Question

(1 point) Compute the outward flux of the vector field F(:,, :) - 2ri + 4y + 4k across the boundary of the right cylinder with radius 5 with bottom edge at height z = 5 and upper edge at 2= 6. Note: The vectors in this field point outwards from the origin, so we would expect the flux across each face of the cylinder to be positive Part 1 - Using a Surface Integral First we parameterize the three faces of the cylinder Note: in the following type 0 as theta. The bottom face: 0(0,0) = (+1(1,8), 31(0,0), 2(1,0) where 0 < & < 2,0 < < 5 and Iir, 0) rcostheta wir, 0) rsintheta zi(r.a) = -5 The top face: 09(0,0) = (22(5,8), y(t,0), 29(r, @) where 0 < 0 < 27.0 <35 and z2(t,0) rcostheta ya(,0) rsintheta 22(t,0) - 6 The lateral face: 0(0, 2) - (23(0,2), ya(0,2), (0, 2) where 0 <e < 24.-55 256 and 5costheta y(@) 5sintheta z3(2) z Then (mind the orientation) х θση 89 dr de S[Funds - -6*6*2 *P(6.). Co ) dr db + * F(0) Conte L" LP(s) 303 + θσ 30 dz de Oz Part 2 - Using the Divergence Theorem SLF Finds divFdV de dr do

Answer 1

our ²5 114,0) = Cucoso, using, -5) = (%, 9, 12) latural face; 83(0, 2) = CS caso, ssino, z) = (23,931 Zz). Flx, y, z) = 2x î + 4yj +42 bottom "face: 7,14,0), 040-21, Top face; 8, 14,0) = ( MOSO, usino, 6) = 122, 92,22) OLO! 27, OEMES z 040421) -54246 2T 5 ss finds = JS froid. ( 20 11 x 21 J dudo j j flog). ( 283 x do 2 dedo 0 27 S + X 0 27 6 + J floa) do 3 dz 2o3ldado. Now first we determim the tums used a bove! Frog)= 1 t Flag Y 1, 2) = 22, it uy, ĝ+ 42, k = 24 caso i +44 sino ŷ +41-552 = 24000 i +44sinoſ -20 ? CS Scanned with CamScanner

Өо. - -3°) 29 1ң (049 , esino, =(-ячino, кох 9 , 0) Әсі Әя 2 ОА Счір 0, чанар, -5) ( ЛВ , sino, 6 ) () ar) t x )) 소 20 Әя Arsino НА 0 0 Сов sino = ? (0)-} Cost Pe (=Hsino-malo) -AR до ) де Х до | Оң Х F(0) :( OM J = CP Acoso? + Чилі по 1 - 20 ) (-я : ) 20. d) low F182) aft , 2 2) оксаВ і t Чи «int it 24 2. Сясо, о, млin 6, ) =tt040, simo,о) 26. а О я cs Scanned with CamScanner
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г., og a. ( 9 105 0 , 4 Jun () t 29 ; ord Әң (-Ниһ0, HC 0 , 0) อด) і 20 Cu50 sino Х - ю CAsing ңux 0 0 : : (0) - to'tk [я его кій, - не. Х Fis): (x 2 ): (змам і чым зво! 12 ч), як 12 ЧЯ ft",) = F(33, 93, 24) 2 13 14 чуз і 473 - 2х5 сеѕѕ t + 4х5 sing i tuz 10 саѕ0 + 20 ѕtіng іt + Чz 2 t әС 07 (scoso, ssino ,2) 07. - (0,01) cs Scanned with CamScanner

drag do (5 caso, ssino 2) = (-540, 560, 0) î 3 do 3 x 23 Oz do = Q c4 0 -ssino scaso Q8 =î (-5caso s-} (5seno) + € (0) = -5caso ê. ssino FCO). Co Bx 8 20 2osinoj az -(10cus O î+ +42ů) • (-5caso - Ssihoĝ 2 (- 50 caso 100 sen = -50 [caso + sino ]-sosino 50 [17 11 sinol (11 5 27 5 Using is, is Finds 2012 did 0+ SI + le -5011+ sin o) d2 do CS Scanned with CamScanner

27 5 IT 201² do S = do ot | 24 4 2 1 D 27 + / -50 (Itsino) -21 do -5 27 27 27 = 10X25 01 + 12x25 -50 x[6+5) (0+ O casosino = 250X2a + 300 × 27 - 550x37 El 500 K & 600 t - 1650 a - - 550T 2) SS finds SSS div F du 25 div F. Jdz dedo 00-5 Jis Jacobi'an determinant - M 0 divf in is Cylindsuial coordinates (94073) ta Cafe) toch are du I do dz t a F2 + " cs Scanned with CamScanner