finalexamsol

4 1 we used the half angle formula cos2 1 cos 2

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Unformatted text preview: l coordinates dV = dz r dr dθ. Thus ��� � 2π � 1 � 2 � M= z dV = z dz r dr dθ = G 0 0 2r 2π 1 � 0 2(1 − r2 ) r dr dθ = 4π · 0 1 = π. 4 b) 1 z= ¯ M ��� 1 z · δ dV = π G � 0 2π 1 � 0 � 2 z 2 dz r dr dθ. 2r c) In spherical coordinates: z = 2 ⇒ ρ cos φ = 2 ⇒ ρ = 2 sec φ. Limits on G: 0 ≤ ρ ≤ 2 sec φ; 0 ≤ φ ≤ tan−1 (1/2); 0 ≤ θ ≤ 2π . In spherical coordinates: dV = ρ2 sin φ dρ dφ dθ and z = ρ cos φ, so ��� � � −1 � 1 1 2π tan (1/2) 2 sec φ z= ¯ z · δ dV = (ρ cos φ)2 ρ2 sin φ dρ dφ dθ. M π0 G 0 0 5 Problem 13. a) We have F = �P, Q, R�, where P = y + y 2 z , Q = x − z + 2xyz , R = −y + xy 2 . ∂P ∂R = y2 = ; ∂z ∂x ∂Q ∂R = −1 + 2xy = ; ∂z ∂y ∂P ∂Q = 1 + 2yz = . ∂y ∂x b) � x1 f (x1 , y1 , z1 ) = � P (x, 0, 0) dx + 0 y1 � Q(x1 , y, 0) dy + 0 z1 Q(x1 , y1 , z ) dz. 0 2 Q(x1 , y, 0) = x1 ; R(x1 , y1 , z ) = −y1 + x1 y1 . � y1 � z1 2 f (x1 , y1 , z1 ) = 0 + x1 dy + (−y1 + x1 y1 ) dz P (x, 0, 0) = 0; 0 0 2 ⇒ f (x1 , y1 , z1 ) = x1 y1 − y1 z1 + x1 y1 z1 ⇒ f (x, y, z ) = xy − y z + xy 2 z + C. � c) F · dr = f (1, −1, 2) − f (2, 2, 1) = −10 + 3 = −7. C Problem 14. a) �2 b) n = 1 �x, y, 0�, F = �x, 0, 0�. 1 � Thus, F · n = x2 and in cylindrical coordinates dS = 2dz dθ. 2 6 On the surface x = 2 cos θ and the limits of inte...
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This note was uploaded on 01/21/2014 for the course MATH 18.02C taught by Professor Denisauroux during the Fall '12 term at MIT.

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