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prac4bsol - MIT OpenCourseWare http://ocw.mit.edu 18.02...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.02 Practice Exam 4B – Solutions Problem 1. � �/2 � 1 � 1 0 0 0 r2 r dz dr d�. � � � � Problem 2. a) sphere: � = 2a cos π. b) plane: � = a sec π. � 2� � �/4 � 2a cos � c) �2 sin π d� dπ d�. 0 0 a sec � Problem 3. � a) � y (2xy + z 3 ) = 2x = � 2 �z (x � � + 2y z ) = 2y = � � � 2 3 2 2 � x (x + 2y z ); � z (2xy + z ) = 3z = � x (y � 2 2 ρ �y (y + 3xz − 1); so F is conservative. + 3xz 2 − 1); So f (x, y , z ) = x2 y + y 2 z + xz 3 − z + c. Method 2: � ρr b) Method 1: f (x, y , z ) = C1 +C2 +C3 F · dρ; �(x1 , y1 , z1 ) � � x1 � x1 3 ρr C1 F · dρ = 0 (2xy + z ) dx = 0 0 dx = 0 (y = 0, z = 0) C1 � �y �y � � C3 ρr F · dρ = 0 1 (x2 + 2y z ) dy = 0 1 x2 dy = x2 y1 (x = x1 , z = 0) 1 1 �C C2 2 � � z �z 2 2 3 ρr F · dρ = 0 1 (y 2 + 3xz 2 − 1) dz = 0 1 (y1 + 3x1 z 2 − 1) dz = y1 z1 + x1 z1 − z1 (x = x1 , y = y1 ) C3 �f 3 �x = 2xy + z , so f (x, y , z ) �f �g �g 2 2 �y = x + � y = x + 2y z , so �y = 2y z . Therefore g (y , z ) = y 2 z + h(z ), and �f �z = x2 y + xz 3 + g (y , z ). f (x, y , z ) = x2 y + xz 3 + y 2 z + h(z ). = 3xz 2 + y 2 + h� (z ) = y 2 + 3xz 2 − 1, so h� (z ) = −1. Therefore h(z ) = −z + c, and f (x, y , z ) = x2 y + xz 3 + y 2 z − z + c. Problem 4. a) S is the graph of z = f (x, y ) = 1 − x2 − y 2 , so n dS = �−fx , −fy , 1� dA = �2x, 2y, 1� dA. ˆ �� �� �� �� 2 2 2 ρ · n dS = Therefore S F ˆ S 2x + 2y + 2(1 − z ) dA = S 4x + S �x, y, 2(1 − z )� · �2x, 2y, 1� dA = 4y 2 dA (since z = 1 − x2 − y 2 ). Shadow = unit disc x2 + y 2 � 1; switching to polar coordinates, we have �� � 2� � 1 2 � 2� � 4 �1 ρˆ r 0 d� = 2� . S F · n dS = 0 0 4r r dr d� = 0 ˆ b) Let T = unit disc in the xy -plane, with normal vector pointing down (n = −k). Then ˆ �� �� �� ˆ ρˆ F · n dS = T �x, y, 2� · (−k) dS = T −2 dS = −2 Area = −2� . By divergence theorem, � �T ��� �� �� ρˆ ρ ρ F · n dS = div F dV = 0, since div F = 1 + 1 − 2 = 0. Therefore S = − T = +2� . S +T D Problem 5� . � �ˆ ˆ k� � ˆ� � ı ρ a) curl F = � θx θy θz � = 2yˆ − 2xˆ. ı � � � � −2xz 0 y 2 � ˆ ρˆ b) On the � nit sphere, n = xˆ + yˆ + z k, so curl F · n = �2y, −2x, 0� · �x, y , z � = 2xy − 2xy = 0; ˆ ı � �u ρ · n dS = 0. therefore R curl F ˆ � �� ρr ρˆ c) By Stokes, C F · dρ = R curl F · n dS , where R is the region delimited by C on the unit sphere. � �� ρr ρˆ Using the result of b), we get C F · dρ = R curl F · n dS = 0. ...
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This note was uploaded on 05/04/2011 for the course MATH 18.02 taught by Professor Auroux during the Spring '08 term at MIT.

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