Answer
$$\frac{{27}}{2}\pi $$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /4} {\int_0^6 {\int_0^{6 - r} {rz} dz} dr} d\theta \cr
& \int_0^{\pi /4} {\int_0^6 {\left[ {\int_0^{6 - r} {rz} dz} \right]} dr} d\theta \cr
& {\text{Integrate with respect to }}z \cr
& \int_0^{6 - r} {rz} dz = r\left[ {\frac{1}{2}{z^2}} \right]_0^{6 - r} \cr
& = \frac{1}{2}r\left[ {{{\left( {6 - r} \right)}^2} - {0^2}} \right] \cr
& = \frac{1}{2}{\left( {6 - r} \right)^2}r \cr
& \int_0^{\pi /4} {\int_0^6 {\left[ {\int_0^{6 - r} {rz} dz} \right]} dr} d\theta = \int_0^{\pi /4} {\int_0^6 {\frac{1}{2}{{\left( {6 - r} \right)}^2}r} dr} d\theta \cr
& {\text{Integrate with respect to }}r \cr
& \int_0^6 {\frac{1}{2}{{\left( {6 - r} \right)}^2}r} dr = - \frac{1}{2}\int_0^6 {{{\left( {6 - r} \right)}^2}\left( { - r} \right)} dr \cr
& = \frac{1}{2}\int_0^6 {\left( {36 - 12r + {r^2}} \right)r} dr \cr
& = \frac{1}{2}\int_0^6 {\left( {36r - 12{r^2} + {r^3}} \right)} dr \cr
& = \frac{1}{2}\left[ {18{r^2} - 4{r^3} + \frac{1}{4}{r^4}} \right]_0^6 \cr
& = \frac{1}{2}\left[ {18{{\left( 6 \right)}^2} - 4{{\left( 6 \right)}^3} + \frac{1}{4}{{\left( 6 \right)}^4}} \right] - \frac{1}{2}\left[ 0 \right] \cr
& = \frac{1}{2}\left( {108} \right) \cr
& = 54 \cr
& \int_0^{\pi /4} {\int_0^6 {\frac{1}{2}{{\left( {6 - r} \right)}^2}r} dr} d\theta = \int_0^{\pi /4} {54} d\theta \cr
& {\text{Integrate}} \cr
& = 54\left[ {\frac{\pi }{4} - 0} \right] \cr
& = \frac{{27}}{2}\pi \cr} $$
