Question

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+ 14) [12] Find the flux of the vector field F across the enclosed surface S. Sketch the surface. F = yi +3x j +4zk, and S is the boundary of the solid region enclosed by z=9-x² - y2 and the plane z=2. (note that this includes two surfaces). Assume outward orientation. Do not use the Divergence Theorem. Evaluate completely. Bonus 4 points Use the Divergence Theorem to solve the previous problem #14.

Answer 1

IZ -7-9-x-y2 9 +2 The surface s consists of two surfaces S: 7=9.x² - y2 a 2: Z=2 F = yî+3 aj +42 Flux = S Fods + SF.ds S2 Consider first integral unit normal n = 7 (9-x_y²-2) 1780 (9-22 - y² =)) (-2ni24jZK) √4x²+4y²+ z² so F. ñ = (71 +3 aj +4242) : (- 2x - 2yj - Z) = -20% -60% -4z² √4x²+61 y² + 2² of R, is the projection of kersfaces, √4x²+45²+ z² on nag-plane then

ds= dady 1 Roñil √4x²+44 +2² dady Z As Thus F .ñds (-2xy-684- 48²) dady Z R, si so as Z=q-x² - y² Fiñas = -(* 8 xy +4 (9-8-2) dady 9-2²-42 The region R, can be described in polar coordinates Neroso, y=&sino, andy=&dodo limits of r=0 to 2 OCOST. - (8 x cm 8 sono +419-03) 9-82 rdodo - (88 Co smo +46 (9-823²) do do 21 2 Thus, o 21 2 9-82 0 0

2 2T 883 caso smo & 40 (9-8 (9-7") as do 9-22 2 2 Now any (982)8-98 (-873 9-82 dr = 8 dr 9-82 6 2 97 8 7- der 5-40 (982) do 9-82 0 2 2 +84 2 = 810 + 9 / ln (978 2 - - 72 +16 8 2 t en こー56 2 मा = 16+ 36 ln caso sino 8 - 567 do SATA en sf] (16+36 din 4-copos 21 - 56 X 21 = - 112T As the unit normal n, on s, is oriented ontward to the value is +1125

for S2 ñ = - Z=2 12 so F n = (yit3nj +424) • (-k). – 47 and note that 7=2 on S2, řña - 4x2=- 8 hence and ds = dndy, They 8 rardo SF. ñ, ds = म 2 87 do 0 1 - 1600 25 - 16 do = -32T. 0 Total shin = +1125-3258 = 801 (Aus.)

Divergence theorem application

of we apply Divergence theoren, then di Fe (9) + (34) + m2 (42) = oto+4. i dir F .=4. Flora = Slav Favo 9-2-42 SIS . こ 4 dz da dy R 2 - SS4 (9-8-7² - 2) andy an polar cooodinates, 2 BUT 4 (782) & do do 0 O 2 do s [ 2882 0 MT 0 21 ( 4o do 80T s [56-16] do 0 0

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