Answer
$$\frac{1}{8}\pi $$
Work Step by Step
$$\eqalign{
& \int_0^{2\pi } {\int_0^{\pi /4} {\int_0^{\cos \phi } {{\rho ^2}\sin \phi } d\rho } d\phi } d\theta \cr
& \int_0^{2\pi } {\int_0^{\pi /4} {\sin \phi \left[ {\int_0^{\cos \phi } {{\rho ^2}} d\rho } \right]} d\phi } d\theta \cr
& {\text{Integrate with respect to }}\rho \cr
& \int_0^{\cos \phi } {{\rho ^2}} d\rho = \left[ {\frac{1}{3}{\rho ^3}} \right]_0^{\cos \phi } \cr
& = \frac{1}{3}{\cos ^3}\phi \cr
& \int_0^{2\pi } {\int_0^{\pi /4} {\sin \phi \left[ {\int_0^{\cos \phi } {{\rho ^2}} d\rho } \right]} d\phi } d\theta = \int_0^{2\pi } {\int_0^{\pi /4} {\sin \phi \left( {\frac{1}{3}{{\cos }^3}\phi } \right)} d\phi } d\theta \cr
& = \frac{1}{3}\int_0^{2\pi } {\int_0^{\pi /4} {{{\cos }^3}\phi \sin \phi } d\phi } d\theta \cr
& {\text{Integrate with respect to }}\phi \cr
& = \frac{1}{3}\int_0^{2\pi } {\left[ { - \frac{{{{\cos }^4}\phi }}{4}} \right]_0^{\pi /4}} d\theta \cr
& = - \frac{1}{{12}}\int_0^{2\pi } {\left[ {{{\cos }^4}\left( {\frac{\pi }{4}} \right) - {{\cos }^4}\left( 0 \right)} \right]} d\theta \cr
& = - \frac{1}{{12}}\int_0^{2\pi } {\left( { - \frac{3}{4}} \right)} d\theta \cr
& = \frac{1}{{16}}\int_0^{2\pi } {d\theta } \cr
& {\text{Integrate }} \cr
& = \frac{1}{{16}}\left( {2\pi - 0} \right) \cr
& = \frac{1}{8}\pi \cr} $$
