Question Number 221400 by Nicholas666 last updated on 02/Jun/25
![∫∫∫_(0≤x≤y≤z≤1) [(y − x)^2 (z − y)^2 (z − x)^2 ] dxdydz](https://www.tinkutara.com/question/Q221400.png)
$$ \\ $$$$\:\:\:\:\:\:\:\int\int\int_{\mathrm{0}\leqslant{x}\leqslant{y}\leqslant{z}\leqslant\mathrm{1}} \:\left[\left({y}\:−\:{x}\right)^{\mathrm{2}} \left({z}\:−\:{y}\right)^{\mathrm{2}} \left({z}\:−\:{x}\right)^{\mathrm{2}} \right]\:{dxdydz}\:\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by MrGaster last updated on 05/Jun/25
![∫∫∫_(0≤x≤y≤z≤1) [(y − x)^2 (z − y)^2 (z − x)^2 ] dxdydz s≜x,t≜y−x,u≜z−y ∣((∂(x,y,z))/(∂(s,t,u)))∣=1 ,s∈[0,1],t∈[0,1−s],u∈[0,1−s−t] (y−x)^2 (z−y)^2 (z−x)^2 =t^2 u^2 (t+u)^2 ∫_(s=0) ^1 ∫_(t=0) ^(1−s) ∫_(u=0) ^(1−s−t) t^2 u^2 (t+u)^2 dudtds ∫_(u=0) ^a u^2 (t+u)^2 du=∫_0 ^a (t^2 u^2 +2tu^3 +u^4 )du=[((t^2 u^3 )/3)+((tu^4 )/2)+(u^5 /5)]_0 ^a =((t^2 a^3 )/3)+((ta^4 )/2)+(a^5 /5),a≜1−s−t ∫_0 ^b t^m (b−t)^n dt=b^(m+n+1) ((m!n!)/((m+n+1)!)) (1/3)∫_0 ^b t^4 (b−t)^3 dt=(1/3)b^8 ((4!3!)/(8!)),(1/2)∫_0 ^b t^3 (b−t)^4 dt=(1/2)b^8 ((3!4!)/8),(1/5)∫_0 ^b t^2 (b−t)^5 dt=(1/5)b^8 ((2!5!)/(8!)) 4!=24 3!=6 2!=2 5!=120 8!=40320 (1/3)∙((24∙6)/(40320))b^8 +(1/2)∙((6∙24)/(40320))b^8 +(1/5)∙((2∙120)/(40320))b^8 =((48)/(40320))b^8 +((72)/(40320))b^8 +((48)/(40320))b^8 =((168)/(40320))b^8 =(1/(240))b^8 w≜1−s,dw=−ds,s=0⇒w=1 ∫_(w=1) ^0 (w^8 /(240))(−dw)=∫_0 ^1 (w^4 /(240))dw=(1/(240))[(w^9 /9)]_0 ^1 =(1/(240))∙(1/9)=(1/(2160))](https://www.tinkutara.com/question/Q221434.png)
$$ \\ $$$$\:\:\:\:\:\:\:\int\int\int_{\mathrm{0}\leqslant{x}\leqslant{y}\leqslant{z}\leqslant\mathrm{1}} \:\left[\left({y}\:−\:{x}\right)^{\mathrm{2}} \left({z}\:−\:{y}\right)^{\mathrm{2}} \left({z}\:−\:{x}\right)^{\mathrm{2}} \right]\:{dxdydz}\:\:\:\:\:\:\: \\ $$$${s}\triangleq{x},{t}\triangleq{y}−{x},{u}\triangleq{z}−{y} \\ $$$$\mid\frac{\partial\left({x},{y},{z}\right)}{\partial\left({s},{t},{u}\right)}\mid=\mathrm{1}\:,{s}\in\left[\mathrm{0},\mathrm{1}\right],{t}\in\left[\mathrm{0},\mathrm{1}−{s}\right],{u}\in\left[\mathrm{0},\mathrm{1}−{s}−{t}\right] \\ $$$$\left({y}−{x}\right)^{\mathrm{2}} \left({z}−{y}\right)^{\mathrm{2}} \left({z}−{x}\right)^{\mathrm{2}} ={t}^{\mathrm{2}} {u}^{\mathrm{2}} \left({t}+{u}\right)^{\mathrm{2}} \\ $$$$\int_{{s}=\mathrm{0}} ^{\mathrm{1}} \int_{{t}=\mathrm{0}} ^{\mathrm{1}−{s}} \int_{{u}=\mathrm{0}} ^{\mathrm{1}−{s}−{t}} {t}^{\mathrm{2}} {u}^{\mathrm{2}} \left({t}+{u}\right)^{\mathrm{2}} {dudtds} \\ $$$$\int_{{u}=\mathrm{0}} ^{{a}} {u}^{\mathrm{2}} \left({t}+{u}\right)^{\mathrm{2}} {du}=\int_{\mathrm{0}} ^{{a}} \left({t}^{\mathrm{2}} {u}^{\mathrm{2}} +\mathrm{2}{tu}^{\mathrm{3}} +{u}^{\mathrm{4}} \right){du}=\left[\frac{{t}^{\mathrm{2}} {u}^{\mathrm{3}} }{\mathrm{3}}+\frac{{tu}^{\mathrm{4}} }{\mathrm{2}}+\frac{{u}^{\mathrm{5}} }{\mathrm{5}}\right]_{\mathrm{0}} ^{{a}} =\frac{{t}^{\mathrm{2}} {a}^{\mathrm{3}} }{\mathrm{3}}+\frac{{ta}^{\mathrm{4}} }{\mathrm{2}}+\frac{{a}^{\mathrm{5}} }{\mathrm{5}},{a}\triangleq\mathrm{1}−{s}−{t} \\ $$$$\int_{\mathrm{0}} ^{{b}} {t}^{{m}} \left({b}−{t}\right)^{{n}} {dt}={b}^{{m}+{n}+\mathrm{1}} \frac{{m}!{n}!}{\left({m}+{n}+\mathrm{1}\right)!} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{{b}} {t}^{\mathrm{4}} \left({b}−{t}\right)^{\mathrm{3}} {dt}=\frac{\mathrm{1}}{\mathrm{3}}{b}^{\mathrm{8}} \frac{\mathrm{4}!\mathrm{3}!}{\mathrm{8}!},\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{b}} {t}^{\mathrm{3}} \left({b}−{t}\right)^{\mathrm{4}} {dt}=\frac{\mathrm{1}}{\mathrm{2}}{b}^{\mathrm{8}} \frac{\mathrm{3}!\mathrm{4}!}{\mathrm{8}},\frac{\mathrm{1}}{\mathrm{5}}\int_{\mathrm{0}} ^{{b}} {t}^{\mathrm{2}} \left({b}−{t}\right)^{\mathrm{5}} {dt}=\frac{\mathrm{1}}{\mathrm{5}}{b}^{\mathrm{8}} \frac{\mathrm{2}!\mathrm{5}!}{\mathrm{8}!} \\ $$$$\mathrm{4}!=\mathrm{24}\:\mathrm{3}!=\mathrm{6}\:\mathrm{2}!=\mathrm{2}\:\mathrm{5}!=\mathrm{120}\:\mathrm{8}!=\mathrm{40320} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\centerdot\frac{\mathrm{24}\centerdot\mathrm{6}}{\mathrm{40320}}{b}^{\mathrm{8}} +\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{6}\centerdot\mathrm{24}}{\mathrm{40320}}{b}^{\mathrm{8}} +\frac{\mathrm{1}}{\mathrm{5}}\centerdot\frac{\mathrm{2}\centerdot\mathrm{120}}{\mathrm{40320}}{b}^{\mathrm{8}} =\frac{\mathrm{48}}{\mathrm{40320}}{b}^{\mathrm{8}} +\frac{\mathrm{72}}{\mathrm{40320}}{b}^{\mathrm{8}} +\frac{\mathrm{48}}{\mathrm{40320}}{b}^{\mathrm{8}} =\frac{\mathrm{168}}{\mathrm{40320}}{b}^{\mathrm{8}} =\frac{\mathrm{1}}{\mathrm{240}}{b}^{\mathrm{8}} \\ $$$${w}\triangleq\mathrm{1}−{s},{dw}=−{ds},{s}=\mathrm{0}\Rightarrow{w}=\mathrm{1} \\ $$$$\int_{{w}=\mathrm{1}} ^{\mathrm{0}} \frac{{w}^{\mathrm{8}} }{\mathrm{240}}\left(−{dw}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{w}^{\mathrm{4}} }{\mathrm{240}}{dw}=\frac{\mathrm{1}}{\mathrm{240}}\left[\frac{{w}^{\mathrm{9}} }{\mathrm{9}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{240}}\centerdot\frac{\mathrm{1}}{\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{2160}} \\ $$
Commented by Nicholas666 last updated on 05/Jun/25
$$\mathrm{beautifull} \\ $$