Question Number 195995 by pticantor last updated on 15/Aug/23
$$\Delta=\left\{\left(\bar {{x}}\:{y}\:{z}\right),\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant\mathrm{1},\:{x}\geqslant\mathrm{0},\mathrm{0}<{z}<{y}+\mathrm{1}\right\} \\ $$$${calculer}\:\boldsymbol{{I}}=\int\int\int_{\Delta} {xyzdxdydz} \\ $$$${please}\:{i}\:{need}\:{help} \\ $$
Answered by aleks041103 last updated on 27/Aug/23
![I=∫_0 ^1 x∫_(−(√(1−x^2 ))) ^( (√(1−x^2 ))) y∫_0 ^( y+1) zdzdydx ∫_0 ^( y+1) zdz=(((y+1)^2 )/2) ∫_(−(√(1−x^2 ))) ^( (√(1−x^2 ))) y∫_0 ^( y+1) zdzdy=(1/2)∫_(−(√(1−x^2 ))) ^( (√(1−x^2 ))) y(y+1)^2 dy= =(1/2)∫_(−(√(1−x^2 ))) ^( (√(1−x^2 ))) (y^3 +2y^2 +y)dy= =(1/2)[(y^4 /4)+((2y^3 )/3)+(y^2 /2)]_(−s) ^s = =(2/3)s^3 =(2/3)(1−x^2 )^(3/2) ⇒I=(1/3)∫_0 ^1 (1−x^2 )^(3/2) 2xdx= =(1/3)∫_0 ^( 1) (1−t)^(3/2) dt= =(1/3)∫_0 ^1 r^(3/2) dr=(1/3) (r^(5/2) /(5/2))=(2/(15)) ⇒I=(2/(15))](https://www.tinkutara.com/question/Q196536.png)
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}\int_{−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} ^{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {y}\int_{\mathrm{0}} ^{\:{y}+\mathrm{1}} {zdzdydx} \\ $$$$\int_{\mathrm{0}} ^{\:{y}+\mathrm{1}} {zdz}=\frac{\left({y}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\int_{−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} ^{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {y}\int_{\mathrm{0}} ^{\:{y}+\mathrm{1}} {zdzdy}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} ^{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {y}\left({y}+\mathrm{1}\right)^{\mathrm{2}} {dy}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} ^{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \left({y}^{\mathrm{3}} +\mathrm{2}{y}^{\mathrm{2}} +{y}\right){dy}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{y}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mathrm{2}{y}^{\mathrm{3}} }{\mathrm{3}}+\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\right]_{−{s}} ^{{s}} = \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}{s}^{\mathrm{3}} =\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} \mathrm{2}{xdx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\mathrm{3}/\mathrm{2}} {dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} {r}^{\mathrm{3}/\mathrm{2}} {dr}=\frac{\mathrm{1}}{\mathrm{3}}\:\frac{{r}^{\mathrm{5}/\mathrm{2}} }{\mathrm{5}/\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{15}} \\ $$$$\Rightarrow{I}=\frac{\mathrm{2}}{\mathrm{15}}\: \\ $$