Question

Verify the Divergence Theorem by evaluating [ SF F. Nds as a surface integral and as a triple integral. F(x, y, z) = 2xi - 2yj + z2k S: cube bounded by the planes x = 0, x = 3, y = 0, y = 3, z = 0, z = 3

Answer 1

Solna Given, vector field F(x, y, z) = 2x i-2y7 + Z2 R. Surface se Cube bounded by the planes 230, 233, Y=0,=3, 730,7=3 Z. to,0,3) 3 si 56. S& ss → 3 y 52 Sz Solid Required Region (0,3,0) D (319,0) to According Divergence theorem: If s be a smooth Orientable surface that encloses a solid Din R. if Continous vector field whose Components have Continous partial derivative in an Open set segin fields is a in

Date Page then, Containing D If fonds - Afflivf divf dv. S D Where N is cut ward unit Normal field for the surface s ay a 2 2 + 2z. divĚ = (6743, 743. *).(2x7=24j+z+ Jolive ez fff ez dzdy den clivf dv Ass 3 3 3 e z dzdy da x=0 yzo 2=0 3 73 9. =ff 2 Lely dy dt o x=0 y 50 =f3 g dy dre x = y=0

3 o ly) 3 de 20 3 27. () 0 divf dv al 2 Now, S = Sit S₂ + Szt Sq & Set 56. And ff fonds s si S2 SSF.nds & ff F Nedsat SS Fonds t f f F . Nds 4 Sands th S3 S4 ff Fonds 6. 5 S6 S4 (3) si cil for Surface Unit outwar normal is N=th xi- fflezi-a3j+2E).ads SS z2 ds, on S, Z=3.

Date Page f Futels g ds, SI Projection of ds, on 1-y-Р lane gives ds, dady dady two Ry 1 Eoff dxdy رک 3 3 9 dy da. X=0 y=o 9*3*3 SS Finds, (4) (1) on Surface Sz. Outward unit Normal is N = Sf Fondse Normal is Ñ - ff (2 x 1" - 2y + 2?) k ds ss - z2 ds. S2 S2 Sz on surface Sz z=0

There fore. ISS F in der (5) S2 DO) lilu) On Surface S3. Out ward SS Funds, Unit Normal Ñ = + P. ff 2x ds. Sz S3. On Sz x= 3. = SS 6 dsz. sz on Ч-2 Р Cane gives. Projection of ds3 dsz dydz W.pl dydz of de dy = 6 Y=ogo = 6x3x3. Finds = 54 (6)

lives On Surface Sy Ñ = -2 f Findsa - - 2x dsq. S4 54 On S4 x=0 ff Findse с 34 (v) On Surface 55 + ] A N = ISAN doo f - ey ds 55 S5 On 55 y = 3 ko is ds5 2-2 plane Projection of ds 5 an gives d5 dadz = dadz

Date Page SSF on ds =-off da de da dz. 3 S5 6 dla dz X=02=0 - 6 X 3X3. a) xvi) on S6. Sf Endss-54 Surface -j. SS F. Nds6 = Sey Surface S6 y = 0 N = If zy dsb 56 56. on = 0 ff Fonds6 -9) S Putting values from 144(5), (6), (7), (01 camel (9) in (3) we get Fonds 81+0+54 +0-54 to Fonds. = S བ། 10. SS. Aurt fedente, from ean tetanch tro) Divergence theoren -