# 20 20 40 x 25 25 50 y 30 30 60 z γ 20 20 x 25 25 50

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Unformatted text preview: -20 20 40 x-25 25 50 y-30 30 60 z Γ-20 20 x-25 25 50 y-20 20 40 x-25 25 50 y-30 30 60 z S Γ Greater-20 20 x-25 25 50 y No orientation is given so we assume counterclockwise as viewed from above; this means that S must be oriented by the upward pointing unit normal, e 3 = (0 , , 1). Since ω is defined throughout R 3 , we can apply Stokes’ Theorem giving integraldisplay γ ω = integraldisplay S dω . From the geometry, dy dz and dz dx are zero, so we are only concerned with the “ dxdy ” component of dω ; hence, dω = * dy dz + * dz dx + ( e z- e z ) dxdy . Therefore, integraldisplay γ ω = integraldisplay S dxdy = 0. 9. (a) We know that G will be conservative; i.e., G = ∇ g , for some g : R 3 → R , if integraldisplay σ G · d s = 0 for all closed curves σ . If every closed loop γ bounds an orientable surface S we can apply Stokes’ Theorem ( since F is defined throughout all of R 3 ). Therefore, integraldisplay γ F · d s = integraldisplay S (curl F ) · d S = integraldisplay S · d S = 0 ( we are given that curl F = ). We now conclude that F is conservative. (b) curl G = parenleftbigg ∂G 3 ∂y- ∂G 2 ∂z , ∂G 1 ∂z- ∂G 3 ∂x , ∂G 2 ∂x- ∂G 1 ∂y parenrightbigg = parenleftbigg- ∂ ∂z parenleftbigg x x 2 + y 2 parenrightbigg , ∂ ∂z parenleftbigg- y x 2 + y 2 parenrightbigg- , ∂ ∂x parenleftbigg x x 2 + y 2 parenrightbigg- ∂ ∂y parenleftbigg- y x 2 + y 2 parenrightbiggparenrightbigg = parenleftbigg , , y 2- x 2 ( x 2 + y 2 ) 2- y 2- x 2 ( x 2 + y 2 ) 2 parenrightbigg = (0 , , 0) = . For G to be a gradient field we must have integraldisplay γ G · d s = 0 for all closed loops; in particular, if γ is a circle around the origin. Let γ ( t ) = (cos t, sin t, 0), 0 ≤ t ≤ 2 π , then integraldisplay γ G · d s = MATB42H Solutions # 11 page 5 integraldisplay 2 π parenleftbigg- sin t 1 , cos t 1 , parenrightbigg · (- sin t, cos t, 0) dt = integraldisplay 2 π (sin 2 t + cos 2 t ) dt = integraldisplay 2 π dt = 2 π negationslash = 0. Hence we conclude that G can not be a gradient field. 10. (a) We use Green’s Theorem: 1 2 integraldisplay ∂R 2 xdy- y dx = 1 2 integraldisplay R 2 parenleftbigg ∂x ∂x- ∂ (- y ) ∂y parenrightbigg dA = 1 2 integraldisplay R 2 2 dA = integraldisplay R 2 dA = area of R 2 ....
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• Winter '10
• EricMoore

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