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 ⊛)