Question

verify the divergence theorem for the field A=Z X3 Z and the volume region given by 0

Answer 1

Answer:

Using spherical coordinates withρ=a,dS=a2sinφdφdθ;LHS=∫∫SF·ndS=∫∫Sx2z2dS=∫2π0∫π0(asinφcosθ)2(acosφ)2a2sinφdφdθ=a6∫2π0cos2θdθ∫π0sin3φcos2φdφThe first integral is evaluated by the power-reducing formula:cos2θ=1+cos2θ2In the second integral, we substituteu=cosφ,du=−sinφdφso that∫sin3φcos2φdφ=∫(1−cos2φ)cos2φsinφdφ=−∫(1−u2)u2du=∫u4−u2du=u55−u33+CLHS=a6[12θ+14sin2θ]2π0[cos5φ5−cos3φ3]π0=a6(π)(−15+13−15+13)=415πa6The surfaceSis a level surface ofg(x,y,z)=x2+y2+z2.Its unit normal vector is±∇g|∇g|=±〈2x,2y,2z〉√4x2+4y2+4z2=±〈xa,ya,za〉To ensure the outward normal orientation, we pick the “+” sign:n=〈xa,ya,za〉.An example of a vector fieldFsuch thatF·n=x2z2isF=〈axz2,0,0〉.divF=∂∂x(axz2)+∂∂y(0)+∂∂z(0)=az2Using spherical coordinates withdV=ρ2sinφdρdφdθwe haveRHS=∫∫∫VdivFdV=a∫2π0∫π0∫a0(ρcosφ)2ρ2sinφdρdφdθ=a∫2π0dθ∫π0cos2φsinφdφ∫a0ρ4dρUsing substitution in the middle integralu=cosφ,we obtainRHS=a[θ]2π0[−cos3φ3]π0[ρ55]a0=a(2π)(13+13)(a55)=415πa6