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Question Number 128837    Answers: 0   Comments: 1

Question Number 128823    Answers: 0   Comments: 1

Question Number 128691    Answers: 0   Comments: 1

find∫∫_S ▽×F .n^ ds where F^→ =y^2 i^ +yj^ −xzk^ and S is the upper half of the sphere x^2 +y^2 +z^2 =a^2

$${find}\int\int_{{S}} \bigtriangledown×{F}\:.\hat {{n}ds} \\ $$$${where}\:{F}^{\rightarrow} ={y}^{\mathrm{2}} \hat {{i}}+{y}\hat {{j}}−{xz}\hat {{k}} \\ $$$${and}\:{S}\:{is}\:{the}\:{upper}\:{half}\:{of}\:{the}\:{sphere} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$

Question Number 128688    Answers: 0   Comments: 0

find the flux of the vector field F=xi^ +yj^ +(√(x^2 +y^2 −1 k^ )) through outer side of hyper−boloide z=(√(x^2 +y^2 −1)) bounded by the planes z=0 and z=(√3)

$${find}\:{the}\:{flux}\:{of}\:{the}\:{vector}\:{field} \\ $$$${F}={x}\hat {{i}}+{y}\hat {{j}}+\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}\:\hat {{k}}} \\ $$$${through}\:{outer}\:{side}\:{of}\:{hyper}−{boloide} \\ $$$${z}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}} \\ $$$${bounded}\:{by}\:{the}\:{planes}\: \\ $$$${z}=\mathrm{0}\:{and}\:{z}=\sqrt{\mathrm{3}} \\ $$

Question Number 128592    Answers: 1   Comments: 0

∫∫∫_V ▽×F dV where F=(x+2y)i^ −3zj^ +xk^ and V is the closed region in first octant by the plane 2x+2y+2z=4

$$\int\int\int_{{V}} \bigtriangledown×{F}\:{dV}\:{where}\:{F}=\left({x}+\mathrm{2}{y}\right)\hat {{i}}−\mathrm{3}{z}\hat {{j}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{x}\hat {{k}} \\ $$$${and}\:{V}\:{is}\:{the}\:{closed}\:{region}\:{in}\:{first}\:{octant} \\ $$$${by}\:{the}\:{plane}\:\mathrm{2}{x}+\mathrm{2}{y}+\mathrm{2}{z}=\mathrm{4} \\ $$$$ \\ $$

Question Number 128591    Answers: 0   Comments: 1

verify the gauss divergence theorem f=(x^2 −yz)i^ +(y^2 −zx)j^ +(z^2 −xy)k^ over the region R bounded by the parallelepiped 0≤x≤a,0≤y≤b, 0≤z≤c

$${verify}\:{the}\:{gauss}\:{divergence}\:{theorem} \\ $$$${f}=\left({x}^{\mathrm{2}} −{yz}\right)\hat {{i}}+\left({y}^{\mathrm{2}} −{zx}\right)\hat {\mathrm{j}}+\left({z}^{\mathrm{2}} −{xy}\right)\hat {{k}} \\ $$$${over}\:{the}\:{region}\:{R}\:{bounded}\:{by}\:{the}\: \\ $$$$ \\ $$$${parallelepiped}\:\mathrm{0}\leqslant{x}\leqslant{a},\mathrm{0}\leqslant{y}\leqslant{b}, \\ $$$$\mathrm{0}\leqslant{z}\leqslant{c} \\ $$

Question Number 128282    Answers: 1   Comments: 0

Question Number 128234    Answers: 0   Comments: 0

Question Number 126646    Answers: 1   Comments: 1

Question Number 126572    Answers: 1   Comments: 0

Question Number 126443    Answers: 0   Comments: 0

Question Number 125672    Answers: 1   Comments: 0

N=x32y in base 10. N≡0[3] and N≡0[4]. N>8329 N has four digits. Determinate N.

$${N}={x}\mathrm{32}{y}\:{in}\:{base}\:\mathrm{10}.\: \\ $$$${N}\equiv\mathrm{0}\left[\mathrm{3}\right]\:{and}\:{N}\equiv\mathrm{0}\left[\mathrm{4}\right]. \\ $$$${N}>\mathrm{8329} \\ $$$${N}\:{has}\:{four}\:{digits}. \\ $$$${Determinate}\:{N}. \\ $$

Question Number 124682    Answers: 0   Comments: 0

Given { ((u_0 =5)),(( u_(n+1) =3u_n −4)) :} 1. show that ∀ n∈N, u_n =2+3^u_(n+1) a. Deduct that u_n is odd. c. Show that GCD(u_n ;u_(n+1) )=1 d. Deduct GCD(6+3^(1002) ;6+3^(1003) ). GCD=greatest common divisor^

$${Given}\:\begin{cases}{{u}_{\mathrm{0}} =\mathrm{5}}\\{\:{u}_{{n}+\mathrm{1}} =\mathrm{3}{u}_{{n}} −\mathrm{4}}\end{cases} \\ $$$$\mathrm{1}.\:{show}\:{that}\:\forall\:{n}\in\mathbb{N},\:{u}_{{n}} =\mathrm{2}+\mathrm{3}^{{u}_{{n}+\mathrm{1}} } \:\:\: \\ $$$${a}.\:{Deduct}\:{that}\:{u}_{{n}} \:{is}\:{odd}. \\ $$$${c}.\:{Show}\:{that}\:{GCD}\left({u}_{{n}} ;{u}_{{n}+\mathrm{1}} \right)=\mathrm{1} \\ $$$${d}.\:{Deduct}\:{GCD}\left(\mathrm{6}+\mathrm{3}^{\mathrm{1002}} ;\mathrm{6}+\mathrm{3}^{\mathrm{1003}} \right). \\ $$$$ \\ $$$${GCD}={greatest}\:{common}\:{divisor}^{} \\ $$

Question Number 122358    Answers: 1   Comments: 0

Given a, b ∈[0;4] 309a+15c=226b 1) show that 2b≡0[3] and 3a≡1[5]

$${Given}\:{a},\:{b}\:\in\left[\mathrm{0};\mathrm{4}\right] \\ $$$$\mathrm{309}{a}+\mathrm{15}{c}=\mathrm{226}{b} \\ $$$$\left.\mathrm{1}\right)\:{show}\:{that}\:\mathrm{2}{b}\equiv\mathrm{0}\left[\mathrm{3}\right]\:{and}\:\mathrm{3}{a}\equiv\mathrm{1}\left[\mathrm{5}\right] \\ $$$$ \\ $$

Question Number 122189    Answers: 0   Comments: 0

Given that the forces F_1 , F_2 and F_3 have position vectors r_1 , r_2 and r_3 . Where, F_1 = (2i + 3i + k) N r_1 = (i + 2k) m F_2 = (3i + 2i + 5k) N r_2 = (2i − 4j + k) m F_3 = (−5i−5j −6k) N r_3 = (4i + 3j + 2k) m (a) Find the moment of F_1 about (2i + 3j + 4k) m. (b) Show that this system of forces reduces to a couple, G, and find the magnitude of this couple. (c) Check if F_1 and F_2 are concurrent.

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{forces}\:\boldsymbol{\mathrm{F}}_{\mathrm{1}} ,\:\boldsymbol{\mathrm{F}}_{\mathrm{2}} \:\mathrm{and}\:\boldsymbol{\mathrm{F}}_{\mathrm{3}} \:\mathrm{have}\:\mathrm{position}\:\mathrm{vectors} \\ $$$$\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} ,\:\boldsymbol{\mathrm{r}}_{\mathrm{2}} \:\mathrm{and}\:\boldsymbol{\mathrm{r}}_{\mathrm{3}} .\:\mathrm{Where}, \\ $$$$\:\boldsymbol{\mathrm{F}}_{\mathrm{1}} \:=\:\left(\mathrm{2}\boldsymbol{\mathrm{i}}\:+\:\mathrm{3}\boldsymbol{\mathrm{i}}\:+\:\boldsymbol{\mathrm{k}}\right)\:\mathrm{N}\:\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} \:=\:\left(\boldsymbol{\mathrm{i}}\:+\:\mathrm{2}\boldsymbol{\mathrm{k}}\right)\:\mathrm{m} \\ $$$$\:\:\boldsymbol{\mathrm{F}}_{\mathrm{2}} \:=\:\left(\mathrm{3}\boldsymbol{\mathrm{i}}\:+\:\mathrm{2}\boldsymbol{\mathrm{i}}\:+\:\mathrm{5}\boldsymbol{\mathrm{k}}\right)\:\mathrm{N}\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{2}} \:=\:\left(\mathrm{2}\boldsymbol{\mathrm{i}}\:−\:\mathrm{4}\boldsymbol{\mathrm{j}}\:+\:\boldsymbol{\mathrm{k}}\right)\:\mathrm{m} \\ $$$$\:\boldsymbol{\mathrm{F}}_{\mathrm{3}} \:=\:\left(−\mathrm{5}\boldsymbol{\mathrm{i}}−\mathrm{5}\boldsymbol{\mathrm{j}}\:−\mathrm{6}\boldsymbol{\mathrm{k}}\right)\:\mathrm{N}\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{3}} \:=\:\left(\mathrm{4}\boldsymbol{\mathrm{i}}\:+\:\mathrm{3}\boldsymbol{\mathrm{j}}\:+\:\mathrm{2}\boldsymbol{\mathrm{k}}\right)\:\mathrm{m} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{moment}\:\mathrm{of}\:\boldsymbol{\mathrm{F}}_{\mathrm{1}} \:\mathrm{about}\:\left(\mathrm{2}\boldsymbol{\mathrm{i}}\:+\:\mathrm{3}\boldsymbol{\mathrm{j}}\:+\:\mathrm{4}\boldsymbol{\mathrm{k}}\right)\:\mathrm{m}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{this}\:\mathrm{system}\:\mathrm{of}\:\mathrm{forces}\:\mathrm{reduces}\:\mathrm{to}\:\mathrm{a}\:\mathrm{couple},\:\boldsymbol{\mathrm{G}},\: \\ $$$$\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{this}\:\mathrm{couple}. \\ $$$$\left(\mathrm{c}\right)\:\mathrm{Check}\:\mathrm{if}\:\boldsymbol{\mathrm{F}}_{\mathrm{1}} \:\mathrm{and}\:\boldsymbol{\mathrm{F}}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{concurrent}. \\ $$

Question Number 121188    Answers: 0   Comments: 0

N is written 158b687a in base 10. with a<b. show that N≡2+a[4]

$${N}\:{is}\:{written}\:\mathrm{158}{b}\mathrm{687}{a}\:{in}\:{base}\:\mathrm{10}. \\ $$$${with}\:\:{a}<{b}. \\ $$$${show}\:{that}\:{N}\equiv\mathrm{2}+{a}\left[\mathrm{4}\right] \\ $$

Question Number 119184    Answers: 0   Comments: 0

Nama : Attila Abiem Farhan Kelas : XI BKP Tugas KD 3.18 MTK 1. jawab : u^→ = (((−3)),((−4)),(( 12)) ) ∣u^→ ∣=(√((−3^2 )+(−4)^2 +12^2 )) =(√(9+16+144)) =(√(169))=13 2. jawab : a^→ = ((2),((−1)),(3) ) b^→ = ((7),((13)),(5) ) a^→ +b^→ = ((9),((12)),(8) ) ∣a^→ +b^→ ∣=(√(9^2 +12^2 +8^2 )) =(√(81+144+64)) =(√(289)) 3. jawab u^→ = ((7),(9),((−17)) ) v^→ = (((−2)),((−3)),((19)) ) u^→ −v^→ = ((9),((12)),((36)) ) ∣u^→ +v^→ ∣=(√(9^2 +12^2 +36^2 )) =(√(81+144+1296)) =(√(1521)) 4. jawab : A. 2a^→ = (((−6)),((−8)),((−24)) ) ∣2a^→ ∣=(√((−6)^2 +(−8)^2 +(−24)^2 )) =(√(36+64+576)) =(√(676)) =26 (1/2)b^→ = ((3),(4),((12)) ) ∣(1/2)b^→ ∣=(√(3^2 +4^2 +12^2 ))=(√(9+16+144))=(√(169))=13 B. 4a^→ +b^→ = (((−12)),((−16)),((−48)) ) + ((6),(8),((24)) ) = (((−6)),((−8)),((−24)) ) ∣4a^→ +b^→ ∣=(√((−6)^2 +(−8)^2 +(−24)^2 )) =(√(36+64+576)) =(√(676))=26 5. jawab : u^→ =8i+3k v^→ =5j−9k u^→ .v^→ =(8.0)i+(0.5)j+(3.−9)k =0+0−27k u^→ .v^→ =−27k

$${Nama}\::\:{Attila}\:{Abiem}\:{Farhan} \\ $$$${Kelas}\::\:{XI}\:{BKP} \\ $$$${Tugas}\:{KD}\:\mathrm{3}.\mathrm{18}\:{MTK} \\ $$$$\mathrm{1}.\:{jawab}\:: \\ $$$$\:\overset{\rightarrow} {{u}}=\begin{pmatrix}{−\mathrm{3}}\\{−\mathrm{4}}\\{\:\:\mathrm{12}}\end{pmatrix}\: \\ $$$$\mid\overset{\rightarrow} {{u}}\mid=\sqrt{\left(−\mathrm{3}^{\mathrm{2}} \right)+\left(−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:=\sqrt{\mathrm{9}+\mathrm{16}+\mathrm{144}} \\ $$$$\:\:\:\:\:\:=\sqrt{\mathrm{169}}=\mathrm{13} \\ $$$$\mathrm{2}.\:{jawab}\:: \\ $$$$\overset{\rightarrow} {{a}}=\begin{pmatrix}{\mathrm{2}}\\{−\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:\overset{\rightarrow} {{b}}=\begin{pmatrix}{\mathrm{7}}\\{\mathrm{13}}\\{\mathrm{5}}\end{pmatrix}\: \\ $$$$\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}=\begin{pmatrix}{\mathrm{9}}\\{\mathrm{12}}\\{\mathrm{8}}\end{pmatrix} \\ $$$$\mid\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}\mid=\sqrt{\mathrm{9}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{81}+\mathrm{144}+\mathrm{64}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{289}} \\ $$$$\mathrm{3}.\:{jawab} \\ $$$$\overset{\rightarrow} {{u}}=\begin{pmatrix}{\mathrm{7}}\\{\mathrm{9}}\\{−\mathrm{17}}\end{pmatrix}\:\overset{\rightarrow} {{v}}=\begin{pmatrix}{−\mathrm{2}}\\{−\mathrm{3}}\\{\mathrm{19}}\end{pmatrix} \\ $$$$\overset{\rightarrow} {{u}}−\overset{\rightarrow} {{v}}=\begin{pmatrix}{\mathrm{9}}\\{\mathrm{12}}\\{\mathrm{36}}\end{pmatrix} \\ $$$$\mid\overset{\rightarrow} {{u}}+\overset{\rightarrow} {{v}}\mid=\sqrt{\mathrm{9}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} +\mathrm{36}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{81}+\mathrm{144}+\mathrm{1296}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{1521}} \\ $$$$\mathrm{4}.\:{jawab}\::\: \\ $$$${A}.\:\:\:\mathrm{2}\overset{\rightarrow} {{a}}=\begin{pmatrix}{−\mathrm{6}}\\{−\mathrm{8}}\\{−\mathrm{24}}\end{pmatrix}\: \\ $$$$\:\:\:\:\:\:\:\mid\mathrm{2}\overset{\rightarrow} {{a}}\mid=\sqrt{\left(−\mathrm{6}\right)^{\mathrm{2}} +\left(−\mathrm{8}\right)^{\mathrm{2}} +\left(−\mathrm{24}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{36}+\mathrm{64}+\mathrm{576}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{676}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{26} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\overset{\rightarrow} {{b}}=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{4}}\\{\mathrm{12}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\mid\frac{\mathrm{1}}{\mathrm{2}}\overset{\rightarrow} {{b}}\mid=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\sqrt{\mathrm{9}+\mathrm{16}+\mathrm{144}}=\sqrt{\mathrm{169}}=\mathrm{13} \\ $$$$ \\ $$$${B}.\:\:\:\:\mathrm{4}\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}=\begin{pmatrix}{−\mathrm{12}}\\{−\mathrm{16}}\\{−\mathrm{48}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{6}}\\{\mathrm{8}}\\{\mathrm{24}}\end{pmatrix}\:=\:\begin{pmatrix}{−\mathrm{6}}\\{−\mathrm{8}}\\{−\mathrm{24}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\mid\mathrm{4}\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}\mid=\sqrt{\left(−\mathrm{6}\right)^{\mathrm{2}} +\left(−\mathrm{8}\right)^{\mathrm{2}} +\left(−\mathrm{24}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{36}+\mathrm{64}+\mathrm{576}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{676}}=\mathrm{26} \\ $$$$\mathrm{5}.\:{jawab}\::\: \\ $$$$\overset{\rightarrow} {{u}}=\mathrm{8}{i}+\mathrm{3}{k} \\ $$$$\overset{\rightarrow} {{v}}=\mathrm{5}{j}−\mathrm{9}{k} \\ $$$$\overset{\rightarrow} {{u}}.\overset{\rightarrow} {{v}}=\left(\mathrm{8}.\mathrm{0}\right){i}+\left(\mathrm{0}.\mathrm{5}\right){j}+\left(\mathrm{3}.−\mathrm{9}\right)\mathrm{k} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{0}+\mathrm{0}−\mathrm{27}{k} \\ $$$$\overset{\rightarrow} {{u}}.\overset{\rightarrow} {{v}}=−\mathrm{27}{k} \\ $$

Question Number 119054    Answers: 2   Comments: 0

Show by recurence that: 5^n ≥1+4n ; n∈N

$${Show}\:{by}\:{recurence}\:{that}: \\ $$$$\mathrm{5}^{{n}} \geqslant\mathrm{1}+\mathrm{4}{n}\:;\:{n}\in\mathbb{N} \\ $$

Question Number 118747    Answers: 4   Comments: 0

find the distance of point (2,1,−2) to plane passing through points (−1,2,−3); (0,−4,−2) and (1,3,4).

$${find}\:{the}\:{distance}\:{of}\:{point}\:\left(\mathrm{2},\mathrm{1},−\mathrm{2}\right)\:{to}\:{plane} \\ $$$${passing}\:{through}\:{points}\:\left(−\mathrm{1},\mathrm{2},−\mathrm{3}\right); \\ $$$$\left(\mathrm{0},−\mathrm{4},−\mathrm{2}\right)\:{and}\:\left(\mathrm{1},\mathrm{3},\mathrm{4}\right). \\ $$

Question Number 118017    Answers: 0   Comments: 0

Question Number 117493    Answers: 1   Comments: 0

∫_c (3xy−e^(sin x) )dx+(7x+(√(y^4 +1)) )dy C : triangle with vertex (0,0),(0,1) and (1,0)

$$\int_{\mathrm{c}} \left(\mathrm{3xy}−\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} \right)\mathrm{dx}+\left(\mathrm{7x}+\sqrt{\mathrm{y}^{\mathrm{4}} +\mathrm{1}}\:\right)\mathrm{dy} \\ $$$$\mathrm{C}\::\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{vertex}\:\left(\mathrm{0},\mathrm{0}\right),\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\mathrm{and}\:\left(\mathrm{1},\mathrm{0}\right) \\ $$

Question Number 116094    Answers: 1   Comments: 0

Question Number 114043    Answers: 1   Comments: 0

Σ_(n=1) ^∞ (1/(n^3 +1))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} +\mathrm{1}} \\ $$

Question Number 110320    Answers: 3   Comments: 0

find the point of intersection of the line r^→ =(1−2t,3+4t,t) and the plane 3x−2y+5z=15

$${find}\:{the}\:{point}\:{of}\:{intersection} \\ $$$${of}\:{the}\:{line}\:\overset{\rightarrow} {{r}}=\left(\mathrm{1}−\mathrm{2}{t},\mathrm{3}+\mathrm{4}{t},{t}\right) \\ $$$${and}\:{the}\:{plane}\:\mathrm{3}{x}−\mathrm{2}{y}+\mathrm{5}{z}=\mathrm{15}\: \\ $$

Question Number 108104    Answers: 1   Comments: 0

Question Number 104060    Answers: 1   Comments: 0

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

$${evaluate}\:\int\int_{{s}} \left({xz}+{y}^{\mathrm{2}} \right){dS}\:{where} \\ $$$${S}\:{is}\:{the}\:{surface}\:{described}\:{by}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{16} \\ $$$$,\:\mathrm{0}\leqslant{z}\leqslant\mathrm{3} \\ $$

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