prac3asol

# prac3asol - 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 3 A – Solutions 1. a) The area of the triangle is 2, so y = ¯ b) By symmetry x = 0. ¯ 2. � = |x| = r| cos �|. I0 = � 2� 0 1 2 � 1 � 0 2−2y y dxdy. 2y −2 � 1 0 �� r2 � rdrd� = D r |r cos �|rdrd� = 4 2 � 0 � /2 � 1 0 r cos �drd� = 4 4 � � /2 0 1 4 cos �d� = 5 5 3. a) Nx = 6x2 + by 2 , My = ax2 + 3y 2 . Nx = My provided a = 6 and b = 3. b) fx = 6x2 y + y 3 + 1 =� f = 2x3 y + xy 3 + x + c(y ). Therefore, fy = 2x3 + 3xy 2 + c� (y ). Setting this equal to N , we have 2x3 + 3xy 2 + c� (y ) = 2x3 + 3xy 2 + 2 so c� (y ) = 2 and c = 2y . So f = 2x3 y + xy 3 + x + 2y (+constant). � �r c) C starts at (1, 0) and ends at (−e� , 0), so F · d� = f (−e� , 0) − f (1, 0) = −e−� − 1. C 4. � �� � ux uy � � 2x/y −x2 /y 2 � �=� 5. a) � vx vy � � y x � 4� 5 � yx dx + y dy = C 3 2 � 1 x x dx + (x ) (2xdx) = 23 22 0 dudv = (3x2 /y )dxdy = 3u dxdy � � � = 3x2 /y . Therefore, � =� � 1 3x5 dx = 1/2. 0 dxdy = 1 dudv. 3u �4 1 1 2 b) dudv = ln 5 dv = ln 5. 3 2 1 3u 23 �� � 6. a) M dx = −My dA. C R 1 b) We want M such that −My = (x + y )2 . Use M = − (x + y )3 . 3 � 1 � x3 �� �1 1 � 7. a) div F = 2y , so 2y dA = 2y dydx = x6 dx = . 7 R 0 0 0 �ˆ ˆ � b) For the ﬂux through C1 , n = −ˆ implies F · n = −(1 + y 2 ) = −1 where y = 0. The length of C1 is 1, so the total ﬂux through C1 is −1. �ı ˆı The ﬂux through C2 is zero because n = ˆ and F � ˆ. �� � � � 8 1 �ˆ � �ˆ �ˆ div F dA − c) F · nds = F · nds − F · nds = − (−1) − 0 = . 7 7 C3 R C1 C2 ...
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