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x-2-y-2-z-2-1-1-1-x-2-y-2-z-2-2-dxdydz-

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Question Number 220892 by Nicholas666 last updated on 20/May/25
∫∫∫_(x^2 + y^2 + z^2 ≤ 1) (1/((1 + x^2 +y^2 + z^2 )^2 )) dxdydz
$$ \\ $$$$\:\:\:\:\int\int\int_{\boldsymbol{{x}}^{\mathrm{2}} \:+\:\boldsymbol{{y}}^{\mathrm{2}} \:+\:\boldsymbol{{z}}^{\mathrm{2}} \:\:\leqslant\:\mathrm{1}} \:\frac{\mathrm{1}}{\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dxdydz}\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by SdC355 last updated on 21/May/25
∫_( R^3 ) (1/((1+x^2 +y^2 +z^2 )^2 )) da da^� =r^2 sin(θ)drdθdρ cus ∣J∣=∣((∂(x,y,z))/(∂(r,θ,ρ)))∣drdθdρ ∫_( R^3 ) ((r^2 sin(θ))/((r^2 +1)^2 )) da^� → 4π∫_0 ^( 1) (r^2 /((r^2 +1)^2 )) dr=π((1/2)π−1)
$$\int_{\:\mathbb{R}^{\mathrm{3}} } \:\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\mathrm{da} \\ $$$$\mathrm{d}\hat {\mathrm{a}}={r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)\mathrm{d}{r}\mathrm{d}\theta\mathrm{d}\rho\:\mathrm{cus}\:\mid\boldsymbol{{J}}\mid=\mid\frac{\partial\left({x},{y},{z}\right)}{\partial\left({r},\theta,\rho\right)}\mid\mathrm{d}{r}\mathrm{d}\theta\mathrm{d}\rho \\ $$$$\int_{\:\mathcal{R}^{\mathrm{3}} } \frac{{r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)}{\left({r}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{d}\hat {\mathrm{a}}\:\rightarrow\:\mathrm{4}\pi\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{{r}^{\mathrm{2}} }{\left({r}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{d}{r}=\pi\left(\frac{\mathrm{1}}{\mathrm{2}}\pi−\mathrm{1}\right) \\ $$
Answered by breniam last updated on 20/May/25
After we switch into spherical coordinates 2∫_0 ^1 ∫_0 ^(π/2) ∫_0 ^(2π) ρ^2 sin(θ)(1/((1+ρ^2 )^2 ))dφdθdρ= 4π∫_0 ^1 (ρ^2 /((1+ρ^2 )^2 ))dρ∫_0 ^(π/2) sin(θ)dθ= 4π∫_0 ^1 ρ(ρ/((1+ρ^2 )^2 ))dρ=−2π∫_0 ^1 ρ((1/(1+ρ^2 )))′dρ= =−π+2π∫_0 ^1 (dρ/(ρ^2 +1))=(π^2 /2)−π
$$\mathrm{After}\:\mathrm{we}\:\mathrm{switch}\:\mathrm{into}\:\mathrm{spherical}\:\mathrm{coordinates} \\ $$$$\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\rho^{\mathrm{2}} \mathrm{sin}\left(\theta\right)\frac{\mathrm{1}}{\left(\mathrm{1}+\rho^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{d}\phi\mathrm{d}\theta\mathrm{d}\rho= \\ $$$$\mathrm{4}\pi\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\rho^{\mathrm{2}} }{\left(\mathrm{1}+\rho^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{d}\rho\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{sin}\left(\theta\right)\mathrm{d}\theta= \\ $$$$\mathrm{4}\pi\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\rho\frac{\rho}{\left(\mathrm{1}+\rho^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{d}\rho=−\mathrm{2}\pi\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\rho\left(\frac{\mathrm{1}}{\mathrm{1}+\rho^{\mathrm{2}} }\right)'\mathrm{d}\rho= \\ $$$$=−\pi+\mathrm{2}\pi\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{d}\rho}{\rho^{\mathrm{2}} +\mathrm{1}}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\pi \\ $$

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