Question Number 104060 by bramlex last updated on 19/Jul/20

evaluate ∫∫_s (xz+y^2 )dS where S is the surface described by x^2 +y^2 =16 , 0≤z≤3

Answered by john santu last updated on 19/Jul/20

∫∫_s f(x,y,z)dS = ∫∫_R f(x,y,z)∣r_u ^→ ×r_v ^→ ∣dudv ⇒x=4cos θ , y=4sin θ r^→ (θ,z) = (((4cos θ)),((4sin θ)),(( z)) ) ,0≤θ≤2π , 0≤z≤3 . r_θ ^→ = (((−4sin θ)),(( 4 cos θ)),(( 0)) ) r_z ^→ = ((0),(0),(1) ) ; r_θ ^→ ×r_z ^→ = determinant (((−4sin θ 4cos θ 0)),(( 0 0 1)))= (4cos θ , 4sin θ ,0 ) ; ∣r_θ ^→ ×r_z ^→ ∣ = 4 S= ∫_0 ^(2π) ∫_0 ^3 [4z cos θ+16sin ^2 θ] 4 dz dθ S= ∫_0 ^(2π) 4{(2z^2 cos θ+16z.sin ^2 θ)}_0 ^3 dθ S= ∫_0 ^(2π) 4(18cos θ+48sin ^2 θ )dθ S = 24∫_0 ^(2π) (3cos θ+8sin ^2 θ) dθ S = 24∫_0 ^(2π) (3cos θ + 4−4cos 2θ) dθ S= 24 { 3sin θ+4θ−2sin 2θ} _0^(2π) S = 192π . (JS ⊛)