Exempelsamling Vektoranalys

7879139. Integralen =Alltså:∮ ∮⎧⎪= (r1 2 − 2r 1 · r 2 + r2 2 )dr e1 · dr 2 = [1] =⎨ 1 = (xe x − ye y )/ √ x 2 + y 2C1 C2e∮ ∮2 = (ye x + xe y )/ √ x 2 + y 2⎪ ⎩= − 2(r 1 · r 2 )dr 1 · dr 2 =e 3 = e zC1 C2∮ ∮Uppenbart gäller e i · e j = δ ij .= −2 dr 1 · (r 1 · r 2 )dr 2 = [2] =b)C1 C2∮ ∫∫1= −2 dr 1 · ˆn 2 × ∇ 2 (r 1 · r 2 )dS 2 = [3] =h 1 =2 √ x 2 + y = 12 2(u 2 1 + 4u2 2 )1/4C1 S2∮ ∫∫1= −2 dr 1 · ˆn 2 × r 1 dS 2 =h 2 =(u 2 1 C1 S2∮ ( ∫∫ ) + 4u2 2 )1/4h 3 = 1= 2 dr 1 · r 1 × ˆn 2 dS 2 =Alltså:C1S2( ∮ ) ∫∫= 2 dr 1 × r 1 · ˆn 2 dS 2 = [4] =divA =√(C1S2∫∫ ∫∫= 2 u 2 1= −4 ˆn 1 dS 1 · ˆn 2 dS + ∂ A 14u2 22∂u 1 (u 2 1 + + ∂ A 24u2 2 )1/4 ∂u 2 2(u 2 1 + + 4u2 2))1/4 S1S2∮ ∮∮ ∮[1] : r1dr 2 1 · dr 2 = r1dr 2 1 · dr 2 = 0+ ∂ A 3∂u 3 2 √ u 2 C1 C2C1 C21∮+ 4u2 2[2] : Integralsatsen φdr = . . . har använts.C[3] : ∇ 2 opererar bara på r 2 .143. a)⎧ √ √x2[4] : Resultatet av ex. 73 har använts.u = + y ⎪⎨2 + z 2 + z 0 ≤ u < ∞√ √x2v = + y 2 + z 2 − z 0 ≤ v < ∞140. e 1 = (1, 1, 2)/h 1 ,e 2 = (1, −3, 1)/h 2 ,e 3 = (7, 1, −4)/h 3 , h 1 = √ 6, h 2 = √ ⎪⎩√ 11, h 3 =tanϕ = y/x0 ≤ ϕ < 2π66b)141. a) a 3 bc4π/15z = 1 (u 2 0 − x2 + y 2 )2 u 2 , rotationsparaboloider0b) 594/5z = 1 ( x 2 + y 2 )2 v 2 − v02 , rotationsparaboloider0y = xtanϕ 0 , halvplan genom z-axeln142. a)⎧d)u ⎪⎨ 1 = x 2 − y 2(1 ∂φgradφ = √u 2 = xyu2 + v 2 ∂u e u + ∂φ )∂v e v + 1 ∂φuv ∂ϕ e ϕ⎪⎩ u 3 = ze)ger1e u = √ (v cosϕ, v sinϕ, u)u2 + v2 ⎧⎪ e ⎨ 1 /h 1 = ∇u 1 = 2(xe x − ye y )1ee 2 /h 2 = ∇u 2 = ye x + xe v = √ (u cosϕ, u sinϕ, −v)yu2 + v2 ⎪ ⎩ e 3 /h 3 = ∇u 3 = e ze ϕ = (− sinϕ, cosϕ, 0)