Problem 4 a s is the graph of z f x y 1 x2 y 2

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Unformatted text preview: d 2: ∂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 � · n dS = Therefore S F ˆ S �x, y, 2(1 − z )� · �2x, 2y, 1� dA = S 2x + 2y + 2(1 − z ) dA = S 4x + 4y 2 dA (since z = 1 − x2 − y 2 ). Shadow = unit disc x2 + y 2 ≤ 1; switching to polar coordinates, we have �� � 2π � 1 2 � 2π � 4 �1 �ˆ r 0 dθ = 2π . S F · n dS = 0 0 4r r dr dθ = 0 ˆ b) Let T = unit disc in the xy -plane, with normal vector pointing down (n = −k). Then ˆ �� �� �� ˆ �ˆ F · n dS = T �x, y, 2� · (−k) dS = T −2 dS = −2 Area = −2π . By divergence theorem, ��� �� �� ��T �ˆ � � F · n dS = div F dV = 0, since div F = 1 + 1 − 2 = 0. Therefore S = − T = +2π . S +T D Problem 5� . � �ˆ ˆ k� j ˆ� � ı � a) curl F = � ∂x ∂y ∂z � = 2yˆ...
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