20E - HW 7 Solutions

20E - HW 7 Solutions - Math 20E Homework #8 24 February...

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Unformatted text preview: Math 20E Homework #8 24 February 2005 Section 7.5, Problem 6: Verify that in spherical coordinates, on a sphere of radius R , k T T k d d = R 2 sin d d . Solution. The spherical coordinates for = R are ( x , y , z ) = ( R cos sin , R sin sin , R cos ) , so we get T = (- R sin sin , R cos sin , 0 ) and T = ( R cos cos , R sin cos ,- R sin ) . Hence! We have T T = (- R 2 cos sin 2 ,- R 2 sin sin 2 ,- R 2 sin cos ) , and so k T T k d d = q R 4 cos 2 sin 4 + R 4 sin 2 sin 4 + R 4 sin 2 cos 2 d d = R 2 q ( cos 2 + sin 2 ) sin 4 + sin 2 cos 2 d d = R 2 q sin 2 ( sin 2 + cos 2 ) d d = R 2 sin d d . Section 7.5, Problem 15: Let : D R 2 R 3 be a parameterization of a surface S described by x = x ( u , v ) , y = y ( u , v ) , z = z ( u , v ) . (a) Let u = x u , y u , z u and v = x v , y v , z v , and set E = u 2 , F = u v , G = v 2 . Show that EG- F 2 = k T u T v k , and that the surface area of S is A ( S ) = ZZ D p EG- F 2 du dv . In this notation, how can we express RR S f dS for a general function f ? (b) What does the formula for A ( S ) become if the vectors / u and / v are orthogonal? (c) Use parts (a) and (b) to compute the surface area of a sphere of radius a . Solution. (a) For the rest of this problem, we will use the shorthand notation x u = x / u , and so on. k T u T v k 2 = ( y u z v- y v z u ) 2 + ( z u x v- z v x u ) 2 + ( x u y v- x v y u ) 2 = x 2 u ( y 2 v + z 2 v ) + y 2 u ( x 2 v + z 2 v ) + z 2 u ( x 2 v + y 2 v )- 2 ( x u x v y u y v + x u x v z u z v + y u y v z u z v ) = ( x 2 u + y 2 u + z 2 u )( x 2 v + y 2 v + z 2 v )- ( x 2 u x 2 v + y 2 u y 2 v + z 2 u z 2 v + 2 x u x v y u y v + 2 x u x v z u z v + 2 y u y v z u z v ) = EG- ( x u x v + y u y v + z u z v ) 2 = EG- F 2 , 1 Math 20E Homework #8 24 February 2005 as desired....
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20E - HW 7 Solutions - Math 20E Homework #8 24 February...

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