# We now find g f 1 3 x 2 z 2 2 xy 3 g x so g x y z

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We now find g . F 1 = 3 x 2 z 2 + 2 xy 3 = ∂g ∂x , so g ( x, y, z ) = integraldisplay ( 3 x 2 z 2 + 2 xy 3 ) dx + h ( y, z ) = x 3 z 2 + x 2 y 3 + h ( y, z ). F 2 = 3 x 2 y 2 + z cos y = ∂g ∂y = 3 x 2 y 2 + ∂h ∂y , so ∂h ∂y = z cos y = h ( y, z ) = integraldisplay z cos y dy + k ( z ) = z sin y + k ( z ) giving g ( x, y, z ) = x 3 z 2 + x 2 y 3 + z sin y + k ( z ). F 3 = 2 x 3 z + sin y + e z = ∂g ∂z = 2 x 3 z + sin y + dk dz , so dk dz = e z = k ( z ) = integraldisplay e z dz = e z + C . Hence, g ( x, y, z ) = x 3 z 2 + x 2 y 3 + z sin y + e z + C , where C is some constant. 4. We note that, if we rotate both F and n by π 2 radians in the counterclockwise direction we will not change the value of the dot product F · n . Rotating n by π 2 radians in the coun- terclockwise direction gives T = γ prime ( t ) bardbl γ prime ( t ) bardbl , the unit tangent vector. Rotating F by π 2 radi- ans in the counterclockwise direction gives the vec- tor field G = ( - F 2 , F 1 ). Now integraldisplay γ F · n ds = integraldisplay γ G · T ds Green prime s = T heorem integraldisplayintegraldisplay R parenleftbigg ∂F 1 ∂x + ∂F 2 ∂y parenrightbigg dA . n T R Γ 5. (a) We parametrize by Φ ( r, θ ) = ( r, cos θ, sin θ ) , 0 θ 2 π , - 1 r 3. Now φ r = (1 , 0 , 0) and φ θ = ( 0 , - sin θ, cos θ ) so φ r × φ θ = ( 0 , - cos θ, - sin θ ) . A unit nor- mal is φ r × φ θ bardbl φ r × φ θ bardbl = 1 1 ( 0 , - cos θ, - sin θ ) = ( 0 , - cos θ, - sin θ ) . -1 1 3 x 0 1 y 0 1 z

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MATB42H Solutions # 6 page 3 (b) We want the piece of the plane z = x + y + 5 over the unit disk so we parametrize by Φ ( r, θ ) = ( r cos θ, r sin θ, r cos θ + r sin θ + 5 ) , 0 θ 2 π , 0 r 1. Then φ r = ( cos θ, sin θ, cos θ + sin θ ) and φ θ = ( - r sin θ, r cos θ, - r sin θ + r cos θ ) . Hence φ r × φ θ = ( - r sin 2 θ + r sin θ cos θ - r cos 2 θ - r sin θ cos θ, - r sin θ cos θ - r sin 2 θ + r sin θ cos θ - r cos 2 θ, r cos 2 θ + r sin 2 θ ) = ( - r (sin 2 θ + cos 2 θ ) , - r (sin 2 θ + cos 2 θ ) , r (sin 2 θ + cos 2 θ ) ) = ( - r, - r, r ) = 0 1 x 0 1 y 4 5 6 z r ( - 1 , - 1 , 1 ) . A unit normal is φ r × φ θ bardbl φ r × φ θ bardbl = 1 3 ( - 1 , - 1 , 1 ) . (c) The sphere meets the plane z = 1 when cos ϕ = 1 2 or ϕ = π 3 (using spherical polars). Hence we can parametrize the part of the sphere above z = 1 by Φ ( u, v ) = ( 2 cos u sin v, 2 sin u sin v, 2 cos v ) , 0 u 2 π , 0 v π 3 . We know from class that φ u × φ v = ( - 2 sin v ) Φ ( u, v ) and bardbl φ u × φ v bardbl = 4 | sin v | = 4 sin v , since v bracketleftbigg 0 , π 3 bracketrightbigg . A unit normal could be φ u × φ v bardbl φ u × φ v bardbl = ( - 2 sin v ) Φ ( u, v ) 4 sin v = - 1 2 Φ ( u, v ) = ( - 1) ( cos u sin v, sin u sin v, cos v ) .
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