Question Number 80369 by M±th+et£s last updated on 02/Feb/20
Commented by mathmax by abdo last updated on 03/Feb/20
![let A =∫∫∫_([0,1]^3 ) (1+u^2 +v^2 +w^2 )^(−2) dudvdw we use the diffeomrphisme (r,θ,ϕ)→ϕ(r,θ,ϕ) =(ϕ_1 ,ϕ_2 ,ϕ_3 )=(u,v,w) with u=rsinθ cosϕ v=rsinθ sinϕ w=rcosθ we have o≤u^2 ≤1 ,o≤v^2 ≤1 ,o≤w^2 ≤1 ⇒1≤u^(2 ) +v^2 +w^2 ≤3 ⇒ 0≤r^2 sin^2 θ cos^2 ϕ +r^2 sin^2 θsin^2 ϕ +r^2 cos^2 θ ≤3 ⇒ 0≤r^2 ≤3 ⇒0≤r≤(√3) A =∫∫∫_(0≤r≤(√3) and 0≤θ≤π and 0≤ϕ≤2π) (1+r^2 )^(−2) r^2 sinθ dr dθ dϕ =∫_0 ^(√3) (r^2 /((1+r^2 )^2 ))dr ∫_0 ^π sinθ dθ ∫_0 ^(2π) dϕ =2π[cosθ]_0 ^π .∫_0 ^(√3) ((r^2 dr)/((r^2 +1)^2 )) =−4π ∫_0 ^(√3) ((r^2 +1−1)/((r^2 +1)^2 ))dr =−4π ∫_0 ^(√3) (dr/(r^2 +1)) +4π ∫_0 ^(√3) (dr/((r^2 +1)^2 )) ∫_0 ^(√3) (dr/(r^2 +1)) =[arctanr]_0 ^(√3) =(π/3) ∫_0 ^(√3) (dr/((r^2 +1)^2 )) =_(r=tant) ∫_0 ^(π/3) (((1+tan^2 t)dt)/((1+tan^2 t)^2 ))=∫_0 ^(π/3) (dt/(1+tan^2 t)) =∫_0 ^(π/3) cos^2 t dt =∫_0 ^(π/3) ((1+cos(2t))/2)dt=(π/6) +(1/2) ∫_0 ^(π/3) cos(2t)dt =(π/6) +(1/4)[sin(2t)]_0 ^(π/3) =(π/6) +(1/4){sin(((2π)/3))}=(π/6)+(1/4)×((√3)/2) ⇒ A =−(4/3)π^2 +4π{(π/6) +((√3)/8)} =−((4π^2 )/3) +((2π^2 )/3) +((π(√3))/2) =((π(√3))/2)−((2π^2 )/3)](https://www.tinkutara.com/question/Q80534.png)
$${let}\:{A}\:=\int\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{3}} } \:\:\:\left(\mathrm{1}+{u}^{\mathrm{2}} \:+{v}^{\mathrm{2}} \:+{w}^{\mathrm{2}} \right)^{−\mathrm{2}} {dudvdw} \\ $$$${we}\:{use}\:{the}\:{diffeomrphisme}\:\left({r},\theta,\varphi\right)\rightarrow\varphi\left({r},\theta,\varphi\right) \\ $$$$=\left(\varphi_{\mathrm{1}} ,\varphi_{\mathrm{2}} ,\varphi_{\mathrm{3}} \right)=\left({u},{v},{w}\right)\:{with}\:{u}={rsin}\theta\:{cos}\varphi \\ $$$${v}={rsin}\theta\:{sin}\varphi\:\:{w}={rcos}\theta\:\:\:\:\:\:{we}\:{have} \\ $$$${o}\leqslant{u}^{\mathrm{2}} \leqslant\mathrm{1}\:,{o}\leqslant{v}^{\mathrm{2}} \leqslant\mathrm{1}\:,{o}\leqslant{w}^{\mathrm{2}} \leqslant\mathrm{1}\:\Rightarrow\mathrm{1}\leqslant{u}^{\mathrm{2}\:} +{v}^{\mathrm{2}} \:+{w}^{\mathrm{2}} \leqslant\mathrm{3}\:\Rightarrow \\ $$$$\mathrm{0}\leqslant{r}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\:{cos}^{\mathrm{2}} \varphi\:+{r}^{\mathrm{2}} \:{sin}^{\mathrm{2}} \theta{sin}^{\mathrm{2}} \varphi\:+{r}^{\mathrm{2}} \:{cos}^{\mathrm{2}} \theta\:\leqslant\mathrm{3}\:\Rightarrow \\ $$$$\mathrm{0}\leqslant{r}^{\mathrm{2}} \leqslant\mathrm{3}\:\Rightarrow\mathrm{0}\leqslant{r}\leqslant\sqrt{\mathrm{3}} \\ $$$${A}\:=\int\int\int_{\mathrm{0}\leqslant{r}\leqslant\sqrt{\mathrm{3}}\:\:{and}\:\:\mathrm{0}\leqslant\theta\leqslant\pi\:{and}\:\:\mathrm{0}\leqslant\varphi\leqslant\mathrm{2}\pi} \:\:\left(\mathrm{1}+{r}^{\mathrm{2}} \right)^{−\mathrm{2}} {r}^{\mathrm{2}} \:{sin}\theta\:\:{dr}\:{d}\theta\:{d}\varphi \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \frac{{r}^{\mathrm{2}} }{\left(\mathrm{1}+{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{dr}\:\int_{\mathrm{0}} ^{\pi} \:{sin}\theta\:{d}\theta\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{d}\varphi \\ $$$$=\mathrm{2}\pi\left[{cos}\theta\right]_{\mathrm{0}} ^{\pi} .\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\:\frac{{r}^{\mathrm{2}} {dr}}{\left({r}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:=−\mathrm{4}\pi\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\:\frac{{r}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\left({r}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dr} \\ $$$$=−\mathrm{4}\pi\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\frac{{dr}}{{r}^{\mathrm{2}} +\mathrm{1}}\:+\mathrm{4}\pi\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\:\frac{{dr}}{\left({r}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\:\frac{{dr}}{{r}^{\mathrm{2}} \:+\mathrm{1}}\:=\left[{arctanr}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} =\frac{\pi}{\mathrm{3}} \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\:\frac{{dr}}{\left({r}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:=_{{r}={tant}} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:\frac{{dt}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} {cos}^{\mathrm{2}} {t}\:{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt}=\frac{\pi}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:{cos}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\frac{\pi}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \:=\frac{\pi}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{4}}\left\{{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right\}=\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{4}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow \\ $$$${A}\:=−\frac{\mathrm{4}}{\mathrm{3}}\pi^{\mathrm{2}} +\mathrm{4}\pi\left\{\frac{\pi}{\mathrm{6}}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\right\}\:=−\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{3}}\:+\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}}\:+\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}} \\ $$