{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

prac4bsol

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online