Chapter 13 - Multiple Integration - 13.4 Triple Integrals - 13.4 Exercises - Page 1003: 14

$$\frac{3}{{64}}$$

$$\eqalign{ & \iiint\limits_D {xyz{e^{ - {x^2} - {y^2}}}}dV \cr & D = \left\{ {\left( {x,y,z} \right):0 \leqslant x \leqslant \sqrt {\ln 2} ,{\text{ 0}} \leqslant y \leqslant \sqrt {\ln 4} ,{\text{ }}0 \leqslant z \leqslant 1} \right\} \cr & {\text{Substituting the region }}D \cr & = \int_0^1 {\int_0^{\sqrt {\ln 4} } {\int_0^{\sqrt {\ln 2} } {xyz{e^{ - {x^2} - {y^2}}}} dx} dy} dz \cr & {\text{Rewrite the integrand}} \cr & = \int_0^1 {\int_0^{\sqrt {\ln 4} } {\left[ {\int_0^{\sqrt {\ln 2} } {xyz{e^{ - {x^2}}}{e^{ - {y^2}}}} dx} \right]} dy} dz \cr & = \int_0^1 {\int_0^{\sqrt {\ln 4} } {{e^{ - {y^2}}}yz\left[ {\int_0^{\sqrt {\ln 2} } {x{e^{ - {x^2}}}} dx} \right]} dy} dz \cr & {\text{Integrate with respect to }}x \cr & = \int_0^1 {\int_0^{\sqrt {\ln 4} } {{e^{ - {y^2}}}yz\left[ { - \frac{1}{2}{e^{ - {x^2}}}} \right]_0^{\sqrt {\ln 2} }} dy} dz \cr & = - \frac{1}{2}\int_0^1 {\int_0^{\sqrt {\ln 4} } {{e^{ - {y^2}}}yz\left[ {{e^{ - {{\left( {\sqrt {\ln 2} } \right)}^2}}} - {e^0}} \right]} dy} dz \cr & = - \frac{1}{2}\int_0^1 {\int_0^{\sqrt {\ln 4} } {{e^{ - {y^2}}}yz\left( { - \frac{1}{2}} \right)} dy} dz \cr & = \frac{1}{4}\int_0^1 {\int_0^{\sqrt {\ln 4} } {{e^{ - {y^2}}}yz} dy} dz \cr & = \frac{1}{4}\int_0^1 {z\int_0^{\sqrt {\ln 4} } {{e^{ - {y^2}}}y} dy} dz \cr & {\text{Integrate with respect to }}z \cr & = \frac{1}{4}\int_0^1 {z\left[ { - \frac{1}{2}{e^{ - {y^2}}}} \right]_0^{\sqrt {\ln 4} }} dz \cr & = - \frac{1}{8}\int_0^1 {z\left[ {{e^{ - {y^2}}}} \right]_0^{\sqrt {\ln 4} }} dz \cr & = - \frac{1}{8}\int_0^1 {z\left[ {{e^{ - {{\left( {\sqrt {\ln 4} } \right)}^2}}} - {e^0}} \right]} dz \cr & = - \frac{1}{8}\int_0^1 {z\left( { - \frac{3}{4}} \right)} dz \cr & = \frac{3}{{32}}\int_0^1 z dz \cr & {\text{Integrate with respect to }}z \cr & = \frac{3}{{32}}\left[ {\frac{{{z^2}}}{2}} \right]_0^1 \cr & = \frac{3}{{32}}\left( {\frac{1}{2}} \right) \cr & = \frac{3}{{64}} \cr} $$