Exempelsamling Vektoranalys
Exempelsamling Vektoranalys Exempelsamling Vektoranalys
101e = e x ,e y ,e z⇒∫∫S(A × ˆn)dS = 0197. 197(rotB) 2 = (rotA) 2 + (rotD) 2 + 2 rotA · rotD (1)div(D × rotA) = rotD · rotA − DrotrotA (2)rotrotA = graddivA − ∇ 2 A = −∇ 2 A = 0 (3)D × ˆn = 0 ty A × ˆn = B × ˆn = C (4)∫∫∫=====V∫∫∫(rotB) 2 dV − (rotA) 2 dV = {(1)} =V∫∫∫((rotD) 2 + 2 rotA · rotD)dV = {(2)} =V∫∫∫((rotD) 2 + 2 div(D × rotA) + 2D · rotrotA)dV = {(3)} =V∫∫∫∫∫(rotD) 2 dV + 2 ○ (D × rotA) · ˆn dS =VS∫∫∫∫∫(rotD) 2 dV + 2 ○ (ˆn × D) · rotAdS = {(4)} =VS∫∫∫(rotD) 2 dV ≥ 0V198. 198rotA = αA ⇒ α 2 A = rotrotADessutom ärrotrotA = graddivA − ∇ 2 Avilket medför att( ) 1∇ 2 A + rotrotA = (∇ 2 + α 2 )A = graddivA = gradα div rotA = 0199. 199a)∇ 2 ∇ 2 u = 1 (dρ d ( ( 1 dρ du )))= 0ρ dρ dρ ρ dρ dρ∇ 2 u = 1 (dρ du )= a lnρ + bρ dρ dρu= Aρ 2 lnρ + Bρ 2 + C lnρ + D
102b)∇ 2 ∇ 2 u =∇ 2 u =u =1 (dr 2 dr1r 2 ddrr 2 d ( 1 ddr r 2 dr(r 2 du )= a dr r + bAr 2 + Br + C r + D(r 2 du )))= 0dr200. 200 Sätt B = rotC.∫∫∫A · rotC dV =V∫∫∫= (div(C × A) + C · rotA} {{ })dV = {enl. Gauss’ sats} =V=0∫∫∫∫= ○ (C × A) · ˆn dS = ○ (ˆn × C) · A dS = 0SS201. 201a) Man får∇ · ∇φ = 1 rd 2dr 2 rφ = 1 d 2r dr 2 e−λr = λ2r e−λrför r ≠ 0. För r = 0 är ∇ 2 φ ej definierad. Utesluts origo genom att betraktavolymen mellan r = R och r = ε ger Gauss’ sats:∫∫ ∫∫∫∫∫ 1○ ∇φ · dS + ∇φ · dS = λ 2S εr e−λr dVVi får alltsåI(R) =Sλ 2 4π−∫ Rε∫ 2π ∫ π00e −λr r dr −(e −λε − 1 ε 2 − λ )e r · (−e r ε 2 )sin θ dθ dϕ =ε= 4π(e −λε (1 + ελ) − e −λR (1 + Rλ) − e −λε (1 + λε))= −4πe −λR (1 + Rλ)b)( 1∇φ = −r 2 + λ )e −λr e rrdS =e r R 2 sin θ dθ dϕgerI(R) = −(1 + Rλ)e −λR 4π
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- Page 57 and 58: 5661. 61 Eftersom rotA = (1, 1, 1),
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- Page 61 and 62: 6072. 7212 ○ ∫∫SdS × (a × r
- Page 63 and 64: 6283. 83a) På C ärr = a(cosϕ, si
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- Page 67 and 68: 66(4, 5) → (3) ger slutligen:dvs.
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- Page 81 and 82: 80139. 139 Integralen ==∮C 1∮
- Page 83 and 84: 82d)e)f)gradφ =(1 ∂φ√u2 + v 2
- Page 85 and 86: a) Ur ⎧⎪ ⎨⎪ ⎩ξ 2 = ρ
- Page 87 and 88: 86148. 148r = (r cosϕsin θ, r sin
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101e = e x ,e y ,e z⇒∫∫S(A × ˆn)dS = 0197. 197(rotB) 2 = (rotA) 2 + (rotD) 2 + 2 rotA · rotD (1)div(D × rotA) = rotD · rotA − DrotrotA (2)rotrotA = graddivA − ∇ 2 A = −∇ 2 A = 0 (3)D × ˆn = 0 ty A × ˆn = B × ˆn = C (4)∫∫∫=====V∫∫∫(rotB) 2 dV − (rotA) 2 dV = {(1)} =V∫∫∫((rotD) 2 + 2 rotA · rotD)dV = {(2)} =V∫∫∫((rotD) 2 + 2 div(D × rotA) + 2D · rotrotA)dV = {(3)} =V∫∫∫∫∫(rotD) 2 dV + 2 ○ (D × rotA) · ˆn dS =VS∫∫∫∫∫(rotD) 2 dV + 2 ○ (ˆn × D) · rotAdS = {(4)} =VS∫∫∫(rotD) 2 dV ≥ 0V198. 198rotA = αA ⇒ α 2 A = rotrotADessutom ärrotrotA = graddivA − ∇ 2 Avilket medför att( ) 1∇ 2 A + rotrotA = (∇ 2 + α 2 )A = graddivA = gradα div rotA = 0199. 199a)∇ 2 ∇ 2 u = 1 (dρ d ( ( 1 dρ du )))= 0ρ dρ dρ ρ dρ dρ∇ 2 u = 1 (dρ du )= a lnρ + bρ dρ dρu= Aρ 2 lnρ + Bρ 2 + C lnρ + D
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