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


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

$e v a l u a t e \int \int_{s} (x z + y^{2}) d S w h e r e$ $S i s t h e s u r f a c e d e s c r i b e d b y x^{2} + y^{2} = 16$ $, 0 ⩽ z ⩽ 3$
	Answered by john santu last updated on 19/Jul/20
	$∫∫_s f(x,y,z)dS = ∫∫_R f(x,y,z)∣r_u ^→ ×r_v ^→ ∣dudv ⇒x=4cos θ , y=4sin θ r^→ (θ,z) = (((4cos θ)),((4sin θ)),(( z)) ) ,0≤θ≤2π , 0≤z≤3 . r_θ ^→ = (((−4sin θ)),(( 4 cos θ)),(( 0)) ) r_z ^→ = ((0),(0),(1) ) ; r_θ ^→ ×r_z ^→ = determinant (((−4sin θ 4cos θ 0)),(( 0 0 1)))= (4cos θ , 4sin θ ,0 ) ; ∣r_θ ^→ ×r_z ^→ ∣ = 4 S= ∫_0 ^(2π) ∫_0 ^3 [4z cos θ+16sin^2 θ] 4 dz dθ S= ∫_0 ^(2π) 4{(2z^2 cos θ+16z.sin^2 θ)}_0 ^3 dθ S= ∫_0 ^(2π) 4(18cos θ+48sin^2 θ )dθ S = 24∫_0 ^(2π) (3cos θ+8sin^2 θ) dθ S = 24∫_0 ^(2π) (3cos θ + 4−4cos 2θ) dθ S= 24 { 3sin θ+4θ−2sin 2θ} _0^(2π) S = 192π . (JS ⊛)$
	$\int \int_{s} f (x, y, z) d S = \int \int_{R} f (x, y, z) ∣ {\vec{r}}_{u} \times {\vec{r}}_{v} ∣ d u d v$ $\Rightarrow x = 4 \cos θ, y = 4 \sin θ$ $\vec{r} (θ, z) = (\begin{matrix} 4 \cos θ \\ 4 \sin θ \\ z \end{matrix}), 0 ⩽ θ ⩽ 2 π,$ $0 ⩽ z ⩽ 3 . {\vec{r}}_{θ} = (\begin{matrix} - 4 \sin θ \\ 4 \cos θ \\ 0 \end{matrix})$ ${\vec{r}}_{z} = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}); {\vec{r}}_{θ} \times {\vec{r}}_{z} =$ $\| \begin{matrix} - 4 \sin θ 4 \cos θ 0 \\ 0 0 1 \end{matrix} \| = (4 \cos θ$ $, 4 \sin θ, 0); ∣ {\vec{r}}_{θ} \times {\vec{r}}_{z} ∣ = 4$ $S = \underset{0}{\int^{2 π}} \underset{0}{\int^{3}} [4 z \cos θ + 16 \sin^{2} θ] 4 d z d θ$ $S = \underset{0}{\int^{2 π}} 4 {(2 z^{2} \cos θ + 16 z . \sin^{2} θ)}_{0}^{3} d θ$ $S = \underset{0}{\int^{2 π}} 4 (18 \cos θ + 48 \sin^{2} θ) d θ$ $S = 24 \underset{0}{\int^{2 π}} (3 \cos θ + 8 \sin^{2} θ) d θ$ $S = 24 \underset{0}{\int^{2 π}} (3 \cos θ + 4 - 4 \cos 2 θ) d θ$ $S = 24 {3 \sin θ + 4 θ - 2 \sin 2 θ}_{0}^{2 π}$ $S = 192 π . (J S ⊛)$
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