Question Number 215995 by universe last updated on 25/Jan/25
$$\int\int\underset{{D}} {\int}\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\:\mathrm{dv}\:=\:? \\ $$$$\mathrm{D}\:=\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} <{z} \\ $$
Answered by mr W last updated on 25/Jan/25
![D=x^2 +y^2 +(z−a)^2 ≤a^2 with a=(1/2) x=ρ cos θ cos φ y=ρ cos θ sin φ z=a+ρ sin θ I=∫_(−(π/2)) ^(π/2) ∫_0 ^a (√(ρ^2 cos^2 θ+(a+ρ sin θ)^2 ))2πρ cos θ ρdρdθ =2π∫_0 ^a ρ^2 ∫_(−(π/2)) ^(π/2) cos θ(√(ρ^2 +2aρ sin θ+a^2 ))dθdρ =((2π)/(3a))∫_0 ^a ρ[(ρ^2 +2aρ sin θ+a^2 )^(3/2) ]_(−(π/2)) ^(π/2) dρ =((2π)/(3a))∫_0 ^a ρ[(ρ^2 +2aρ+a^2 )^(3/2) −(ρ^2 −2aρ+a^2 )^(3/2) ]dρ =((2π)/(3a))∫_0 ^a ρ[(a+ρ)^3 −(a−ρ)^3 ]dρ =((2π)/(3a))∫_0 ^a ρ(6a^2 ρ+2ρ^3 )dρ =((2π)/(3a))[2a^2 ρ^3 +((2ρ^5 )/5)]_0 ^a =((2π)/3)(2+(2/5))a^4 =((8πa^4 )/5) =(π/(10))](https://www.tinkutara.com/question/Q216019.png)
$${D}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left({z}−{a}\right)^{\mathrm{2}} \leqslant{a}^{\mathrm{2}} \:\:{with}\:{a}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}=\rho\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\phi \\ $$$${y}=\rho\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi \\ $$$${z}={a}+\rho\:\mathrm{sin}\:\theta \\ $$$${I}=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{{a}} \sqrt{\rho^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta+\left({a}+\rho\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }\mathrm{2}\pi\rho\:\mathrm{cos}\:\theta\:\rho{d}\rho{d}\theta \\ $$$$\:\:=\mathrm{2}\pi\int_{\mathrm{0}} ^{{a}} \rho^{\mathrm{2}} \int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\:\theta\sqrt{\rho^{\mathrm{2}} +\mathrm{2}{a}\rho\:\mathrm{sin}\:\theta+{a}^{\mathrm{2}} }{d}\theta{d}\rho \\ $$$$\:\:=\frac{\mathrm{2}\pi}{\mathrm{3}{a}}\int_{\mathrm{0}} ^{{a}} \rho\left[\left(\rho^{\mathrm{2}} +\mathrm{2}{a}\rho\:\mathrm{sin}\:\theta+{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {d}\rho \\ $$$$\:\:=\frac{\mathrm{2}\pi}{\mathrm{3}{a}}\int_{\mathrm{0}} ^{{a}} \rho\left[\left(\rho^{\mathrm{2}} +\mathrm{2}{a}\rho+{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\left(\rho^{\mathrm{2}} −\mathrm{2}{a}\rho+{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]{d}\rho \\ $$$$\:\:=\frac{\mathrm{2}\pi}{\mathrm{3}{a}}\int_{\mathrm{0}} ^{{a}} \rho\left[\left({a}+\rho\right)^{\mathrm{3}} −\left({a}−\rho\right)^{\mathrm{3}} \right]{d}\rho \\ $$$$\:\:=\frac{\mathrm{2}\pi}{\mathrm{3}{a}}\int_{\mathrm{0}} ^{{a}} \rho\left(\mathrm{6}{a}^{\mathrm{2}} \rho+\mathrm{2}\rho^{\mathrm{3}} \right){d}\rho \\ $$$$\:\:=\frac{\mathrm{2}\pi}{\mathrm{3}{a}}\left[\mathrm{2}{a}^{\mathrm{2}} \rho^{\mathrm{3}} +\frac{\mathrm{2}\rho^{\mathrm{5}} }{\mathrm{5}}\right]_{\mathrm{0}} ^{{a}} \\ $$$$\:\:=\frac{\mathrm{2}\pi}{\mathrm{3}}\left(\mathrm{2}+\frac{\mathrm{2}}{\mathrm{5}}\right){a}^{\mathrm{4}} \\ $$$$\:\:=\frac{\mathrm{8}\pi{a}^{\mathrm{4}} }{\mathrm{5}} \\ $$$$\:\:=\frac{\pi}{\mathrm{10}} \\ $$
Commented by universe last updated on 25/Jan/25
$${thank}\:{you}\:{sir} \\ $$