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96. Consider a vector field F(x, y, z) =< x + x cos(yz), 2y - eyz, z- xy > and scalar function f(x, y, z) = xy3e2z. Find the

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* Solution . The given vector field F(21,4,2)=<2+2 cas y2), 24-2, 2-ay>(a) grado=DF li 021 42 -Coslyz) +2y-ént 2-24 ƏH ) = 12 (21+ cos(42) + 24-4X2+2-244 ) +32 1 21 + cos (42) + 2Y - 2² - 2.27) +Ŕa grad F = (1-4) + (-zsin(y2 + 2-ze-415 Toto + (y sinlyz) – Lez² +2) Ř Answer (6) div F = lot 4) ((74 +381 60342)] ê +2y-ek ((c) collf) = axt = (1 + 72 tk 2 ) x 2143 22 -ot because f(nia, 2) = nye22 is a scalar function. And curl is a vector quantityd) curl( 979df) = VX(vf) 24 = 1 2 4 32 2 )(193222) _4 = 12 (443222) 432 (25227 ) (123 22 ) f = 43321 4 399227 今 + 243-22 Then+ t (e) div (cula) = D. (AXIF) OXF = o zno uthe (2280) retre/ hre-zore tre - (Rre-z) te! = [((77) 503 re +re) J+ [(ke-z)-((to 505 re +re) [((2750) le +re ) thg - (zz-R3)Dat [ť +(28) Wys-nte-]sD. (DXF/= a1ty sin(y2 + t e sin(92) 1 - + esin (42) 4420 HY Z cos(42) 1-(xFF y siny2)) Answer

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96. Consider a vector field F(x, y, z) =< x + x cos(yz), 2y - eyz,...
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