MATH 201 Final, Fall 2008

American University of Beirut MATH 201 Calculus and Analytic Geometry III Fall 2008-2009 Final Exam - solution Exercise 1 a. If f(u, v, w) is a differentiable function and if u = x−y, v = y−z, and w = z−x, show that ∂f ∂x + ∂f ∂y + ∂f ∂z = 0 ∂f ∂x = ∂f ∂u. ∂u ∂x + ∂f ∂v . ∂v ∂x + ∂f ∂w.∂w ∂x = ∂f ∂u − ∂f ∂w; ∂f ∂y = −∂f ∂u + ∂f ∂v , and ∂f ∂z = −∂f ∂v + ∂f ∂w, and hence ∂f ∂x + ∂f ∂y + ∂f ∂z = 0 b. Use the method of Lagrange multipliers to find the maximum and minimum values of f(x, y) = 3x − y + 6 on the circle x2 + y2 = 4 ∇f = λ∇g ⇔      3 = 2λx −1 = 2λy x2 + y2 = 4 (g(x, y) = 0) ⇔      3 2λ = x − 1 2λ = y x2 + y2 = 4 x2 + y2 = 4 ⇒ 9 4λ2 + 1 4λ2 = 4 ⇒ λ2 = 10 16 ⇒ λ = ± √ 10 4 for λ = √ 10 4 ; x = 6 √ 10, and y = − 2 √ 10, and f(x, y) = 2 √ 10 + 6 (maximum value of f) for λ = − √ 10 4 ; x = − 6 √ 10, and y = + 2 √ 10, and f(x, y) = −2 √ 10 + 6 (minimum value of f) Exercise 2 Convert to polar coordinates, then evaluate the following integral 6 - x y 1 r=2 sin θ ...................... � 2 0 � 0 −√ 1−(y−1)2 xy2dxdy = � π π/2 � 2 sin θ 0 r4 cos θ sin2 θ drdθ = � π π/2 32 5 cos θ sin7 θ dθ = 4 5 � sin8 θ �π π/2 = −4 5 Exercise 3 � 1 0 � 0 −1 � y2 0 dz dy dx = 1/3 = j x 6 z=y2 y z (0,-1,1) ............... ............... (1,-1,1) (1,-1,0) � 0 −1 � 1 0 � y2 0 dz dx dy = � 1 0 � 1 0 � −√z −1 dy dz dx = � 1 0 � 1 0 � −√z −1 dy dx dz = � 0 −1 � y2 0 � 1 0 dx dz dy = � 1 0 � −√z −1 � 1 0 dx dy dz Exercise 4 Let V be the volume of the region D that is bounded by the paraboloid z = x2+y2, and the plane z = 2y. 6 z z x ) y D x2 + (y − 1)2 = 1 1 z = x2 + y2 y = 2x a) cartesian coordinates: V = � 2 0 � √ 1−(y−1)2 −√ 1−(y−1)2 � 2y x2+y2 dz dx dy b) cylindrical coordinates: V = � π 0 � 2 sin θ 0 � 2r sin θ r2 r dz dr dθ = � π 0 � 2 sin θ 0 (2r2 sin θ − r3) dr dθ = � π 0 �2 3r3 sin θ − r4 4 �2 sin θ r=0 dθ = � π 0 4 3 sin4 θ dθ = 4 3.3π 8 = π 2 Exercise 5 Let V be the volume of the region D that is bounded below by the xy-plane, above by the sphere x2 + y2 + z2 = 4, and on the sides by the cylinder x2 + y2 = 1. 6 - = ......................... x y z 1 2 D a) spherical coordinates: order dρ dφ dθ V = � 2π 0 � π 6 0 � 2 0 ρ2 sin φ dρ dφ dθ + � 2π 0 � π 2 π 6 � 1 sin φ 0 ρ2 sin φ dρ dφ dθ 6 - φ ρ π 6 π 2 ................. ....... 1 2 ρ = 1 sin φ b) spherical coordinates: order dφ dρ dθ V = � 2π 0 � 2 0 � π 6 0 ρ2 sin φ dφ dρ dθ + � 2π 0 � 1 0 � π 2 π 6 ρ2 sin φ dφ dρ dθ + � 2π 0 � 2 1 � sin−1(1/ρ) π 6 ρ2 sin φ dφ dρ dθ the answer of part b) can also be written: � 2π 0 � 1 0 � π 2 0 ρ2 sin φ dφ dρ dθ + � 2π 0 � 2 1 � sin−1(1/ρ) 0 ρ2 sin φ dφ dρ dθ (why??) Exercise 6 a. Find the work done by the force F = xi + y2j + (y − z)k along the straight line from (0, 0, 0) to (1, 1, 1). r(t) = ti + tj + tk , 0 ≤ t ≤ 1; dr dt = i + j + k, F(t) = ti + t2k, and F � dr dt = t + t2, hence W = � 1 0 (t + t2)dt = 5/6 b. Evaluate � (1,π/2,e) (0,0,1) (ln z + ex sin y)dx + ex cos ydy + (x/z − z)dz f(x, y, z) = x ln z + ex sin y − z2 2 + C is a potential function (check it!), hence � (1,π/2,e) (0,0,1) (ln z + ex sin y)dx + ex cos ydy + (x/z − z)dz = � x ln z + ex sin y − z2 2 �(1,π/2,e) (0,0,1) = 3 2 + e − e2 2 c. Find the outward flux of the field F = (y − 2x)i + (x + y)j across the curve C in the first quadrant, bounded by the lines y = 0, y = x and x + y = 1. 6 - x x + y = 1 y C1 y = x C2 C3 R - i) direct calculation: Flux = � C Mdy − Ndx C1 : r1(t) = ti , 0 ≤ t ≤ 1; Mdy − Ndx = −tdt and � C1 Mdy − Ndx = � 1 0 −tdt = −1/2 C2 : r2(t) = (1 − t)i + tj , 0 ≤ t ≤ 1/2; Mdy − Ndx = (1 − 3t)dt and � C2 Mdy − Ndx = � 1/2 0 (1 − 3t)dt = −1/8 C3 : r3(t) = (1 2 − t)i + (1 2 − t)j , 0 ≤ t ≤ 1/2; Mdy − Ndx = (3 2 − 3t)dt, and � C3 Mdy − Ndx = � 1/2 0 (3 2 − 3t)dt = 3/8 Flux(F) = � C Mdy − Ndx = −1/2 − 1/8 + 3/8 = −1/4 ii) Green’s theorem: Flux = � � R divFdA divF = ∂M ∂x + ∂N ∂y = −2 + 1 = −1 � � R divFdA = � 1/2 0 � 1−y y − dxdy = � 1/2 0 (2y − 1)dy = −1/4