Question Number 210601 by mnjuly1970 last updated on 13/Aug/24
$$ \\ $$$$\:\: \\ $$$$\:\:\:\:\:\:\:\:\:{Find}\:\:{the}\:\:{value}\:{of}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \:\int_{\sqrt{\:{x}^{\:\mathrm{2}} \:+{y}^{\:\mathrm{2}} }} ^{\:\sqrt{\mathrm{2}−{x}^{\mathrm{2}} \:−{y}^{\mathrm{2}} }} \:{xy}\:{dz}\:{dy}\:{dx}\:=?\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\: \\ $$
Answered by mr W last updated on 14/Aug/24
![Ω=∫_0 ^1 x∫_0 ^(√(1−x^2 )) y∫_(√(x^2 +y^2 )) ^(√(2−(x^2 +y^2 ))) dzdydx =∫_0 ^1 x∫_0 ^(√(1−x^2 )) y((√(2−(x^2 +y^2 )))−(√(x^2 +y^2 ))dydx =∫_0 ^(π/2) ∫_0 ^1 r^3 cos θ sin θ((√(2−r^2 ))−r)drdθ =(1/2)∫_0 ^(π/2) sin 2θ dθ∫_0 ^1 (r^3 (√(2−r^2 ))−r^4 )dr =(1/2)∫_0 ^1 (r^3 (√(2−r^2 ))−r^4 )dr =(1/2)∫_0 ^1 r^3 (√(2−r^2 ))dr−(1/(10)) =(1/4)∫_0 ^1 r^2 (√(2−r^2 ))d(r^2 )−(1/(10)) =(1/4)∫_0 ^1 t(√(2−t))dt−(1/(10)) =(1/4)∫_0 ^1 [2(√(2−t))−(2−t)^(3/2) ]dt−(1/(10)) =(1/4)[−(4/3)(2−t)^(3/2) +(2/5)(2−t)^(5/2) ]_0 ^1 −(1/(10)) =(1/4)[−(4/3)+(2/5)+((8(√2))/3)−((8(√2))/5)]−(1/(10)) =((4(√2)−5)/(15)) ✓](https://www.tinkutara.com/question/Q210649.png)
$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} {x}\int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {y}\int_{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }} ^{\sqrt{\mathrm{2}−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}} {dzdydx} \\ $$$$\:=\int_{\mathrm{0}} ^{\mathrm{1}} {x}\int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {y}\left(\sqrt{\mathrm{2}−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{dydx}\right. \\ $$$$\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\mathrm{1}} {r}^{\mathrm{3}} \mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\left(\sqrt{\mathrm{2}−{r}^{\mathrm{2}} }−{r}\right){drd}\theta \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}\:\mathrm{2}\theta\:{d}\theta\int_{\mathrm{0}} ^{\mathrm{1}} \left({r}^{\mathrm{3}} \sqrt{\mathrm{2}−{r}^{\mathrm{2}} }−{r}^{\mathrm{4}} \right){dr} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({r}^{\mathrm{3}} \sqrt{\mathrm{2}−{r}^{\mathrm{2}} }−{r}^{\mathrm{4}} \right){dr} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {r}^{\mathrm{3}} \sqrt{\mathrm{2}−{r}^{\mathrm{2}} }{dr}−\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} {r}^{\mathrm{2}} \sqrt{\mathrm{2}−{r}^{\mathrm{2}} }{d}\left({r}^{\mathrm{2}} \right)−\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}\sqrt{\mathrm{2}−{t}}{dt}−\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \left[\mathrm{2}\sqrt{\mathrm{2}−{t}}−\left(\mathrm{2}−{t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]{dt}−\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{4}}\left[−\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{2}−{t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\frac{\mathrm{2}}{\mathrm{5}}\left(\mathrm{2}−{t}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{4}}\left[−\frac{\mathrm{4}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{3}}−\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{5}}\right]−\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\:=\frac{\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{5}}{\mathrm{15}}\:\checkmark \\ $$
Commented by mnjuly1970 last updated on 14/Aug/24
$${grateful}\:{sir}\:{W}\:\:{for}\:{your}\:{effort}\:\underline{\underbrace{\lesseqgtr}} \\ $$
Commented by mnjuly1970 last updated on 14/Aug/24
Commented by mr W last updated on 14/Aug/24
$${i}\:{forgot}\:{a}\:“{r}''.\:{now}\:{it}'{s}\:{correct}. \\ $$
Commented by mnjuly1970 last updated on 14/Aug/24
$${thank}\:{you}\:{so}\:{much} \\ $$