Question Number 211006 by universe last updated on 26/Aug/24
Answered by mr W last updated on 26/Aug/24
$$=\mathrm{0}\:{because}\:{of}\:{odd}\:{function}\:{zy}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 26/Aug/24
$${you}\:{could}\:{also}\:{have}\:{made}\:{it}\:{look} \\ $$$${more}\:{complicated}\:{like} \\ $$$$\int\int\underset{{V}} {\int}{zy}^{\mathrm{2}} \sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{z}^{\mathrm{6}} \right)}{dxdydz}. \\ $$$${but}\:{the}\:{result}\:{is}\:{always}\:\mathrm{0}. \\ $$
Commented by universe last updated on 26/Aug/24
$${sir}\:{what}\:{is}\:{the}\:{value}\:{in}\:{first}\:{octant}??? \\ $$
Commented by mr W last updated on 26/Aug/24
![∫∫∫_D zy^2 dxdydz =∫_0 ^1 ∫_0 ^(√(1−x^2 )) y^2 ∫_0 ^(√(1−x^2 −y^2 )) zdzdydx =(1/2)∫_0 ^1 ∫_0 ^(√(1−x^2 )) y^2 (1−x^2 −y^2 )dydx =(1/2)∫_0 ^1 [(((1−x^2 )^4 )/3)−(((1−x^2 )^5 )/5)]dx =((64)/(945))−((128)/(3465))=((64)/(2079)) ✓](https://www.tinkutara.com/question/Q211042.png)
$$\int\int\int_{{D}} {zy}^{\mathrm{2}} {dxdydz} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {y}^{\mathrm{2}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }} {zdzdydx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {y}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dydx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left[\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{4}} }{\mathrm{3}}−\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{5}} }{\mathrm{5}}\right]{dx} \\ $$$$=\frac{\mathrm{64}}{\mathrm{945}}−\frac{\mathrm{128}}{\mathrm{3465}}=\frac{\mathrm{64}}{\mathrm{2079}}\:\checkmark \\ $$
Commented by universe last updated on 27/Aug/24
$${thank}\:{you}\:{sir} \\ $$