## Calculus with Applications (10th Edition)

$$V = \frac{{308}}{3}$$
\eqalign{ & z = {x^2} + {y^2};\,\,\,\,\,\,\,3 \leqslant x \leqslant 5,\,\,\,\,\,2 \leqslant y \leqslant 4 \cr & {\text{by the equation just given}}{\text{, the volume is}} \cr & V = \iint\limits_R {f\left( {x,y} \right)}dxdy \cr & {\text{replacing the limits for the region }}R{\text{ and }}f\left( {x,y} \right) = z \cr & V = \int_2^4 {\int_3^5 {\left( {{x^2} + {y^2}} \right)dx} dy} \cr & V = \int_2^4 {\left[ {\int_3^5 {\left( {{x^2} + {y^2}} \right)dx} } \right]dy} \cr & {\text{solve the inner integral by treating }}y{\text{ as a constant}} \cr & = \int_3^5 {\left( {{x^2} + {y^2}} \right)dx} \cr & {\text{integrate }} \cr & = \left[ {\frac{{{x^3}}}{3} + x{y^2}} \right]_3^5 \cr & {\text{evaluating the limits in the variable }}x \cr & = \left( {\frac{{{{\left( 5 \right)}^3}}}{3} + \left( 5 \right){y^2}} \right) - \left( {\frac{{{{\left( 3 \right)}^3}}}{3} + \left( 3 \right){y^2}} \right) \cr & {\text{simplifying}} \cr & = \left( {\frac{{125}}{3} + 5{y^2}} \right) - \left( {9 + 3{y^2}} \right) \cr & = \frac{{125}}{3} + 5{y^2} - 9 - 3{y^2} \cr & = \frac{{98}}{3} + 2{y^2} \cr & \cr & V = \int_2^4 {\left[ {\int_3^5 {\left( {{x^2} + {y^2}} \right)dx} } \right]dy} \cr & V = \int_2^4 {\left( {\frac{{98}}{3} + 2{y^2}} \right)dy} \cr & {\text{integrating}} \cr & V = \left( {\frac{{98}}{3}y + \frac{{2{y^3}}}{3}} \right)_2^4 \cr & V = \left( {\frac{{98}}{3}\left( 4 \right) + \frac{{2{{\left( 4 \right)}^3}}}{3}} \right) - \left( {\frac{{98}}{3}\left( 2 \right) + \frac{{2{{\left( 2 \right)}^3}}}{3}} \right) \cr & V = \frac{{520}}{3} - \frac{{212}}{3} \cr & V = \frac{{308}}{3} \cr}