practice3bsolutions

0 b let r be the unit disc inside c f dr nx my da

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Unformatted text preview: trization of the circle C is x = cos t, y = sin t, for 0 � t < 2� ; then dx = − sin tdt, dy = cos tdt and � 2� W= (5 cos t + 3 sin t)(− sin t)dt + (1 + cos(sin t)) cos tdt. 0 b) Let R be the unit disc inside C ; � �� �� F · dr = (Nx − My )dA = (0 − 3)dA = −3 area(R) = −3�. C R � C3 • (1, 4) � 5. a) (0, 4) • C4 � (0, 0) • C2 C1 � • (1, 0) R � ˆ F · n ds = C� � �� div F dxdy R = (y + cos x cos y − cos x cos y )dxdy = R � 4� 1 � 4 = y dxdy = y dy = [y 2 /2]4 = 8. 0 0 0 0 �� y dxdy R ˆ ˆ b) On C4 , x = 0, so F = − sin y ˆ, whereas n = −ˆ. Hence F · n = 0. Therefore the flux of F � ı through C4 equals 0. Thus � � � � ˆ ˆ ˆ ˆ F · n ds = F · n ds − F · n ds = F · n ds ; C1 +C2 +C3 C C4 C and the total flux through C1 + C2 + C3 is equal to the flux through C . � �� � � ux uy � � 2 −1 � � �=� � = 3. 6. Let u = 2x − y and v = x + y − 1. The Jacobian � vx vy � � 1 1 � 1 Hence dudv = 3dxdy and dxdy = dudv , so that 3 V �� = (4 − (2x − y )2 − (x + y − 1)2 ) dxdy (2x−y )2 +(x+y −1)2 <4 = 1 (4 − u2 − v 2 ) dudv 3 �� u2 +v 2 <4 = = � � 2� 0 2� 0 � 2 0 1 (4 − r ) rdrd� = 3 2 4 8� d� = . 3 3 � 2� 0 � 22 1 r − r4 3 12 �2 0 d�...
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