Question

zi+yj + zk 3. Given the vector field in space F(x, y, z) or more conveniently, (x2 + y2 + 22)3/2 f where r = ci + yj + zk and r= |||| = V2 + y2 + z2 (instead of p) 1 F(r) = r2 (a) [10 pts] Find the divergence of F, that is, V.F. (b) (10 pts] Directly evaluate the surface integral lle F.NDS where S is the unit sphere x2 + y2 + z2 = 1 oriented with outward normal vectors. (c) (5 pts) Do these two results from (a) and (b) contradict the Divergence Theorem? Explain. (a ) (b)

Answer 1

meer since giram veuter field ałtujtzie F (aey, 2) = (&?ty? 422 3h fir) = q I 73 Tit= {?. (200x (2²4 2² +22) 772)51) z/ ( by sommetric ) E} ²4y²+ 2², 32 2.34.22 ( x+y+z^^. 2 32 (x²+3² 22 32 = Elchtige y²2 2232 = (x 24 y ²1 22 y 3h - 3 (x²+02+2%) (+y+z2,512 3 3 co (x4y² +2232 (x²+4² +22 32 DoF (219, 2) = D. D. F=o

b SP eet I=SS Fños det si akty²+2²=1 ø= x² + y² + 2², * Jø= acxitohtk) fo, Ñ = DO 2 (1+y+z) √4x²+9y²4922 xl tyú tz 1001 ملح *=Ztehtype in Ñ = ałtoutz drdy and ds - ت didy I w. i 2 Z Si since from eq" ☺ 7 9 I = SE R. Nos Y, 27

ITSSs ( Iz Caltyjt ZR) ds Sss ( xifyjt zal ¥x²+4²472 ? ds = dn dy - andy 2 IN RI 11 S 2²4 4²4 7 ds 8²74²+7² x² + y² +2²=1 x?ty²7z² } y V Se andy (1-x²-y2 Z So andy So Andy ixty +Sese q andy √1-2²42 12 / 1-x²y2 since in the region R (figure) curice, contentation are opposite for our face si and S₂ in the gilken region is same 80, one integral is negativa to other but both case do I=0 surface s is © since according to according to divergence theorem if closed and orelosed volume v SS, F. Êds = SS, (0.F)ow

do from egr by divergence theorem e here. surface x² + y²47²1 I=SIS, (0.6) dw is closed 3V.f Vifo / I = SS och ao ICO is varified by part ① and do divergence theorem