prac3asol

prac3asol - MIT OpenCourseWare http://ocw.mit.edu 18.02...

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

View Full Document Right Arrow Icon
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 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 ...
View Full Document

Ask a homework question - tutors are online