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If-D-x-2-y-2-z-2-1-D-x-2-2y-2-x-2-4y-2-z-2-dxdydz-

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Question Number 210787 by mnjuly1970 last updated on 19/Aug/24
{ (( If, D : x^2 +y^( 2) + z^( 2) ≤1)),(( ⇒∫∫_D^ ∫(( x^2 + 2y^( 2) )/(x^2 + 4y^2 +z^2 )) dxdydz=?)) :}
$$ \\ $$$$\:\begin{cases}{\:\:\mathrm{I}{f},\:\mathrm{D}\::\:{x}^{\mathrm{2}} \:+{y}^{\:\mathrm{2}} \:+\:{z}^{\:\mathrm{2}} \leqslant\mathrm{1}}\\{\:\Rightarrow\int\underset{\overset{} {\mathrm{D}}} {\int}\int\frac{\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{y}^{\:\mathrm{2}} }{{x}^{\mathrm{2}} \:+\:\mathrm{4}{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} }\:{dxdydz}=?}\end{cases} \\ $$$$ \\ $$$$ \\ $$
Answered by BHOOPENDRA last updated on 19/Aug/24
((2π)/3)
$$\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$
Commented by mnjuly1970 last updated on 19/Aug/24
please with explanation λ
$${please}\:\:{with}\:{explanation}\:\underbrace{\lambda} \\ $$
Answered by Berbere last updated on 19/Aug/24
(x,y,z)→(z,y,x) dont Change D by symetrie =V=∫∫∫_D ((x^2 +2y^2 )/(x^2 +4y^2 +z^2 ))dxdydz=∫∫∫_D ((z^2 +2y^2 +x^2 )/(x^2 +4y^2 +z^2 ))dxdydz ⇒2V=∫∫∫_D ((x^2 +4y^2 +z^2 )/(x^2 +4y^2 +z^2 ))dxdydz=2V=∫∫∫_D dxdydz =2V=Volum(D)=(4/3)π∈Spher of radius 1 V=((2π)/3)
$$\left({x},{y},{z}\right)\rightarrow\left({z},{y},{x}\right) \\ $$$${dont}\:{Change}\:{D}\:{by}\:{symetrie} \\ $$$$={V}=\int\int\int_{{D}} \frac{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{dxdydz}=\int\int\int_{{D}} \frac{{z}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} +{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{dxdydz} \\ $$$$\Rightarrow\mathrm{2}{V}=\int\int\int_{{D}} \frac{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{dxdydz}=\mathrm{2}{V}=\int\int\int_{{D}} {dxdydz} \\ $$$$=\mathrm{2}{V}={Volum}\left({D}\right)=\frac{\mathrm{4}}{\mathrm{3}}\pi\in{Spher}\:{of}\:{radius}\:\mathrm{1} \\ $$$${V}=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$
Commented by mnjuly1970 last updated on 19/Aug/24
⋛
$$\:\cancel{\lesseqgtr} \\ $$
Answered by BHOOPENDRA last updated on 19/Aug/24
Commented by BHOOPENDRA last updated on 21/Aug/24
x^2 +y^2 +z^2 =1 f(x,y)=((x^2 +2y^2 )/(x^2 +4y^2 +z^2 ))
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{1} \\ $$$${f}\left({x},{y}\right)=\frac{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} } \\ $$
Answered by BHOOPENDRA last updated on 19/Aug/24
Answered by BHOOPENDRA last updated on 19/Aug/24
Given that the region D is a sphere x^2 +y^2 +z^2 ≤1 V=∫∫∫ ((x^2 +2y^2 )/(x^2 +4y^2 +z^2 )) dxdydz if we apply the transformation (x,y,z)→(z,y,x) integral becomes V=∫∫∫_D ((z^2 +2y^2 )/(z^2 +4y^2 +x^2 )) dxdydz By symmetry swapping x and z does not change the value of the integral because integrand and the limits of integration remain the same. This leads to 2V=∫∫∫_D (((x^2 +2y^2 )/(x^2 +4y^2 +z^2 ))+((z^2 +2y^2 )/(z^2 +4y^2 +x^2 )))dxdydz Now ∫∫∫_D 1dxdydz 2V=volume of D {Vol(D)} =(4/3)π r^3 =(4/3)π1^3 =(4/3)π V=(2/3)π
$${Given}\:{that}\:{the}\:{region}\:{D}\:{is} \\ $$$$\:{a}\:{sphere}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \leqslant\mathrm{1} \\ $$$${V}=\int\int\int\:\frac{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\:{dxdydz} \\ $$$${if}\:{we}\:{apply}\:{the}\:{transformation} \\ $$$$\left({x},{y},{z}\right)\rightarrow\left({z},{y},{x}\right)\:{integral}\:{becomes} \\ $$$${V}=\int\int\int_{{D}} \frac{{z}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{{z}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:{dxdydz} \\ $$$${By}\:{symmetry}\:{swapping}\:{x}\:{and}\:{z}\:{does} \\ $$$${not}\:{change}\:{the}\:{value}\:{of}\:{the}\:{integral} \\ $$$${because}\:{integrand}\:{and}\:{the}\:{limits} \\ $$$${of}\:{integration}\:{remain}\:{the}\:{same}. \\ $$$${This}\:{leads}\:{to}\: \\ $$$$\mathrm{2}{V}=\int\int\int_{{D}} \left(\frac{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{{z}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{x}^{\mathrm{2}} }\right){dxdydz} \\ $$$$ \\ $$$${Now}\: \\ $$$$\int\int\int_{{D}} \mathrm{1}{dxdydz} \\ $$$$\mathrm{2}{V}={volume}\:{of}\:{D}\:\left\{{Vol}\left({D}\right)\right\}\: \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{4}}{\mathrm{3}}\pi\:{r}^{\mathrm{3}} =\frac{\mathrm{4}}{\mathrm{3}}\pi\mathrm{1}^{\mathrm{3}} =\frac{\mathrm{4}}{\mathrm{3}}\pi \\ $$$${V}=\frac{\mathrm{2}}{\mathrm{3}}\pi \\ $$$$ \\ $$
Commented by BHOOPENDRA last updated on 19/Aug/24
Commented by BHOOPENDRA last updated on 21/Aug/24
plot when x^2 +y^2 +z^2 =1 f(z,y)=((z^2 +2y^2 )/(z^2 +4y^2 +z^2 ))
$${plot} \\ $$$${when} \\ $$$${x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{1} \\ $$$${f}\left({z},{y}\right)=\frac{{z}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{{z}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} } \\ $$
Commented by mnjuly1970 last updated on 19/Aug/24
Answered by mr W last updated on 19/Aug/24
I=∫∫∫((x^2 +4y^2 +z^2 −2y^2 −z^2 )/(x^2 +4y^2 +z^2 ))dxdydz =∫∫∫(1−((2y^2 +z^2 )/(x^2 +4y^2 +z^2 )))dxdydz =V−I_1 with I_1 =∫∫∫(((2y^2 +z^2 )/(x^2 +4y^2 +z^2 )))dxdydz I_1 =∫∫∫(((2y^2 +x^2 )/(z^2 +4y^2 +x^2 )))dzdydx I_1 +I_1 =∫∫∫(((2y^2 +z^2 )/(x^2 +4y^2 +z^2 ))+((2y^2 +x^2 )/(x^2 +4y^2 +z^2 )))dxdydz =∫∫∫(((x^2 +4y^2 +z^2 )/(x^2 +4y^2 +z^2 )))dxdydz =∫∫∫dxdydz =V ⇒I_1 =(V/2) ⇒I=V−(V/2)=(V/2)=(1/2)×((4π×1^3 )/3)=((2π)/3)
$${I}=\int\int\int\frac{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} −{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{dxdydz} \\ $$$$\:=\int\int\int\left(\mathrm{1}−\frac{\mathrm{2}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\right){dxdydz} \\ $$$$\:={V}−{I}_{\mathrm{1}} \\ $$$${with}\:{I}_{\mathrm{1}} =\int\int\int\left(\frac{\mathrm{2}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\right){dxdydz} \\ $$$${I}_{\mathrm{1}} =\int\int\int\left(\frac{\mathrm{2}{y}^{\mathrm{2}} +{x}^{\mathrm{2}} }{{z}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{x}^{\mathrm{2}} }\right){dzdydx} \\ $$$${I}_{\mathrm{1}} +{I}_{\mathrm{1}} =\int\int\int\left(\frac{\mathrm{2}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }+\frac{\mathrm{2}{y}^{\mathrm{2}} +{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\right){dxdydz} \\ $$$$\:\:\:=\int\int\int\left(\frac{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\right){dxdydz} \\ $$$$\:\:\:=\int\int\int{dxdydz} \\ $$$$\:\:\:={V} \\ $$$$\Rightarrow{I}_{\mathrm{1}} =\frac{{V}}{\mathrm{2}} \\ $$$$\Rightarrow{I}={V}−\frac{{V}}{\mathrm{2}}=\frac{{V}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}\pi×\mathrm{1}^{\mathrm{3}} }{\mathrm{3}}=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$
Commented by mnjuly1970 last updated on 20/Aug/24
grateful sir W
$${grateful}\:{sir}\:{W} \\ $$

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