Answer
$$\frac{{5\sqrt 2 }}{{144}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /4} {\int_0^{\pi /4} {\int_0^{\cos \theta } {{\rho ^2}\sin \phi \cos \phi } d\rho } d\theta } d\phi \cr
& \int_0^{\pi /4} {\int_0^{\pi /4} {\left[ {\int_0^{\cos \theta } {{\rho ^2}\sin \phi \cos \phi } d\rho } \right]} d\theta } d\phi \cr
& {\text{Integrate with respect to }}\rho \cr
& \int_0^{\cos \theta } {{\rho ^2}\sin \phi \cos \phi } d\rho = \sin \phi \cos \phi \left[ {\frac{1}{3}{\rho ^3}} \right]_0^{\cos \theta } \cr
& = \frac{1}{3}\sin \phi \cos \phi {\cos ^3}\theta \cr
& = \int_0^{\pi /4} {\int_0^{\pi /4} {\left[ {\int_0^{\cos \theta } {{\rho ^2}\sin \phi \cos \phi } d\rho } \right]} d\theta } d\phi \cr
& = \int_0^{\pi /4} {\int_0^{\pi /4} {\frac{1}{3}\sin \phi \cos \phi {{\cos }^3}\theta } d\theta } d\phi \cr
& = \frac{1}{3}\int_0^{\pi /4} {\sin \phi \cos \phi \left[ {\int_0^{\pi /4} {{{\cos }^3}\theta } d\theta } \right]} d\phi \cr
& {\text{Integrate with respect to }}\theta \cr
& = \int_0^{\pi /4} {{{\cos }^2}\theta \cos \theta } d\theta \cr
& = \int_0^{\pi /4} {\left( {1 - {{\sin }^2}\theta } \right)\cos \theta } d\theta \cr
& = \left[ {\sin \theta - \frac{{{{\sin }^3}\theta }}{3}} \right]_0^{\pi /4} \cr
& = \left[ {\sin \left( {\frac{\pi }{4}} \right) - \frac{{{{\sin }^3}\left( {\pi /4} \right)}}{3}} \right] - \left[ {\sin \left( 0 \right) - \frac{{{{\sin }^3}\left( 0 \right)}}{3}} \right] \cr
& = \left[ {\frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{{12}}} \right] - \left[ 0 \right] \cr
& = \frac{{5\sqrt 2 }}{{12}} \cr
& = \frac{1}{3}\int_0^{\pi /4} {\sin \phi \cos \phi \left[ {\int_0^{\pi /4} {{{\cos }^3}\theta } d\theta } \right]} d\phi \cr
& = \frac{{5\sqrt 2 }}{{36}}\int_0^{\pi /4} {\sin \phi \cos \phi } d\phi \cr
& {\text{Integrate }} \cr
& = \frac{{5\sqrt 2 }}{{36}}\left( {\frac{{{{\sin }^2}\phi }}{2}} \right)_0^{\pi /4} \cr
& = \frac{{5\sqrt 2 }}{{36}}\left( {{{\sin }^2}\left( {\frac{\pi }{4}} \right) - {{\sin }^2}\left( 0 \right)} \right) \cr
& = \frac{{5\sqrt 2 }}{{36}}\left( {\frac{1}{4}} \right) \cr
& = \frac{{5\sqrt 2 }}{{144}} \cr} $$
