3). [WBUT 007]
Introduction to Vector Calculus (59) 0. if 3 3x y y z find. Show that ˆi j ˆ 3kˆ Ans.: 4 at (,, ). [WBUT 004] Ans.: i ˆ 9j ˆ 6kˆ 3 ˆ ˆ F xy z i x j 3xz kˆ is a conservative force field. Find also the scalar 3 potential. [WBUT 006, 003] Ans. : x y z x constant. Evaluates F.n ˆ ds where ˆ F 8x zi y ˆ j yzk ˆ and s is the surface of the cube bounded s by x 0,, y 0,, z 0,.