evaluate-s-xz-y-2-dS-where-S-is-the-surface-described-by-x-2-y-2-16-0-z-3-

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

\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 ⊛)