Answer
$$\frac{1}{6}\pi $$
Work Step by Step
$$\eqalign{
& \int_1^2 {\int_z^2 {\int_0^{\sqrt 3 y} {\frac{y}{{{x^2} + {y^2}}}} } } dxdydz \cr
& = \int_1^2 {\int_z^2 {\left[ {\int_0^{\sqrt 3 y} {\frac{y}{{{x^2} + {y^2}}}} dx} \right]} } dydz \cr
& {\text{Solve the inner integral}}{\text{, integrate with respect to }}x,{\text{ treat }}y{\text{ and }}z{\text{ as constants}} \cr
& \int_0^{\sqrt 3 y} {\frac{y}{{{x^2} + {y^2}}}} dx = \left[ {\arctan \left( {\frac{x}{y}} \right)} \right]_0^{\sqrt 3 y} \cr
& {\text{Evaluate the limits in the variable }}x \cr
& = \arctan \left( {\frac{{\sqrt 3 y}}{y}} \right) - \arctan \left( {\frac{0}{y}} \right) \cr
& = \frac{\pi }{3} - 0 \cr
& {\text{Then}}{\text{,}} \cr
& \int_1^2 {\int_z^2 {\left[ {\int_0^{\sqrt 3 y} {\frac{y}{{{x^2} + {y^2}}}} dx} \right]} } dydz = \int_1^2 {\int_z^2 {\left( {\frac{\pi }{3}} \right)} } dydz \cr
& = \frac{\pi }{3}\int_1^2 {\left( {\int_z^2 {dy} } \right)} dz \cr
& {\text{Integrate with respect to }}y,{\text{ treat }}z{\text{ as a constant}} \cr
& \int_z^2 {dy} = \left( z \right)_z^2 = 2 - z \cr
& \frac{\pi }{3}\int_1^2 {\left( {\int_z^2 {dy} } \right)} dz = \frac{\pi }{3}\int_1^2 {\left( {2 - z} \right)} dz \cr
& {\text{Integrate}} \cr
& = \frac{\pi }{3}\left( {2z - \frac{{{z^2}}}{2}} \right)_1^2 \cr
& {\text{Evaluate}} \cr
& = \frac{\pi }{3}\left( {2\left( 2 \right) - \frac{{{{\left( 2 \right)}^2}}}{2}} \right) - \frac{\pi }{3}\left( {2\left( 1 \right) - \frac{{{{\left( 1 \right)}^2}}}{2}} \right) \cr
& = \frac{\pi }{3}\left( 2 \right) - \frac{\pi }{3}\left( {\frac{3}{2}} \right) \cr
& = \frac{1}{6}\pi \cr} $$
