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Question Number 219340 Answers: 1 Comments: 0

F^→ (x,y,z)=−(x/( (√(x^2 +y^2 +z^2 ))))e_1 ^→ −(y/( (√(x^2 +y^2 +z^2 ))))e_2 ^→ −(z/( (√(x^2 +y^2 +z^2 ))))e_3 ^→ x^2 +y^2 +z^2 =R^2 ∫∫_( S) F^→ ∙dS^→ =??

$$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y},{z}\right)=−\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −\frac{{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} −\frac{{z}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\int\int_{\:\boldsymbol{\mathcal{S}}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}=?? \\ $$

Question Number 219339 Answers: 4 Comments: 0

Question Number 219338 Answers: 0 Comments: 0

F^→ =−xe_1 ^→ −ye_2 ^→ −ze_3 ^→ S^→ (u,v)= { (((2+3sin(u))cos(v))),(((2+3sin(v))sin(u))),((3cos(u))) :} u∈[0,2π] , v∈[0,2π] ∫∫_( S) F^→ ∙dS^→ =???

$$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}=−{x}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −{y}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} −{z}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \\ $$$$\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}\left({u},{v}\right)=\begin{cases}{\left(\mathrm{2}+\mathrm{3sin}\left({u}\right)\right)\mathrm{cos}\left({v}\right)}\\{\left(\mathrm{2}+\mathrm{3sin}\left({v}\right)\right)\mathrm{sin}\left({u}\right)}\\{\mathrm{3cos}\left({u}\right)}\end{cases} \\ $$$${u}\in\left[\mathrm{0},\mathrm{2}\pi\right]\:,\:{v}\in\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$$$\int\int_{\:\boldsymbol{\mathcal{S}}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}=??? \\ $$

Question Number 219337 Answers: 5 Comments: 0

Question Number 219336 Answers: 0 Comments: 0

Question Number 219333 Answers: 0 Comments: 0

F^→ (x,y,z)=−xye_1 ^→ +yze_2 ^→ −xye_3 ^→ S^→ (u,v) { (((2+v∙cos(u))sin(2πv))),((v∙cos(u))),(((2+v∙cos(u))cos(2πv)+2v−2)) :} u∈[−π,π] , v∈[0,(π/2)] ∫∫_( S) F^→ ∙dS^→ =??

$$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y},{z}\right)=−{xy}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} +{yz}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} −{xy}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \\ $$$$\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}\left({u},{v}\right)\begin{cases}{\left(\mathrm{2}+{v}\centerdot\mathrm{cos}\left({u}\right)\right)\mathrm{sin}\left(\mathrm{2}\pi{v}\right)}\\{{v}\centerdot\mathrm{cos}\left({u}\right)}\\{\left(\mathrm{2}+{v}\centerdot\mathrm{cos}\left({u}\right)\right)\mathrm{cos}\left(\mathrm{2}\pi{v}\right)+\mathrm{2}{v}−\mathrm{2}}\end{cases} \\ $$$${u}\in\left[−\pi,\pi\right]\:,\:{v}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right] \\ $$$$\int\int_{\:\boldsymbol{\mathcal{S}}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}=?? \\ $$

Question Number 219323 Answers: 0 Comments: 0

f(s)=(1/(2π)) ∫ e^(−it(s−α)) dt ∫_0 ^( ∞) ∫_(−∞) ^( +∞) e^(−it(s−α)) e^(−sp) dtds=?

$${f}\left({s}\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int\:\:{e}^{−\boldsymbol{{i}}{t}\left({s}−\alpha\right)} \:\mathrm{d}{t}\: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \int_{−\infty} ^{\:+\infty} \:\:{e}^{−\boldsymbol{{i}}{t}\left({s}−\alpha\right)} {e}^{−{sp}} \mathrm{d}{t}\mathrm{d}{s}=? \\ $$

Question Number 219318 Answers: 1 Comments: 0

Question Number 219315 Answers: 1 Comments: 0

Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^4 ))=?

$$ \\ $$$$\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{4}} }=? \\ $$

Question Number 219316 Answers: 1 Comments: 0

Prove; ∫^( ∞) _( 0) (1/(2^2^(⌊x⌋) + {x})) dx = ln2

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{Prove}; \\ $$$$\:\:\:\:\:\:\:\underset{\:\mathrm{0}} {\int}^{\:\infty} \frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}^{\lfloor\boldsymbol{{x}}\rfloor} } \:+\:\left\{{x}\right\}}\:{dx}\:=\:{ln}\mathrm{2}\:\:\: \\ $$

Question Number 219305 Answers: 1 Comments: 0

Prove; I_0 (x) =(1/π)∫_0 ^( π) e^( x cox(θ)) dθ ; x^2 I_0 ^(′′) (x) + xI′_0 (x) − x^2 I_0 (x) = 0;

$$ \\ $$$$\:\:\:\:\:\:{Prove}; \\ $$$$\:\:\:{I}_{\mathrm{0}} \left({x}\right)\:=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\:\pi} \:{e}^{\:{x}\:{cox}\left(\theta\right)} \:{d}\theta\:; \\ $$$$\:\:\:{x}^{\mathrm{2}} {I}_{\mathrm{0}} ^{''} \left({x}\right)\:+\:{xI}'_{\mathrm{0}} \left({x}\right)\:−\:{x}^{\mathrm{2}} {I}_{\mathrm{0}} \left({x}\right)\:=\:\mathrm{0}; \\ $$$$\: \\ $$

Question Number 219304 Answers: 0 Comments: 0

Question Number 219301 Answers: 0 Comments: 0

f(s)=(1/(2π)) ∫_(−∞) ^(+∞) e^(−iω(s−α)) dω ∫_0 ^∞ (1/(2π))[e^(−st) ∫_( −∞) ^( +∞) e^(−iω(s−α)) dω]ds=....?

$${f}\left({s}\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\boldsymbol{{i}}\omega\left({s}−\alpha\right)} \mathrm{d}\omega\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{2}\pi}\left[{e}^{−{st}} \:\int_{\:−\infty} ^{\:+\infty} \:\:{e}^{−\boldsymbol{{i}}\omega\left({s}−\alpha\right)} \mathrm{d}\omega\right]\mathrm{d}{s}=....? \\ $$

Question Number 219293 Answers: 1 Comments: 7

Question Number 219289 Answers: 2 Comments: 1

if a,b,c ∈ Z , and a^2 + b^2 = c^2 , then 3∣(ab) = ?

$$ \\ $$$$\:\:\:\:{if}\:\:\:\:\:\:\:\:\:\:\:{a},{b},{c}\:\in\:\mathbb{Z}\:\:\:, \\ $$$$\:\:\:\:\:{and}\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:=\:\:{c}^{\mathrm{2}} \:\:\:, \\ $$$$\:\:\:\:\:\:\:\:{then}\:\:\:\:\:\:\mathrm{3}\mid\left({ab}\right)\:=\:? \\ $$$$ \\ $$

Question Number 219281 Answers: 1 Comments: 0

Question Number 219283 Answers: 1 Comments: 0

Question Number 219279 Answers: 2 Comments: 0

Question Number 219278 Answers: 0 Comments: 0

Question Number 219268 Answers: 1 Comments: 0

f(x,y)=ln∫_0 ^(x^2 +y^2 ) e^t^2 dt,f(x)(1,2)=?

$${f}\left({x},{y}\right)=\mathrm{ln}\int_{\mathrm{0}} ^{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } {e}^{{t}^{\mathrm{2}} } {dt},{f}\left({x}\right)\left(\mathrm{1},\mathrm{2}\right)=? \\ $$

Question Number 219267 Answers: 1 Comments: 0

Une fonction P est dite quasi polynomiale s′il existe (pour k∈N ) k+1 fonction periodique(c_i )_(i∈[∣0;k∣]) de Z dans R telles que P(n)=Σ_(k=1) ^n c_i (n)n^i (1) Montrez que l′ensemble des fonction quasi polynomiale forme un R−ev(real space vector). (2)Montrez que si P,Q:Z→R sont desfonction quasi polynomiale tel que P(n)=Q(n) ∀n∈N alors P=Q

$${Une}\:{fonction}\:{P}\:{est}\:{dite}\:{quasi}\:{polynomiale}\:{s}'{il}\:{existe}\:\left({pour}\:{k}\in\mathbb{N}\:\right)\:{k}+\mathrm{1}\:{fonction}\:{periodique}\left({c}_{{i}} \right)_{{i}\in\left[\mid\mathrm{0};{k}\mid\right]} {de}\:\mathbb{Z}\:{dans}\:\mathbb{R} \\ $$$$\:{telles}\:{que}\:{P}\left({n}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{c}_{{i}} \left({n}\right){n}^{{i}} \\ $$$$\left(\mathrm{1}\right)\:{Montrez}\:{que}\:{l}'{ensemble}\:{des}\:{fonction}\:{quasi}\:{polynomiale}\:{forme}\:{un}\:\mathbb{R}−{ev}\left({real}\:{space}\:{vector}\right). \\ $$$$\left(\mathrm{2}\right){Montrez}\:{que}\:{si}\:{P},{Q}:\mathbb{Z}\rightarrow\mathbb{R}\:{sont}\:{desfonction}\:{quasi}\:{polynomiale}\:{tel}\:{que}\:{P}\left({n}\right)={Q}\left({n}\right)\:\forall{n}\in\mathbb{N}\:{alors}\:{P}={Q} \\ $$

Question Number 219262 Answers: 0 Comments: 0

E^ lectric field strenth at any point in the space is defined as the force per unit charge at that point. It is a vector quantity whose magnitude is given by Coulomb^(s ) law and diection is in straight line loining the at that point. mathemstically

$$\overset{} {\mathrm{E}lectric}\:\mathrm{field}\:\mathrm{strenth}\:\mathrm{at}\:\mathrm{any}\:\mathrm{point}\:\mathrm{in}\:\mathrm{the}\:\mathrm{space} \\ $$$$\mathrm{is}\:\mathrm{defined}\:\mathrm{as}\:\mathrm{the}\:\mathrm{force}\:\mathrm{per}\:\mathrm{unit}\:\mathrm{charge}\:\mathrm{at}\:\mathrm{that}\:\mathrm{point}. \\ $$$$\:\mathrm{It}\:\mathrm{is}\:\mathrm{a}\:\mathrm{vector}\:\mathrm{quantity}\:\mathrm{whose}\:\mathrm{magnitude}\:\mathrm{is} \\ $$$$\mathrm{given}\:\mathrm{by}\:\mathrm{Coulomb}^{\mathrm{s}\:\:} \:\mathrm{law}\:\mathrm{and}\:\mathrm{diection}\:\mathrm{is}\:\mathrm{in}\: \\ $$$$\mathrm{straight}\:\mathrm{line}\:\mathrm{loining}\:\mathrm{the}\:\mathrm{at}\:\mathrm{that}\:\mathrm{point}. \\ $$$$\mathrm{mathemstically} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 219255 Answers: 6 Comments: 0

Question Number 219254 Answers: 4 Comments: 0

Question Number 219243 Answers: 3 Comments: 2

Question Number 219236 Answers: 1 Comments: 0

f(t) = (1/(2πi)) ∫_( c−i∞) ^( c+i∞) (e^(st) /(s^k )) ds , k ∈C

$$ \\ $$$$\:\:\:\:{f}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\:\int_{\:{c}−{i}\infty} ^{\:{c}+{i}\infty} \:\frac{{e}^{{st}} }{{s}^{{k}} \:}\:\:{ds}\:\:\:,\:\:{k}\:\in\mathbb{C} \\ $$$$\: \\ $$

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