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Unformatted text preview: φ v = parenleftbigg cos u bracketleftbigg cos u 2 cos v 2 cos 2 v sin u 2 bracketrightbigg , sin u bracketleftbigg cos u 2 cos v 2 cos 2 v sin u 2 bracketrightbigg , 2 cos u 2 cos 2 v + cos v sin u 2 parenrightbigg , so φ v ( u, 0) = parenleftbigg cos u bracketleftbigg cos u 2 2 sin u 2 bracketrightbigg , sin u bracketleftbigg cos u 2 2 sin u 2 bracketrightbigg , 2 cos u 2 + sin u 2 parenrightbigg . Now φ u × φ v ( u, 0) = parenleftbigg 3 cos u bracketleftbigg 2 cos u 2 +sin u 2 bracketrightbigg , 3 sin u bracketleftbigg 2 cos u 2 +sin u 2 bracketrightbigg , 6 sin u 2 3 cos u 2 parenrightbigg . Hence we see that φ u × φ v (0 , 0) = ( 6 , , 3 ) while φ u × φ v (2 π, 0) = ( 6 , , 3 ) . After you have traveled once around the u –line at v = 0 you discover that the normal vector is now pointing in the opposite direction. Hence you can conclude that this surface can not be oriented. (This surface is a parametrization of a Klein surface.) 3. (a) We parametrize S by Φ ( u, v ) = ( u, v, 1 u v ), ( u, v ) ∈ projection into the uv –plane; i.e., 0 ≤ v ≤ 1 u , 0 ≤ u ≤ 1. Now φ u × φ v = (1 , 1 , 1) and since the z –component is positive, Φ ( u, v ) is orientation preserving. Hence integraldisplay S F · d S = integraldisplay projection F ( Φ ( u, v )) · φ u × φ v dA = integraldisplay 1 integraldisplay 1 u ( u, v, 1 u v ) · (1 , 1 , 1) dv du = integraldisplay 1 integraldisplay 1 u ( u + v +1 u v ) dv du = integraldisplay 1 integraldisplay 1 u dv du = integraldisplay 1 (1 u ) du = bracketleftbigg u u 2 2 bracketrightbigg 1 = 1 1 2 = 1 2 . MATB42H Solutions # 8 page 3 (b) We parametrize S by Φ ( u, v ) = ( u, v, 4 u 2 v 2 ), 0 ≤ u ≤ 1, 0 ≤ v ≤ 1. Now φ u × φ v = (2 u, 2 v, 1) and Φ ( u, v ) is orientation preservation since the z – component is positive; i.e., pointing outward. Hence integraldisplay S F · d S = integraldisplay 1 integraldisplay 1 F ( Φ ( u, v )) · φ u × φ v dA = integraldisplay 1 integraldisplay 1 ( u, v, 4 u 2 v 2 ) · (2 u, 2 v, 1) du dv = integraldisplay 1 integraldisplay 1 (2 u 2 +2 v 2 +4 u 2 v 2 ) du dv = integraldisplay 1 integraldisplay 1 (4 + u 2 + v 2 ) du dv = integraldisplay 1 parenleftbigg 4 + v 2 + 1 3 parenrightbigg dv = bracketleftbigg 4 v + v 3 3 + v 3 bracketrightbigg 1 = 4 + 1 3 + 1 3 = 14 3 ....
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 Winter '10
 EricMoore
 Multivariable Calculus, Cone

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