Question Number 35018 by ajfour last updated on 14/May/18
$$\int\int\int\frac{{dxdydz}}{\left({x}+{y}+{z}+\mathrm{1}\right)^{\mathrm{3}} }\:\:\:{bounded}\:{by}\:{the} \\ $$$${coordinate}\:{planes}\:{and}\:{the}\:{plane} \\ $$$${x}+{y}+{z}=\mathrm{1}\:. \\ $$
Answered by ajfour last updated on 14/May/18
![∫_0 ^( 1) [∫_0 ^( 1−x) [∫_0 ^( 1−x−y) (dz/((x+y+z+1)^3 ))]dy]dx =∫_0 ^( 1) [∫_0 ^( 1−x) (−(1/(2(x+y+z+1)^2 )))_0 ^(1−x−y) ]dy]dx =(1/2)∫_0 ^( 1) [∫_0 ^( 1−x) ((1/((x+y+1)^2 ))−(1/4))dy]dx =(1/2)∫_0 ^( 1) [((−1)/(x+y+1))−(y/4)]_0 ^(1−x) dx =(1/2)∫_0 ^( 1) ((1/(x+1))−(1/2)−((1−x)/4))dx =(1/2)[ln (x+1)−((3x)/4)+(x^2 /8)]_0 ^1 =((ln 2)/2)−(5/(16)) .](Q35019.png)
$$\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \left[\int_{\mathrm{0}} ^{\:\:\mathrm{1}−{x}} \left[\int_{\mathrm{0}} ^{\:\:\mathrm{1}−{x}−{y}} \frac{{dz}}{\left({x}+{y}+{z}+\mathrm{1}\right)^{\mathrm{3}} }\right]{dy}\right]{dx} \\ $$$$\left.=\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \left[\int_{\mathrm{0}} ^{\:\:\mathrm{1}−{x}} \left(−\frac{\mathrm{1}}{\mathrm{2}\left({x}+{y}+{z}+\mathrm{1}\right)^{\mathrm{2}} }\right)_{\mathrm{0}} ^{\mathrm{1}−{x}−{y}} \right]{dy}\right]{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \left[\int_{\mathrm{0}} ^{\:\:\mathrm{1}−{x}} \left(\frac{\mathrm{1}}{\left({x}+{y}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}}\right){dy}\right]{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \left[\frac{−\mathrm{1}}{{x}+{y}+\mathrm{1}}−\frac{{y}}{\mathrm{4}}\right]_{\mathrm{0}} ^{\mathrm{1}−{x}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \left(\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}−{x}}{\mathrm{4}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\:\left({x}+\mathrm{1}\right)−\frac{\mathrm{3}{x}}{\mathrm{4}}+\frac{{x}^{\mathrm{2}} }{\mathrm{8}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{16}}\:. \\ $$