Question

= and z= 8. Let A be the part of the cylinder x2 + y2 1 between the planes z = 2, where n points away from the z-axis. Let C be the counterclockwise boundary of A. Let F(x, y, z) = (2xz + 2yz, –2xz, x2 + y²). Verify Stokes' Theorem: (a) Evaluate the line integral in Stokes' Theorem. (Hint: C has two separate parts.]

Answer 1

8. F(x, y, z) =<222 + 2yz , -2X27;**+y> A is the part of the cylinder x² + y² = 1 between the planes 2.1 and z=2 . The boundary the from o to 21 ) and the upper with t 21 to boundary & his consists of two parts, botton circle (cost, sint ,!o) with & ranging upper circle (eost, sint, 2) ranging from (as the line integral is, TE 2x2 + 2yz) dm) 1-294, 2 dy, ti6x2 + y2) dz} ${[%cost + y sint) (-sint at) --2cost econt dit į is TUL dot 42*2+ с C 21 + {ktcost 4 sint Desinto af dycost.cost dt dt} 21 21 (-2 sint cost 2sin2t - 2 lost th) dt 4 cos?t) dt + J C-4 sint cost : a úsin?t 2 IT 21 (-sinati - (2) dt 2).dt. + ( sijn 2t y sinat? 4).dt 21 2.Tt Zeth 2 cos 2t cos at 2t 1 ] v+] 2 2 (-4 -41 1/2 ) + (1-1+81) 410

) الداع لالا = (.21 + 2x) + ( 27 - 28 - .2%) (5) อx . 59 22 +292 2²ty امام - 22 ! می 4 L ا (22 - 21 - )... ۔ , و2 + 2 ) جا - 3 2 + اجا

On see, mis + y = -1 《火 a Ole+y-1) by + gering grad (m² + y² - 1) = ( 2 y + 5 2x + zy grad (x² + y²-1) 201 + 2y Igrad (x² + y²_1) | x + 93 **-] 2 Is cuad ² ñ dstangeni S] {(24+2y) îl * 2yî. – 42.X} i cał tyội) ds, S. (232 +2xy + 24²) dsg S [[ { x + y + 2+ 3 4s, to put ne a cos o Ч. a sino ds & do dz 2 = 2 = 1 Í * 2 *25050) 48 da ^ f<2+ sin 20 ) do di [20- = 2=1 =0 21T 2 dz cos 20 2 so 조리 (41 - + 1 / 2 ), dz 41 [2] 245 2 The surface integral is :: 4T