MATHEMATIC
s10

# 1 2 x 1 2 y 4 2 z n t n 1 2 x 4 2 1 2 x 1 2 y 4 2 z n

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Unformatted text preview: 1 2 x 1 2 y-4-2 z N T n 1 2 x-4 2 1 2 x 1 2 y-4-2 z N T n-4 2 We parametrize S by Φ ( θ, r ) = ( r cos θ, r sin θ, r 2- 1), 0 ≤ θ ≤ 2 π , 0 ≤ r ≤ 2. Now dxdy = ∂ ( x, y ) ∂ ( θ, r ) = vextendsingle vextendsingle vextendsingle vextendsingle- r sin θ cos θ r cos θ sin θ vextendsingle vextendsingle vextendsingle vextendsingle =- r < 0, so Φ ( θ, r ) is orientation pre- serving. Now integraldisplay S dω = integraldisplay Φ 5 dy dz + 2 dz dx- dxdy = integraldisplay 2 integraldisplay 2 π parenleftbigg 5 vextendsingle vextendsingle vextendsingle vextendsingle r cos θ sin θ 2 r vextendsingle vextendsingle vextendsingle vextendsingle + 2 vextendsingle vextendsingle vextendsingle vextendsingle 2 r- r sin θ cos θ vextendsingle vextendsingle vextendsingle vextendsingle + r parenrightbigg dθ dr = integraldisplay 2 integraldisplay 2 π (5 (2 r 2 cos θ ) + 2 (2 r 2 sin θ ) + r ) dθ dr = integraldisplay 2 integraldisplay 2 π parenleftbigg a24 a24 a24 a24 a24 a24 a58 =0 10 r 2 cos θ + a24 a24 a24 a24 a24 a58 = 0 4 r 2 sin θ + r parenrightbigg dθ dr = parenleftbigg 2 π parenrightbiggbracketleftbigg 1 2 r 2 bracketrightbigg 2 = 4 π . Inorder to have compatible orientations, ∂S must be parametrized in the clockwise di- rection when viewed from above. We can do this with γ ( t ) = (2 cos t,- 2 sin t, 0), ≤ t ≤ 2 π . Now integraldisplay ∂S ω = integraldisplay γ (2 y- z ) dx + ( x + y 2- z ) dy + (4 y- 3 x ) dz = MATB42H Solutions # 10 page 5 integraldisplay 2 π (- 4 sin t ) (- 2 sin t ) + (2 cos t + 4 sin 2 t ) (- 2 cos t ) + (- 8 sin t ) (- 6 cos t ) (0) dt = integraldisplay 2 π 8 sin 2 t- 4 cos 2 t- 8 sin 2 t cos tdt = integraldisplay 2 π 8- 6 (1 + cos 2 t )- 8 sin 2 t cos tdt = integraldisplay 2 π 2- a24 a24 a24 a24 a24 a58 =0 6 cos 2 t- a24 a24 a24 a24 a24 a24 a24 a58 = 0 8 sin 2 t cos t dt = 4 π . Hence integraldisplay S dω = integraldisplay ∂S ω , as required. 5. ω = 5 z dx + 3 xdy + y dz and dω = 5 dz dx + 3 dxdy + dy dz . LParen1 a RParen1-1 1 1-1 1 x y z Less Γ LParen1 t RParen1 1-1 1 1-1 1 x y z S-1 1 (a) Here Stokes’ Theorem gives integraldisplay γ = ∂S ω = integraldisplay S dω = integraldisplay S 5 dz dx + 3 dxdy + dy dz . We are given γ ( t ) = (sin t, 1 , cos t ), 0 ≤ t ≤ 2 π , so integraldisplay γ ω = integraldisplay 2 π 5 (cos t ) (cos t ) + 3 (sin t ) (0) + (1) (- sin t ) dt = integraldisplay 2 π 5 cos 2 t- a24 a24 a24 a58 = 0 sin t dt = parenleftbigg 5 2 parenrightbigg (2 π ) = 5 π ....
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• Winter '10
• EricMoore

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