Answer
$$8$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {\int_0^2 {\int_0^1 {\left( {{x^2} + {y^2} + {z^2}} \right)} } } dxdydz \cr
& = \int_{ - 1}^1 {\int_0^2 {\left[ {\int_0^1 {\left( {{x^2} + {y^2} + {z^2}} \right)} dx} \right]} } dydz \cr
& {\text{Solve the inner integral}}{\text{, integrate with respect to }}x,{\text{ and treat }}y{\text{ and }}z{\text{ as constants}} \cr
& \int_0^1 {\left( {{x^2} + {y^2} + {z^2}} \right)} dx = \left[ {\frac{{{x^3}}}{3} + x{y^2} + x{z^2}} \right]_0^1 \cr
& {\text{Evaluate the limits in the variable }}x \cr
& = \left( {\frac{{{{\left( 1 \right)}^3}}}{3} + \left( 1 \right){y^2} + \left( 1 \right){z^2}} \right) - \left( {\frac{{{{\left( 0 \right)}^3}}}{3} + \left( 0 \right){y^2} + \left( 0 \right){z^2}} \right) \cr
& = \frac{1}{3} + {y^2} + {z^2} \cr
& {\text{Then}}{\text{,}} \cr
& = \int_{ - 1}^1 {\int_0^2 {\left( {\frac{1}{3} + {y^2} + {z^2}} \right)} } dydz \cr
& = \int_{ - 1}^1 {\left[ {\int_0^2 {\left( {\frac{1}{3} + {y^2} + {z^2}} \right)} dy} \right]} dz \cr
& {\text{Integrate with respect to }}y,{\text{ treat }}z{\text{ as a constant}} \cr
& \int_0^2 {\left( {\frac{1}{3} + {y^2} + {z^2}} \right)} dy = \left( {\frac{1}{3}y + \frac{{{y^3}}}{3} + {z^2}y} \right)_0^2 \cr
& = \left( {\frac{1}{3}\left( 2 \right) + \frac{{{{\left( 2 \right)}^3}}}{3} + {z^2}\left( 2 \right)} \right) - \left( {\frac{1}{3}\left( 0 \right) + \frac{{{{\left( 0 \right)}^3}}}{3} + {z^2}\left( 0 \right)} \right) \cr
& = \frac{2}{3} + \frac{8}{3} + 2{z^2} \cr
& = \frac{{10}}{3} + 2{z^2} \cr
& \int_{ - 1}^1 {\left[ {\int_0^2 {\left( {\frac{1}{3} + {y^2} + {z^2}} \right)} dy} \right]} dz = \int_{ - 1}^1 {\left( {\frac{{10}}{3} + 2{z^2}} \right)} dz \cr
& {\text{Integrate}} \cr
& = \left( {\frac{{10}}{3}z + \frac{{2{z^3}}}{3}} \right)_{ - 1}^1 \cr
& {\text{Evaluate}} \cr
& = \left( {\frac{{10}}{3}\left( 1 \right) + \frac{{2{{\left( 1 \right)}^3}}}{3}} \right) - \left( {\frac{{10}}{3}\left( { - 1} \right) + \frac{{2{{\left( { - 1} \right)}^3}}}{3}} \right) \cr
& = 4 - \left( { - 4} \right) \cr
& = 8 \cr} $$
