## Calculus with Applications (10th Edition)

$$V = \frac{{40}}{3}$$
\eqalign{ & z = {x^2};\,\,\,\,\,\,\,0 \leqslant x \leqslant 2,\,\,\,\,\,0 \leqslant y \leqslant 5 \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_0^5 {\int_0^2 {{x^2}dx} dy} \cr & V = \int_0^5 {\left[ {\int_0^2 {{x^2}dx} } \right]dy} \cr & {\text{solve the inner integral by treating }}y{\text{ as a constant}} \cr & = \int_0^2 {{x^2}dx} \cr & {\text{integrate }} \cr & = \left[ {\frac{{{x^3}}}{3}} \right]_0^2 \cr & {\text{evaluating the limits in the variable }}x \cr & = \frac{1}{3}\left( {{{\left( 2 \right)}^3} - {{\left( 0 \right)}^3}} \right) \cr & {\text{simplifying}} \cr & = \frac{8}{3} \cr & \cr & V = \int_0^5 {\left[ {\int_0^2 {{x^2}dx} } \right]dy} \cr & V = \int_0^5 {\frac{8}{3}dy} \cr & {\text{integrating}} \cr & V = \frac{8}{3}\left( y \right)_0^5 \cr & V = \frac{8}{3}\left( {5 - 0} \right) \cr & V = \frac{{40}}{3} \cr}