finalexamsol

1 3 r 4 problem 11 a net ux out of r will be positive

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Unformatted text preview: = D. �f = 2�x − x0 , y − y0 , z − z0 �, �g = �A, B, C �. 2(x − x0 ) = λA 2(y − y0 ) = λB �f = λ�g, and g = D ⇒ 2(z − z0 ) = λC Ax + B y + C z = D. Problem 8. a) ∂F ∂F ∂x ∂F ∂y ∂F ∂z = + + ∂φ ∂x ∂φ ∂y ∂φ ∂ z ∂ φ ∂x = ρ cos φ cos θ ⇒ xφ (2, π/4, −π/4) = 2 cos(π/4) cos(−π/4) = 1. ∂φ ∂y = ρ cos φ sin θ ⇒ yφ (2, π/4, −π/4) = 2 cos(π/4) sin(−π/4) = −1. ∂φ √ ∂z = −ρ sin φ ⇒ zφ (2, π/4, −π/4) = −2 sin(π/4) = − 2. ∂φ b) NOT Possible �−y, x is not a gradient field. (Test: �−y, x� = �P, Q�: 3 Py = −1 �= Qx = 1.) Problem 9. √ x2 ≤ y ≤ 2 x 0 ≤ x ≤ 2 � a) R = b) � R = y 2 /8 ≤ x ≤ 0 ≤ y ≤ 4 √ y ⇒ � Ru,v : � √ y f dA = R Problem 10. 4 � �� f (x, y ) dx dy. 0 y 2 /8 4≤u≤9 1≤v≤2 � �� � � � � xu xv � � u−2/3 v −1/3 /3 −u1/3 v −4/3 /3 � � � = � −2/3 2/3 � = 2 + 1 u−1/3 v −2/3 = 1 u−1/3 v −2/3 . Jacobian J = � yu yv � � u v /3 2u1/3 v −1/3 /3 � 99 3 � �9 1 f (x, y ) dA = int2 f (u1/3 v −1/3 , u1/3 v 2/3 )( u−1/3 v −2/3 ) du dv. 1 3 R 4 Problem 11. a) Net flux out of R will be positive (more flow out than into R) b) � � C � F · n ds = � −N dx + M dy = C −x dx + x dy C C1 : x = 1 − t, y = t ⇒ dx = −dt, dy = dt � �1 �1 � ⇒ F · n ds = −(1 − t)(−1) + (1 − t)(1) dt = −(1 − t)2 � = 1. 0 C1 0 4 � C2 : x = 0 ⇒ F · n ds = 0. � � C3 : y = 0, dy = 0 ⇒ F · n ds = C2 C3 1 −x dx = − 0 x2 1 =− . 2 2 Thus, � C � � F · n ds = C1 +C2 +C3 1 = 1 + 0 + (−1/2) = . 2 c) �� div(F) = Mx + Ny = 1 ⇒ �� div(F) dA = dA = area(R) = R R 1 1 ·1·1= 2 2 Problem 12. a) Limits on G are 2r ≤ z ≤ 2; 0 ≤ r ≤ 1; 0 ≤ θ ≤ 2π . In cylindrica...
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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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