5 a recall from class lecture 14 that bardbl a b

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Unformatted text preview: 5. (a) Recall from class (lecture 14) that bardbl a × b bardbl 2 = bardbl a bardbl 2 bardbl b bardbl 2 − ( a · b ) 2 , so bardbl φ u × φ v bardbl = radicalbig bardbl φ u bardbl 2 bardbl φ v bardbl 2 − ( φ u · φ v ) 2 = √ E G − F 2 . Hence the area of S is A ( S ) = integraldisplay Φ dS = integraldisplay D bardbl φ u × φ v bardbl dA = integraldisplay D √ E G − F 2 dA . (b) If φ u and φ v are orthogonal we have φ u · φ v = 0. Hence A ( S ) = integraldisplay Φ dS = integraldisplay D bardbl φ u × φ v bardbl dA = integraldisplay D radicalbig bardbl φ u bardbl 2 bardbl φ v bardbl 2 − dA = integraldisplay D bardbl φ u bardblbardbl φ v bardbl dA . (c) We can parametrize the sphere of radius a by Φ ( u,v ) = ( a cos u sin v, a sin u sin v, a cos v ), ≤ u ≤ 2 π , 0 ≤ v ≤ π . The tangents are φ u = ( − a sin u sin v, a cos u sin v, 0) and φ v = ( a cos u cos v, a sin u cos v, a sin v ). Since φ u · φ v = − a 2 sin u cos u sin v cos v + a 2 a 2 cos u sin u sin v cos v = 0, the area is A ( S ) = integraldisplay Φ dS part (c) = integraldisplay D bardbl φ u bardblbardbl φ v bardbl dA = integraldisplay 2 π integraldisplay π ( a sin v ) ( a ) dv du = 2 π a 2 integraldisplay π sin v dv = 2 π a 2 bracketleftbigg − cos v bracketrightbigg π = 4 π a 2 . 6. (a) We parametrize by Φ ( s,t ) = ( 2 cos t sin s, 2 sin t sin s, 2 cos s ) . To de- termine the domain we note that z = 2 cos s = 1 = ⇒ cos s = 1 2 = ⇒ s = π 3 . Hence D = braceleftbigg ( s,t ) vextendsingle vextendsingle vextendsingle vextendsingle ≤ s ≤ π 3 , ≤ t ≤ 2 π bracerightbigg . We al- ready know that bardbl φ s × φ t bardbl = 4 | sin s | = 4 sin s . Now integraldisplay S z dS = integraldisplay D √ 2 cos s bardbl φ s × φ t bardbl dA = 4 √ 2 integraldisplay 2 π integraldisplay π 3 √ cos s, sin sdsdt =-1 1 x-1 1 y 1 2 z-1 1 x-1 1 ( 4 √ 2 )( 2 π ) bracketleftbigg − 2 3 cos 2 3 t bracketrightbigg π 3 = 16 √ 2 π 3 parenleftbigg 1 − parenleftbigg 1 2 parenrightbigg 3 2 parenrightbigg . MATB42H Solutions # 7 page 4 (b) We parametrize by Φ ( u,v ) = ( u, 4 − u 2 , v ), ≤ u ≤ 2, 0 ≤ v ≤ 1. Now φ u = (1 , − 2 u, 0), φ v = (0 , , 1), φ u × φ v = ( − 2 u, − 1 , 0) and bardbl φ u × φ v bardbl = √ 4 u 2 + 1. The surface in- tegral is integraldisplay S xdS = integraldisplay D ( u ) bardbl φ u × φ v bardbl dA = integraldisplay 2 integraldisplay 1 u √ 4 u 2 + 1 du dv = integraldisplay 2 u √ 4 u 2 + 1 du = parenleftbigg 1 8 parenrightbiggparenleftbigg 2 3 parenrightbiggbracketleftbigg 4 u 2 + 1 bracketrightbigg 2 = 1 12 bracketleftbig 17 3 2 − 1 bracketrightbig ....
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