Exempelsamling Vektoranalys

ger ⎧⎪ ⎨⎪ ⎩e 1 /h 1 = ∇u 1 = 2(xe x − ye y )81e 2 /h 2 = ∇u 2 = ye x + xe ye 3 /h 3 = ∇u 3 = e zAlltså: ⎧⎪ ⎨⎪ ⎩e 1 = (xe x − ye y )/ √ x 2 + y 2e 2 = (ye x + xe y )/ √ x 2 + y 2e 3 = e zb)Uppenbart gäller e i · e j = δ ij .h 1 =12 √ x 2 + y = 12 2(u 2 1 + 4u2 2 )1/4h 2 =1(u 2 1 + 4u2 2 )1/4h 3 = 1Alltså:divA =√(= 2 u 2 1 + ∂ A 14u2 2∂u 1 (u 2 1 + + ∂ A 24u2 2 )1/4 ∂u 2 2(u 2 1 + + 4u2 2))1/4+ ∂ A 3∂u 3 2 √ u 2 1 + 4u2 2143. 143a)⎧⎪⎨⎪⎩√ √x2u = + y 2 + z 2 + z 0 ≤ u < ∞√ √x2v = + y 2 + z 2 − z 0 ≤ v < ∞tan ϕ = y/x0 ≤ ϕ < 2πb)(u 2 0 − x2 + y 2), rotationsparaboloiderz = 1 2 u 2 0z = 1 ( x 2 + y 22 v02 − v02y = xtan ϕ 0 , halvplan genom z-axeln), rotationsparaboloider