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Question Number 213139 by Spillover last updated on 31/Oct/24
Answered by MrGaster last updated on 31/Oct/24
let u=(√x)+(√y)+z⇒dy=(1/(2(√x)))dx+(1/(2(√y)))dy+(1/(2(√z)))dx ⇒0≤u≤((3(√π))/4) ⇒dxdydz=8u^2 du ⇔∫_0 ^((3(√π))/4) sin(u)∙8u^2 du⇒8∫_0 ^((3(√π))/4) u^2 sin(u)du let u=u^2 ,du=2udu,dw=sin(u)du,w=−cos(u) ⇒∫u^2 sin(u)du=−u^2 cos(u)+2∫u cos(u) let v=u,dv=du,dw=cos(u)du,w=sin(u) ⇒∫u cos(u)du=u sin(u)du=u sin(u)+cos(u) ⇒∫u^2 sin(u)du=−u^2 cos(u)+2(u sin(u)+cos(u)) 8−[−(((3(√π))/4))cos(((3(√π))/4))+((3(√π))/2)sin(((3(√π))/4))+2 cos(((3(√π))/4))+2 cos((3(√π))/4)] ⇒8[−((9π)/(16))cos(((3(√π))/4))+((3(√π))/2)sin(((3(√π))/4))+2cos(((3(√π))/4))] ⇔8[(2−((9π)/(16)))cos(((3(√π))/4))+((3(√π))/2)sin(((3(√π))/4))]
$${let}\:{u}=\sqrt{{x}}+\sqrt{{y}}+{z}\Rightarrow{dy}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}{dx}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{y}}}{dy}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{z}}}{dx} \\ $$$$\Rightarrow\mathrm{0}\leq{u}\leq\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}} \\ $$$$\Rightarrow{dxdydz}=\mathrm{8}{u}^{\mathrm{2}} {du} \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}} \mathrm{sin}\left({u}\right)\centerdot\mathrm{8}{u}^{\mathrm{2}} {du}\Rightarrow\mathrm{8}\int_{\mathrm{0}} ^{\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}} {u}^{\mathrm{2}} \mathrm{sin}\left({u}\right){du} \\ $$$${let}\:{u}={u}^{\mathrm{2}} ,{du}=\mathrm{2}{udu},{dw}=\mathrm{sin}\left({u}\right){du},{w}=−\mathrm{cos}\left({u}\right) \\ $$$$\Rightarrow\int{u}^{\mathrm{2}} \mathrm{sin}\left({u}\right){du}=−{u}^{\mathrm{2}} \mathrm{cos}\left({u}\right)+\mathrm{2}\int{u}\:\mathrm{cos}\left({u}\right) \\ $$$${let}\:{v}={u},{dv}={du},{dw}=\mathrm{cos}\left({u}\right){du},{w}=\mathrm{sin}\left({u}\right) \\ $$$$\Rightarrow\int{u}\:\mathrm{cos}\left({u}\right){du}={u}\:\mathrm{sin}\left({u}\right){du}={u}\:\mathrm{sin}\left({u}\right)+\mathrm{cos}\left({u}\right) \\ $$$$\Rightarrow\int{u}^{\mathrm{2}} \mathrm{sin}\left({u}\right){du}=−{u}^{\mathrm{2}} \mathrm{cos}\left({u}\right)+\mathrm{2}\left({u}\:\mathrm{sin}\left({u}\right)+\mathrm{cos}\left({u}\right)\right) \\ $$$$\mathrm{8}−\left[−\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)\mathrm{cos}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)+\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{2}}\mathrm{sin}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)+\mathrm{2}\:\mathrm{cos}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)+\mathrm{2}\:\mathrm{cos}\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right] \\ $$$$\Rightarrow\mathrm{8}\left[−\frac{\mathrm{9}\pi}{\mathrm{16}}\mathrm{cos}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)+\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{2}}\mathrm{sin}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)+\mathrm{2cos}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)\right] \\ $$$$\Leftrightarrow\mathrm{8}\left[\left(\mathrm{2}−\frac{\mathrm{9}\pi}{\mathrm{16}}\right)\mathrm{cos}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)+\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{2}}\mathrm{sin}\left(\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}\right)\right] \\ $$
Commented by Frix last updated on 31/Oct/24
No.
$$\mathrm{No}. \\ $$
Commented by MathematicalUser2357 last updated on 02/Nov/24
=21.097872
$$=\mathrm{21}.\mathrm{097872} \\ $$
Answered by Frix last updated on 31/Oct/24
Using (1) ∫sin (α+(√t)) dt =^([u=(√t)]) 2∫usin (α+u) du =^([by parts.]) =2(sin (α+u) −ucos (α+u))= =2(sin (α+(√t)) −(√t)cos (α+(√t))) (2) Similar ∫cos (α+(√t)) dt=2(cos (α+(√t)) +(√t)sin (α+(√t))) We get 6(4(√π)cos (√π)+(π−4)sin(√π))− −12((√π)cos ((√π)/2) −2sin ((√π)/2))+ +(√π)(π−12)cos ((3(√π))/2) −2(3π−4)sin ((3(√π))/2)
$$\mathrm{Using} \\ $$$$\left(\mathrm{1}\right) \\ $$$$\int\mathrm{sin}\:\left(\alpha+\sqrt{{t}}\right)\:{dt}\:\overset{\left[{u}=\sqrt{{t}}\right]} {=}\:\mathrm{2}\int{u}\mathrm{sin}\:\left(\alpha+{u}\right)\:{du}\:\overset{\left[\mathrm{by}\:\mathrm{parts}.\right]} {=} \\ $$$$=\mathrm{2}\left(\mathrm{sin}\:\left(\alpha+{u}\right)\:−{u}\mathrm{cos}\:\left(\alpha+{u}\right)\right)= \\ $$$$=\mathrm{2}\left(\mathrm{sin}\:\left(\alpha+\sqrt{{t}}\right)\:−\sqrt{{t}}\mathrm{cos}\:\left(\alpha+\sqrt{{t}}\right)\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Similar} \\ $$$$\int\mathrm{cos}\:\left(\alpha+\sqrt{{t}}\right)\:{dt}=\mathrm{2}\left(\mathrm{cos}\:\left(\alpha+\sqrt{{t}}\right)\:+\sqrt{{t}}\mathrm{sin}\:\left(\alpha+\sqrt{{t}}\right)\right) \\ $$$$\mathrm{We}\:\mathrm{get} \\ $$$$\mathrm{6}\left(\mathrm{4}\sqrt{\pi}\mathrm{cos}\:\sqrt{\pi}+\left(\pi−\mathrm{4}\right)\mathrm{sin}\sqrt{\pi}\right)− \\ $$$$\:\:\:\:\:−\mathrm{12}\left(\sqrt{\pi}\mathrm{cos}\:\frac{\sqrt{\pi}}{\mathrm{2}}\:−\mathrm{2sin}\:\frac{\sqrt{\pi}}{\mathrm{2}}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:+\sqrt{\pi}\left(\pi−\mathrm{12}\right)\mathrm{cos}\:\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{2}}\:−\mathrm{2}\left(\mathrm{3}\pi−\mathrm{4}\right)\mathrm{sin}\:\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{2}} \\ $$
Answered by lepuissantcedricjunior last updated on 01/Nov/24
k=∫_0 ^(𝛑/4) ∫_0 ^(𝛑/4) ∫_0 ^(𝛑/4) sin((√x)+(√y)+(√z))dxdydz =Im∫_0 ^(𝛑/4) ∫_0 ^(𝛑/4) ∫_0 ^(𝛑/4) e^(i((√x)+(√y)+(√z))) dxdydz =Im(∫_0 ^(𝛑/4) e^(i(√x)) dx∫_0 ^(𝛑/4) e^(i(√y)) dy∫_0 ^(𝛑/4) e^(i(√z)) dz) posons x=n^2 ;y=p^2 ;z=q^2 =Im[{∫^(√(𝛑/4)) _0 2ne^(in) dn}{∫_0 ^(√(𝛑/4)) 2pe^(ip) dp}{∫_0 ^(√(𝛑/4)) 2qe^(iq) dq}] =Im{[((2ne^(in) )/i)+2e^(in) +((2pe^(ip) )/i)+2e^(ip) +((2qe^(iq) )/i)+2e^(iq) ]_0 ^((√𝛑)/2) =Im[3((√𝛑)/i)e^(i((√𝛑)/2)) −6]=Im(−6+3(√𝛑)e^(i((((√𝛑)−𝛑)/2))) ) i=3(√𝛑)sin(((√𝛑)/2)−(𝛑/2))=−3(√π)cos(((√𝛑)/2))
$$\boldsymbol{{k}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \boldsymbol{{sin}}\left(\sqrt{\boldsymbol{{x}}}+\sqrt{\boldsymbol{{y}}}+\sqrt{\boldsymbol{{z}}}\right)\boldsymbol{{dxdydz}} \\ $$$$\:\:\:=\boldsymbol{{Im}}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \boldsymbol{{e}}^{\boldsymbol{{i}}\left(\sqrt{\boldsymbol{{x}}}+\sqrt{\boldsymbol{{y}}}+\sqrt{\boldsymbol{{z}}}\right)} \boldsymbol{{dxdydz}} \\ $$$$\:\:\:=\boldsymbol{{Im}}\left(\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \boldsymbol{{e}}^{\boldsymbol{{i}}\sqrt{\boldsymbol{{x}}}} \boldsymbol{{dx}}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \boldsymbol{{e}}^{\boldsymbol{{i}}\sqrt{\boldsymbol{{y}}}} \boldsymbol{{dy}}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \boldsymbol{{e}}^{\boldsymbol{{i}}\sqrt{\boldsymbol{{z}}}} \boldsymbol{{dz}}\right) \\ $$$$\boldsymbol{{posons}}\:\boldsymbol{{x}}=\boldsymbol{{n}}^{\mathrm{2}} ;\boldsymbol{{y}}=\boldsymbol{{p}}^{\mathrm{2}} ;\boldsymbol{{z}}=\boldsymbol{{q}}^{\mathrm{2}} \\ $$$$=\boldsymbol{{Im}}\left[\left\{\underset{\mathrm{0}} {\int}^{\sqrt{\frac{\boldsymbol{\pi}}{\mathrm{4}}}} \mathrm{2}\boldsymbol{{ne}}^{\boldsymbol{{in}}} \boldsymbol{{dn}}\right\}\left\{\int_{\mathrm{0}} ^{\sqrt{\frac{\boldsymbol{\pi}}{\mathrm{4}}}} \mathrm{2}\boldsymbol{{pe}}^{\boldsymbol{{ip}}} \boldsymbol{{dp}}\right\}\left\{\int_{\mathrm{0}} ^{\sqrt{\frac{\boldsymbol{\pi}}{\mathrm{4}}}} \mathrm{2}\boldsymbol{{qe}}^{\boldsymbol{{iq}}} \boldsymbol{{dq}}\right\}\right] \\ $$$$=\boldsymbol{{Im}}\left\{\left[\frac{\mathrm{2}\boldsymbol{{ne}}^{\boldsymbol{{in}}} }{\boldsymbol{{i}}}+\mathrm{2}\boldsymbol{{e}}^{\boldsymbol{{in}}} +\frac{\mathrm{2}\boldsymbol{{pe}}^{\boldsymbol{{ip}}} }{\boldsymbol{{i}}}+\mathrm{2}\boldsymbol{{e}}^{\boldsymbol{{ip}}} +\frac{\mathrm{2}\boldsymbol{{qe}}^{\boldsymbol{{iq}}} }{\boldsymbol{{i}}}+\mathrm{2}\boldsymbol{{e}}^{\boldsymbol{{iq}}} \right]_{\mathrm{0}} ^{\frac{\sqrt{\boldsymbol{\pi}}}{\mathrm{2}}} \right. \\ $$$$=\boldsymbol{{I}}{m}\left[\mathrm{3}\frac{\sqrt{\boldsymbol{\pi}}}{\boldsymbol{{i}}}\boldsymbol{{e}}^{\boldsymbol{{i}}\frac{\sqrt{\boldsymbol{\pi}}}{\mathrm{2}}} −\mathrm{6}\right]={I}\boldsymbol{{m}}\left(−\mathrm{6}+\mathrm{3}\sqrt{\boldsymbol{\pi}}\boldsymbol{{e}}^{\boldsymbol{{i}}\left(\frac{\sqrt{\boldsymbol{\pi}}−\boldsymbol{\pi}}{\mathrm{2}}\right)} \right) \\ $$$$\boldsymbol{{i}}=\mathrm{3}\sqrt{\boldsymbol{\pi}}\boldsymbol{{sin}}\left(\frac{\sqrt{\boldsymbol{\pi}}}{\mathrm{2}}−\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)=−\mathrm{3}\sqrt{\pi}\boldsymbol{{cos}}\left(\frac{\sqrt{\boldsymbol{\pi}}}{\mathrm{2}}\right) \\ $$
Commented by Frix last updated on 01/Nov/24
But it′s wrong.
$$\mathrm{But}\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong}. \\ $$

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