## Calculus with Applications (10th Edition)

$$\frac{4}{{15}}\left( {33 - {3^{5/2}} - {2^{5/2}}} \right)$$
\eqalign{ & \iint\limits_R {\sqrt {x + y} }dydx;\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \leqslant x \leqslant 3,\,\,\,\,\,\,\,0 \leqslant y \leqslant 1 \cr & {\text{replacing the limits for the region }}R \cr & \int_1^3 {\int_0^1 {\sqrt {x + y} dy} dx} \cr & = \int_1^3 {\left[ {\int_0^1 {\sqrt {x + y} dy} } \right]dx} \cr & {\text{solve the inner integral, treating }}x{\text{ as a constant}} \cr & = \int_0^1 {\sqrt {x + y} dy} \cr & = \int_0^1 {{{\left( {x + y} \right)}^{1/2}}dy} \cr & {\text{integrate using the power rule}} \cr & = \left[ {\frac{{{{\left( {x + y} \right)}^{3/2}}}}{{3/2}}} \right]_0^1 \cr & = \frac{2}{3}\left[ {{{\left( {x + y} \right)}^{3/2}}} \right]_0^1 \cr & {\text{evaluating the limits in the variable }}y \cr & = \frac{2}{3}\left[ {{{\left( {x + 1} \right)}^{3/2}} - {{\left( {x + 0} \right)}^{3/2}}} \right] \cr & = \frac{2}{3}\left[ {{{\left( {x + 1} \right)}^{3/2}} - {x^{3/2}}} \right] \cr & {\text{then}} \cr & \cr & \int_1^3 {\left[ {\int_0^1 {\sqrt {x + y} dy} } \right]dx} = \frac{2}{3}\int_1^3 {\left[ {{{\left( {x + 1} \right)}^{3/2}} - {x^{3/2}}} \right]dx} \cr & {\text{integrating}} \cr & = \frac{2}{3}\left[ {\frac{{{{\left( {x + 1} \right)}^{5/2}}}}{{5/2}} - \frac{{{x^{5/2}}}}{{5/2}}} \right]_1^3 \cr & = \frac{2}{3}\left[ {\frac{{2{{\left( {x + 1} \right)}^{5/2}}}}{5} - \frac{{2{x^{5/2}}}}{5}} \right]_1^3 \cr & = \frac{4}{{15}}\left[ {{{\left( {x + 1} \right)}^{5/2}} - {x^{5/2}}} \right]_1^3 \cr & {\text{evaluating}} \cr & = \frac{4}{{15}}\left[ {{{\left( {3 + 1} \right)}^{5/2}} - {3^{5/2}}} \right] - \frac{4}{{15}}\left[ {{{\left( {1 + 1} \right)}^{5/2}} - {1^{5/2}}} \right] \cr & {\text{simplifying}} \cr & = \frac{4}{{15}}\left[ {32 - {3^{5/2}}} \right] - \frac{4}{{15}}\left[ {{2^{5/2}} - 1} \right] \cr & = \frac{4}{{15}}\left( {32 - {3^{5/2}} - {2^{5/2}} + 1} \right) \cr & = \frac{4}{{15}}\left( {33 - {3^{5/2}} - {2^{5/2}}} \right) \cr}