Answer
$\frac{9}{2}\pi $
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\int_0^3 {\int_0^{4 - z} {zdrdzd\theta } } } \cr
& = \int_0^{\pi /2} {\int_0^3 {\left[ {\int_0^{4 - z} {zdr} } \right]} dzd\theta } \cr
& {\text{Integrate with respect to }}r \cr
& \int_0^{4 - z} {zdr} = \left[ {zr} \right]_0^{4 - z} \cr
& \int_0^{4 - z} {zdr} = \left[ {z\left( {4 - z} \right) - z\left( 0 \right)} \right] \cr
& \int_0^{4 - z} {zdr} = 4z - {z^2} \cr
& = \int_0^{\pi /2} {\int_0^3 {\left( {4z - {z^2}} \right)} dzd\theta } \cr
& {\text{Integrate with respect to }}z \cr
& = \int_0^{\pi /2} {\left[ {2{z^2} - \frac{1}{3}{z^3}} \right]} _0^3d\theta \cr
& = \int_0^{\pi /2} {\left[ {2{{\left( 3 \right)}^2} - \frac{1}{3}{{\left( 3 \right)}^3}} \right]} d\theta \cr
& = \int_0^{\pi /2} {\left( 9 \right)} d\theta \cr
& {\text{Integrate}} \cr
& {\text{ = }}9\left[ {\frac{\pi }{2} - 0} \right] \cr
& {\text{ = }}\frac{9}{2}\pi \cr} $$
