Solutions #12 ( M. y 3 + cos(x) ) dx + ( sin(y) + z 2) dy + xdz = 3π 4. The surface M is parametrized by σ : [0, 1] [0, 2π] R 3 with.

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1 Solutions #1 1. a Show that the path γ : [, π] R 3 defined by γt : cost ı sint j sint k lies on the surface z xy. b valuate y 3 cosx dx siny z dy xdz where is the closed curve parametrized by γ. Solution. a The identity sint sint cost implies that γt lies on the surface z xy for all t [, π]. b Let F x, y, z : y 3 cosx ı siny z j x k and let denote the part of the surface z xy bounded by. Stokes Theorem implies that y 3 cost dx siny z dy xdz F d r F d S The surface is parametrized by σ : [, 1] [, π] R 3 with Since and we have F d S π 1 π 1 1 π 1 1 σr, θ : r cosθ ı r sinθ j r sinθ cosθ k. σ r σ θ det ı j k cosθ sinθ 4r sinθ cosθ r sinθ r cosθ r cos θ r sin θ r sinθ ı r cosθ j r k ı j k F det z ı j 3y k, x y z y 3 cosx siny z x F σr, θ σ r σ θ dr dθ 4r sinθ cosθ ı j 3r sin θ k r sinθ ı r cosθ j r k drdθ 8r 4 sin θ cosθ r cosθ 3r 3 sin θ dθ dr [ 8 3 r4 sin 3 θ r sinθ 3 r3 sinθ cosθ 3 r3 θ ] π dr πr 3 dr π [ 1 4 r4] 1 3π 4. Therefore, we have y 3 cosx dx siny z dy xdz 3π 4. ATH 8: page 1 of 5

2 Solutions #1 ATH 8: page of 5 Observe that integration by parts gives sin tdt sint cost cos tdt sint cost dt sin tdt so sin tdt 1 sint cost t. Alternative Solution for 1b. Since γ t sint ı cost j cost k, we have y 3 cost dx siny z π dy xdz F d r F γt γ t dt π sin 4 t cos cost sint sin sint cost sin t cost cost costdt [sin cost cos sint ] π π π π π sin 4 t 4 sin t 1 sin t cost dt sin 4 t sin t 4 sin 4 t cost dt sin 4 t dt [ sint 3 sin3 t 4 ] π 5 sin5 t [ 1 4 sin3 t cost 3 8 sint cost 3 8 t sin 4 t 4 sin t cos 3 t sin t costdt ] π Observe that integration by parts gives sin 4 tdt sin 3 t cost sin 3 t cost so sin 4 tdt π π 4. 3 sin t cos tdt [ ] π 1 3 sin3 t 3 sin t 3 sin 4 tdt sin 3 t cost 3 sint cost 3t. a. valuate the circulation of the vector field Gx, y, z : xy ız j3y k around a square of side length 6, centered at the origin lying in the yz-plane, and oriented counterclockwise viewed from the positive x-axis. Solution. Let be the square of side 6 centered at the origin lying in the yz-plane with normal vector ı. Since we have G det ı j k ı x k, x y z xy z 3y

3 ATH 8: page 3 of 5 Solutions #1 Stokes Theorem yields G d r G d S G ı da ı x k ı d ds Area 6 7. Alternative Solution. Assuming that the sides of the square are parallel to the coordinate axes, is parametrized by σ : [, 3] [, 3] R 3 where σu, v : u j v k. Hence, σ u σ v det ı j k 1 ı 1 and G d S u σ 3 3 dv du dv du v du G, u, v σ 3 dv Another Solution. Parametrizing the four sides of the square again assuming that they are parallel to the coordinate axes gives 3 3 G d r G, y, j dy G, 3, z k dz 3 3 G, y, 3 j dy dy 3 9 dz 3 3 dy G,, z k dz 3 9 dy b. Let Hx, y, z : y z ı x z j xy k and let be the circle of radius 3 centered at, 1, in the xy-plane oriented counterclockwise when viewed from above. ompute H d r. Is H path-independent? xplain. Solution. Let be the disk of radius 3 centered at, 1, in the xy-plane with normal vector k. Since we have H det ı j k x 1 ı y 1 j, x y z y z x z xy

4 Solutions #1 ATH 8: page 4 of 5 Stokes Theorem gives H d r H d S H k da x 1 ı y 1 j k da da. However, the vector field H is not path-independent because H. Alternative Solution. The circle is cut out by x y 1 3, z and is parametrized by γ : [, π] R 3 where γt : 3 cost ı 3 sint 1 j. Hence, we have π H d r H 3 cost, 3 sint 1, sint ı 3 cost j dt π π π [ 18 3 sint 1 sint 3 cost 3 cost dt 9 sin t 3 sint 9 cos t 6 cost dt 18 sin t 3 sint 9 6 cost dt ] π sint cost 9t 3 cost 9t 6 sint. On the other hand, if is the circle cut out by y z 1, x then it is parametrized by β : [, π] R 3 where β : cost j sint k. In this case, we have π H d r H, cost, sint sint j cost k dt π sin t dt 1 [ ] π sint cost t π Since H d r, the vector field H is path-dependent. 3. Water in a bathtub has velocity vector field near the drain given, for x, y, z in cm, by V x, y, z : y ı x j y j zx ı k z 1 z 1 z 1 y xz yz x 1 z 1 ı z 1 j z 1 k cm s 1. a The drain in the bathtub is a disk in the xy-plane with center at the origin and radius 1 cm. Find the rate at which the water is leaving the bathtub. b Find the flux of the water through the hemisphere of radius 1, centered at the origin, lying below the xy-plane and oriented downward.

5 ATH 8: page 5 of 5 Solutions #1 c onsider the vector field Ux, y, z : 1 y z 1 ı x z 1 j x y z 1 k. ompute U d r where is the edge of the drain oriented clockwise when viewed from above. d alculate U and explain why your answers in parts a and c are equal. Solution. a Let be the disk representing the drain oriented downward. The rate at which the water is leaving the bathtub is the flux of the water flowing out of the drain: Flux out of V d S V k da da Area π cm 3 s 1. b Let W be the closed region bounded by the hemisphere of radius 1 lying below the xy-plane and oriented downward and the disk in the xy-plane representing the drain. Since both and are oriented downward, we have W and the ivergence Theorem gives V dv V d S V d S V d S V d S π. W W Therefore, V d S π cm 3 s 1. c Since is oriented clockwise, we parameterize the circle by γ : [, π] R 3 with γt : sint ı cost j. On the drain, we have z and 1 U d r y ı x j x y k d r 1 d We have π cost ı sint j 1 π k cost ı sint j dt U det ı j k x y z y z 1 x z 1 x y z 1 y xz yz x z 1 ı z 1 j 1 z 1 k cm s 1. Since U V, Stokes Theorem implies that U d r U d S V d S. dt π cm 3 s 1.

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