We parametrize s by φ r θ r sin θ r 2 r cos θ 0

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Unformatted text preview: We parametrize S by Φ ( r, θ ) = ( r sin θ, r 2 , r cos θ ), 0 ≤ θ ≤ 2 π , 0 ≤ r ≤ 1. The order “ r, θ ” is correct since the parametrization of γ requires the normal for S to have positive y –component for compatibility. Now integraldisplay S dω = integraldisplay 2 π integraldisplay 1 5 vextendsingle vextendsingle vextendsingle vextendsingle cos θ- r sin θ sin θ r cos θ vextendsingle vextendsingle vextendsingle vextendsingle + 3 vextendsingle vextendsingle vextendsingle vextendsingle sin θ r cos θ 2 r vextendsingle vextendsingle vextendsingle vextendsingle + vextendsingle vextendsingle vextendsingle vextendsingle 2 r cos θ- r sin θ vextendsingle vextendsingle vextendsingle vextendsingle dr dθ = integraldisplay 2 π bracketleftbigg 5 2 r 2- 2 r 3 cos θ- 2 3 r 3 sin θ bracketrightbigg 1 dθ = integraldisplay 2 π 5 2- a24 a24 a24 a24 a58 =0 2 cos θ- a26 a26 a26 a26 a62 = 0 2 3 sin θ dθ = 5 2 θ vextendsingle vextendsingle vextendsingle vextendsingle 2 π = 5 π . (b) S prime is another surface with γ as boundary, so we again use Stokes’ Theorem giv- ing integraldisplay γ = ∂S prime ω = integraldisplay S prime dω . For compatibility the normal will again require posi- tive y –component. Hence we parametrize by Φ prime ( r, θ ) = ( r sin θ, 1 , r cos θ ), 0 ≤ MATB42H Solutions # 10 page 6 θ ≤ 2 π , 0 ≤ r ≤ 1. This time we have integraldisplay S prime dθ = integraldisplay 2 π integraldisplay 1 5 vextendsingle vextendsingle vextendsingle vextendsingle cos θ- r sin θ sin θ r cos θ vextendsingle vextendsingle vextendsingle vextendsingle + 3 vextendsingle vextendsingle vextendsingle vextendsingle sin θ r cos θ vextendsingle vextendsingle vextendsingle vextendsingle + vextendsingle vextendsingle vextendsingle vextendsingle cos θ- r sin θ vextendsingle vextendsingle vextendsingle vextendsingle dr dθ = integraldisplay 2 π integraldisplay 1 (5 r (cos 2 θ + sin 2 θ ) + 0 + 0) dr dθ = integraldisplay 2 π 5 2 r 2 vextendsingle vextendsingle vextendsingle vextendsingle 1 dθ = 5 2 integraldisplay 2 π dθ = 5 π . LParen1 b RParen1-1 1 1-1 1 x y z S’-1 LParen1 c RParen1-1 1 1-1 1 x y z S Intersection S' 1 (c) Together S and S prime enclose a region R . The special case of Stokes’ Theorem with n = 3 , k = 3 (Divergence Theorem) gives integraldisplay S ∪ S prime dω = integraldisplay R d ( dω ) = 0, since d 2 ω = 0. Here S ∪ S prime would be oriented with the outward unit normal. This agrees with the normal in part (b) but is opposite to the normal in (a). When the normal is corrected we have 0 = integraldisplay S ∪ S prime dω =- integraldisplay S dω + integraldisplay S prime dω or integraldisplay S dω = integraldisplay S prime dω ....
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