Problem 3.6 Given B = ˆxz 3y) +ŷx 3z) ẑx+y), find a unit vector parallel to B at point P =,0, ). Solution: At P =,0, ), B = ˆx )+ŷ+3) ẑ) = ˆx+ŷ5 ẑ, ˆb = B B = ˆx+ŷ5 ẑ = ˆx+ŷ5 ẑ. +5+ 7
Problem 3.4 Convert the coordinates of the following points from spherical to cylindrical coordinates: a) P = 5,0,0), b) P = 5,0,π), c) P 3 = 3,π/,0). Solution: a) P = r,φ,z) = Rsinθ,φ,Rcosθ) = 5sin0,0,5cos0) b) P = r,φ,z) = 5sin0,π,5cos0) = 0,π,5). c) P 3 = r,φ,z) = 3sin π,0,3cos π ) = 3,0,0). = 0,0,5).
Problem 3.7 A section of a sphere is described by 0 R, 0 θ 90, and 30 φ 90. Find: a) the surface area of the spherical section, b) the enclosed volume. Also sketch the outline of the section. Solution: z y x φ=30 o Figure P3.7: Outline of section. π/ S = φ=π/6 π/ θ=0 R sinθ dθ dφ R= π = 4 π )[ ] cosθ π/ 0 = 4 π 6 3 = 4π 3 π/ π/ v = R sinθ dr dθ dφ = R3 3 R=0 φ=π/6 π 0 π 6 θ=0 ) [ cosθ π/ 0 ] = 8 π 3 3 = 8π 9 m ), m 3 ).
Problem 3.30 Given vectors A = ˆrcosφ + 3z) ˆφr+ 4sinφ)+ẑr z), B = ˆrsinφ + ẑcosφ, find a) θ AB at,π/,0), b) a unit vector perpendicular to both A and B at,π/3,). 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. a) At,π/,0), A = ˆφ8+ẑ and B = ˆr. From Eq. 3.8), θ AB = cos A B AB ) = cos 0 AB ) = 90. b) At,π/3,), A = ˆr 7 ˆφ4+ 3) and B = ˆr 3+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+ 3)) ) ˆφ 7 ) ) ẑ4+ 3)) 3) + 3)) + 7 4 ) +3+ 3) ˆr0.487+ ˆφ0.8+ẑ0.843).
Problem 3.34 Transform the following vectors into cylindrical coordinates and then evaluate them at the indicated points: a) A = ˆxx+y) at P =,,3), b) B = ˆxy x)+ŷx y) at P =,0,), c) C = ˆxy /x + y ) ŷx /x + y )+ẑ4 at P 3 =,,), d) D = ˆRsinθ + ˆθcosθ + ˆφcos φ at P 4,π/,π/4), e) E = ˆRcosφ + ˆθsinφ + ˆφsin θ at P 5 = 3,π/,π). Solution: From Table 3-: a) b) A = ˆrcosφ ˆφsinφ)r cosφ + r sinφ) = ˆrr cosφcosφ + sinφ) ˆφr sinφcosφ + sinφ), P = +,tan /),3) = 5,63.4,3), AP ) = ˆr0.447 ˆφ0.894) 5.447+.894) = ˆr.34 ˆφ.68. B = ˆrcosφ ˆφsinφ)r sinφ r cosφ)+ˆφcosφ + ˆrsinφ)r cosφ r sinφ) = ˆrrsinφ cosφ )+ ˆφrcos φ sin φ) = ˆrrsinφ )+ ˆφr cosφ, P = + 0,tan 0/),) =,0,), BP ) = ˆr+ ˆφ. c) d) C = ˆrcosφ ˆφsinφ) r sin φ ˆφcosφ + ˆrsinφ) r cos φ + ẑ4 r = ˆrsinφ cosφsinφ cosφ) ˆφsin 3 φ + cos 3 φ)+ẑ4, P 3 = + ),tan /),) =, 45,), CP 3 ) = ˆr0.707+ẑ4. D = ˆrsinθ + ẑcosθ)sinθ +ˆrcosθ ẑsinθ)cosθ + ˆφcos φ = ˆr+ ˆφcos φ, P 4 = sinπ/),π/4,cosπ/)) =,45,0), DP 4 ) = ˆr+ ˆφ. r
e) E = ˆrsinθ + ẑcosθ)cosφ +ˆrcosθ ẑsinθ)sinφ + ˆφsin θ, P 5 = 3, π ),π, EP 5 ) = ˆrsin π + ẑcos π ) cosπ + ˆrcos π ẑsin π ) sinπ + ˆφsin π = ˆr+ ˆφ.
Problem 3.36 Find the gradient of the following scalar functions: a) T = 3/x + z ), b) V = xy z 4, c) U = zcosφ/+r ), d) W = e R sinθ, e) S = 4x e z + y 3, f) N = r cos φ, g) M = Rcosθ sinφ. Solution: a) From Eq. 3.7), b) From Eq. 3.7), c) From Eq. 3.8), d) From Eq. 3.83), e) From Eq. 3.7), f) From Eq. 3.8), 6x T = ˆx x + z ) ẑ 6z x + z ). V = ˆxy z 4 + ŷxyz 4 + ẑ4xy z 3. U = ˆr rzcosφ zsinφ ˆφ +r ) r+r ) + ẑ cosφ +r. W = ˆRe R sinθ + ˆθe R /R)cosθ. S = 4x e z + y 3, S = ˆx S x + ŷ S y + ẑ S z = ˆx8xe z + ŷ3y ẑ4x e z. N = r cos φ, N = ˆr N r + ˆφ N r φ + ẑ N z = ˆrr cos φ ˆφr sinφ cosφ. g) From Eq. 3.83), M = Rcosθ sinφ, M M = ˆR R + ˆθ M R θ + ˆφ Rsinθ M φ = ˆRcosθ sinφ ˆθsinθ sinφ + ˆφ cosφ tanθ.
Problem 3.40 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). Solution: The directional derivative is given by Eq. 3.75) as dv/dl = V â l, where the unit vector in the direction of A is given by Eq. 3.): â l = ˆx ŷz +z, and the gradient of V in Cartesian coordinates is given by Eq. 3.7): V = ˆxy + ŷxy ẑz. Therefore, by Eq. 3.75), At P =,,4), dv dv dl dl ),,4) = y xyz +z. = 9 7 =.8.
Problem 3.47 For the vector field E = ˆr0e r ẑ3z, verify the divergence theorem for the cylindrical region enclosed by r =, z = 0, and z = 4. Solution: E ds = + + = 0+ π r=0 φ=0 π 4 φ=0 z=0 π r=0 φ=0 π 4 φ=0 ˆr0e r ẑ3z) ẑr dr dφ) ) z=0 z=0 ˆr0e r ẑ3z) ˆrr dφ dz) ) r= ˆr0e r ẑ3z) ẑr dr dφ) ) z=4 π 0e dφ dz+ r dr dφ r=0 φ=0 = 60πe 48π 8.77, 4 π 0e r ) r) E dv = 3 r dφ dr dz z=0 r=0 φ=0 r = 8π 0e r r) 3r) dr r=0 ) = 8π 0e r + 0e r +r) 3r r=0 = 60πe 48π 8.77.
Problem 3.5 Verify Stokes s theorem for the vector field B = ˆrr cosφ + ˆφsinφ) by evaluating: a) B dl over the semicircular contour shown in Fig. P3.5a), and C b) B) ds over the surface of the semicircle. S y y L - L 3 0 L a) x 0 L3 L L 4 L b) x Figure P3.5: Contour paths for a) Problem 3.5 and b) Problem 3.53. Solution: a) B dl = B dl+ B dl+ B dl, L L L 3 B dl = ˆrr cosφ + ˆφsinφ) ˆr dr+ ˆφr dφ + ẑ dz) = r cosφ dr+ r sinφ dφ, 0 B dl = r cosφ dr) + r sinφ dφ) L r=0 φ=0, z=0 φ=0 z=0 = r) r=0 + 0 =, π B dl = r cosφ dr) + r sinφ dφ) L r= z=0 φ=0 r=, z=0 = 0+ cosφ) π φ=0 = 4, 0 π B dl = r cosφ dr) + r sinφ dφ) L 3 r= φ=π,z=0 φ=π z=0 = r) 0 r= + 0 =, B dl = +4+ = 8.
b) B = ˆrr cosφ + ˆφsinφ) = ˆr r φ 0 ) z sinφ) π B ds = + ˆφ z r cosφ) ) r 0 ) + ẑ r r rsinφ)) r cosφ) φ = ˆr0+ ˆφ0+ẑ r sinφ +r sinφ)) = ẑsinφ = φ=0 r=0 π φ=0 r=0 ẑsinφ + )) ẑr dr dφ) r sinφr+ ) dr dφ = + r ), cosφ r + r) ) r=0 ) π φ=0 = 8.
Problem 3.58 a) V = 0r 3 sinφ b) V = /R )cosθ sinφ Solution: a) Find the Laplacian of the following scalar functions: V = r r V r r ) + V r φ + V z )+ r r r 0r3 sinφ) = r r φ 0r3 sinφ)+0 = r r 30r3 sinφ) r 0r3 )4sinφ = 90r sinφ 40r sinφ = 50r sinφ. b) V = R R V ) + R R R sinθ V ) + sinθ θ θ = R R )) cosθ sinφ R R R + R sinθ )) cosθ sinφ sinθ θ θ R ) + R sin θ φ cosθ sinφ R = 4 R 4 cosθ sinφ 4 R 4 cosθ sinφ cosθ R 4 sin θ sinφ = cosθ sinφ R 4 sin. θ R sin θ V φ