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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
$let A =∫∫∫_([0,1]^3 ) (1+u^2 +v^2 +w^2 )^(−2) dudvdw we use the diffeomrphisme (r,θ,ϕ)→ϕ(r,θ,ϕ) =(ϕ_1 ,ϕ_2 ,ϕ_3 )=(u,v,w) with u=rsinθ cosϕ v=rsinθ sinϕ w=rcosθ we have o≤u^2 ≤1 ,o≤v^2 ≤1 ,o≤w^2 ≤1 ⇒1≤u^(2 ) +v^2 +w^2 ≤3 ⇒ 0≤r^2 sin^2 θ cos^2 ϕ +r^2 sin^2 θsin^2 ϕ +r^2 cos^2 θ ≤3 ⇒ 0≤r^2 ≤3 ⇒0≤r≤(√3) A =∫∫∫_(0≤r≤(√3) and 0≤θ≤π and 0≤ϕ≤2π) (1+r^2 )^(−2) r^2 sinθ dr dθ dϕ =∫_0 ^(√3) (r^2 /((1+r^2 )^2 ))dr ∫_0 ^π sinθ dθ ∫_0 ^(2π) dϕ =2π[cosθ]_0 ^π .∫_0 ^(√3) ((r^2 dr)/((r^2 +1)^2 )) =−4π ∫_0 ^(√3) ((r^2 +1−1)/((r^2 +1)^2 ))dr =−4π ∫_0 ^(√3) (dr/(r^2 +1)) +4π ∫_0 ^(√3) (dr/((r^2 +1)^2 )) ∫_0 ^(√3) (dr/(r^2 +1)) =[arctanr]_0 ^(√3) =(π/3) ∫_0 ^(√3) (dr/((r^2 +1)^2 )) =_(r=tant) ∫_0 ^(π/3) (((1+tan^2 t)dt)/((1+tan^2 t)^2 ))=∫_0 ^(π/3) (dt/(1+tan^2 t)) =∫_0 ^(π/3) cos^2 t dt =∫_0 ^(π/3) ((1+cos(2t))/2)dt=(π/6) +(1/2) ∫_0 ^(π/3) cos(2t)dt =(π/6) +(1/4)[sin(2t)]_0 ^(π/3) =(π/6) +(1/4){sin(((2π)/3))}=(π/6)+(1/4)×((√3)/2) ⇒ A =−(4/3)π^2 +4π{(π/6) +((√3)/8)} =−((4π^2 )/3) +((2π^2 )/3) +((π(√3))/2) =((π(√3))/2)−((2π^2 )/3)$
$l e t A = \int \int \int_{{[0, 1]}^{3}} {(1 + u^{2} + v^{2} + w^{2})}^{- 2} d u d v d w$ $w e u s e t h e d i f f e o m r p h i s m e (r, θ, φ) \to φ (r, θ, φ)$ $= (φ_{1}, φ_{2}, φ_{3}) = (u, v, w) w i t h u = r s i n θ c o s φ$ $v = r s i n θ s i n φ w = r c o s θ w e h a v e$ $o ⩽ u^{2} ⩽ 1, o ⩽ v^{2} ⩽ 1, o ⩽ w^{2} ⩽ 1 \Rightarrow 1 ⩽ u^{2} + v^{2} + w^{2} ⩽ 3 \Rightarrow$ $0 ⩽ r^{2} {s i n}^{2} θ {c o s}^{2} φ + r^{2} {s i n}^{2} θ {s i n}^{2} φ + r^{2} {c o s}^{2} θ ⩽ 3 \Rightarrow$ $0 ⩽ r^{2} ⩽ 3 \Rightarrow 0 ⩽ r ⩽ \sqrt{3}$ $A = \int \int \int_{0 ⩽ r ⩽ \sqrt{3} a n d 0 ⩽ θ ⩽ π a n d 0 ⩽ φ ⩽ 2 π} {(1 + r^{2})}^{- 2} r^{2} s i n θ d r d θ d φ$ $= \int_{0}^{\sqrt{3}} \frac{r^{2}}{{(1 + r^{2})}^{2}} d r \int_{0}^{π} s i n θ d θ \int_{0}^{2 π} d φ$ $= 2 π {[c o s θ]}_{0}^{π} . \int_{0}^{\sqrt{3}} \frac{r^{2} d r}{{(r^{2} + 1)}^{2}} = - 4 π \int_{0}^{\sqrt{3}} \frac{r^{2} + 1 - 1}{{(r^{2} + 1)}^{2}} d r$ $= - 4 π \int_{0}^{\sqrt{3}} \frac{d r}{r^{2} + 1} + 4 π \int_{0}^{\sqrt{3}} \frac{d r}{{(r^{2} + 1)}^{2}}$ $\int_{0}^{\sqrt{3}} \frac{d r}{r^{2} + 1} = {[a r c t a n r]}_{0}^{\sqrt{3}} = \frac{π}{3}$ $\int_{0}^{\sqrt{3}} \frac{d r}{{(r^{2} + 1)}^{2}} =_{r = t a n t} \int_{0}^{\frac{π}{3}} \frac{(1 + {t a n}^{2} t) d t}{{(1 + {t a n}^{2} t)}^{2}} = \int_{0}^{\frac{π}{3}} \frac{d t}{1 + {t a n}^{2} t}$ $= \int_{0}^{\frac{π}{3}} {c o s}^{2} t d t = \int_{0}^{\frac{π}{3}} \frac{1 + c o s (2 t)}{2} d t = \frac{π}{6} + \frac{1}{2} \int_{0}^{\frac{π}{3}} c o s (2 t) d t$ $= \frac{π}{6} + \frac{1}{4} {[s i n (2 t)]}_{0}^{\frac{π}{3}} = \frac{π}{6} + \frac{1}{4} {s i n (\frac{2 π}{3})} = \frac{π}{6} + \frac{1}{4} \times \frac{\sqrt{3}}{2} \Rightarrow$ $A = - \frac{4}{3} π^{2} + 4 π {\frac{π}{6} + \frac{\sqrt{3}}{8}} = - \frac{4 π^{2}}{3} + \frac{2 π^{2}}{3} + \frac{π \sqrt{3}}{2}$ $= \frac{π \sqrt{3}}{2} - \frac{2 π^{2}}{3}$
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