## Calculus with Applications (10th Edition)

$$V = \frac{2}{{15}}\left( {{2^{5/2}} - 2} \right)$$
\eqalign{ & z = x\sqrt {{x^2} + y} ;\,\,\,\,\,\,\,0 \leqslant x \leqslant 1,\,\,\,\,\,0 \leqslant y \leqslant 1 \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_0^1 {\int_0^1 {x\sqrt {{x^2} + y} dx} dy} \cr & V = \int_0^1 {\left[ {\int_0^1 {x\sqrt {{x^2} + y} dx} } \right]dy} \cr & {\text{solve the inner integral by treating }}y{\text{ as a constant}} \cr & = \int_0^1 {x\sqrt {{x^2} + y} dx} \cr & = \frac{1}{2}\int_0^1 {\sqrt {{x^2} + y} \left( {2x} \right)dx} \cr & {\text{integrate by using the power rule }} \cr & = \frac{1}{2}\left[ {\frac{{{{\left( {{x^2} + y} \right)}^{3/2}}}}{{3/2}}} \right]_0^1 \cr & = \frac{1}{3}\left[ {{{\left( {{x^2} + y} \right)}^{3/2}}} \right]_0^1 \cr & {\text{evaluating the limits in the variable }}x \cr & = \frac{1}{3}\left[ {{{\left( {{1^2} + y} \right)}^{3/2}} - {{\left( {{0^2} + y} \right)}^{3/2}}} \right] \cr & {\text{simplifying}} \cr & = \frac{1}{3}{\left( {1 + y} \right)^{3/2}} - \frac{1}{3}{y^{3/2}} \cr & \cr & V = \int_0^1 {\left[ {\int_0^1 {x\sqrt {{x^2} + y} dx} } \right]dy} \cr & V = \int_0^1 {\left( {\frac{1}{3}{{\left( {1 + y} \right)}^{3/2}} - \frac{1}{3}{y^{3/2}}} \right)dy} \cr & {\text{integrating}} \cr & V = \frac{1}{3}\left[ {\frac{{{{\left( {1 + y} \right)}^{5/2}}}}{{5/2}} - \frac{{{y^{5/2}}}}{{5/2}}} \right]_0^1 \cr & V = \frac{1}{3}\left[ {\frac{{2{{\left( {1 + y} \right)}^{5/2}}}}{5} - \frac{{2{y^{5/2}}}}{5}} \right]_0^1 \cr & V = \frac{2}{{15}}\left[ {{{\left( {1 + y} \right)}^{5/2}} - {y^{5/2}}} \right]_0^1 \cr & {\text{evaluating}} \cr & = \frac{2}{{15}}\left[ {{{\left( {1 + 1} \right)}^{5/2}} - {1^{5/2}}} \right] - \frac{2}{{15}}\left[ {{{\left( {0 + 1} \right)}^{5/2}} - {0^{5/2}}} \right] \cr & = \frac{2}{{15}}\left( {{2^{5/2}} - 1} \right) - \frac{2}{{15}} \cr & = \frac{2}{{15}}\left( {{2^{5/2}} - 1 - 1} \right) \cr & V = \frac{2}{{15}}\left( {{2^{5/2}} - 2} \right) \cr}