Question Number 189374 by mathocean1 last updated on 15/Mar/23
$$\Delta=\left\{\left({x},{y},{z}\right)\:\in\:\mathbb{R}^{\mathrm{3}} :{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant\mathrm{1}\:,\:\:{x}\geqslant\mathrm{0},\:\mathrm{0}\leqslant{z}\leqslant\mathrm{1}+{y}\right\}. \\ $$$${calculate}: \\ $$$$\int\int\int_{\Delta} {dxdydz}. \\ $$
Answered by Ar Brandon last updated on 15/Mar/23
![=∫_0 ^1 ∫_0 ^(√(1−x^2 )) ∫_0 ^(1+y) dzdydx=∫_0 ^1 ∫_0 ^(√(1−x^2 )) (1+y)dydx =∫_0 ^1 [y+(y^2 /2)]_0 ^(√(1−x^2 )) dx=∫_0 ^1 ((√(1−x^2 ))+((1−x^2 )/2))dx](https://www.tinkutara.com/question/Q189376.png)
$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \int_{\mathrm{0}} ^{\mathrm{1}+\mathrm{y}} {dzd}\mathrm{y}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \left(\mathrm{1}+\mathrm{y}\right){d}\mathrm{y}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\mathrm{y}+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} {dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}}\right){dx} \\ $$