Chapter 9 - Multiveriable Calculus - Chapter Review - Review Exercises - Page 519: 74

$$\frac{4}{{15}}\left( {{{11}^{5/2}} - {7^{5/2}} - {8^{5/2}} + 32} \right)$$

$$\eqalign{ & \iint\limits_R {\sqrt {2x + y} }dxdy;\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \leqslant x \leqslant 3,\,\,\,\,\,\,\,2 \leqslant y \leqslant 5 \cr & \cr & {\text{Replacing the limits for the region }}R \cr & = \int_2^5 {\int_1^3 {\sqrt {2x + y} dx} dy} \cr & = \int_2^5 {\left[ {\int_1^3 {\sqrt {2x + y} dx} } \right]dy} \cr & {\text{solve the inner integral treat }}y{\text{ as a constant}} \cr & = \int_1^3 {{{\left( {2x + y} \right)}^{1/2}}dx} \cr & = \frac{1}{2}\int_1^3 {{{\left( {2x + y} \right)}^{1/2}}\left( 2 \right)dx} \cr & {\text{integrate by using the power rule}} \cr & = \left( {\frac{{{{\left( {2x + y} \right)}^{3/2}}}}{{3/2}}} \right)_1^3 \cr & = \frac{2}{3}\left( {{{\left( {2x + y} \right)}^{3/2}}} \right)_1^3 \cr & {\text{evaluating the limits in the variable }}x \cr & = \frac{2}{3}\left( {{{\left( {2\left( 3 \right) + y} \right)}^{3/2}}} \right) - \frac{2}{3}\left( {{{\left( {2\left( 1 \right) + y} \right)}^{3/2}}} \right) \cr & {\text{simplifying}} \cr & = \frac{2}{3}{\left( {6 + y} \right)^{3/2}} - \frac{2}{3}{\left( {2 + y} \right)^{3/2}} \cr & \cr & \int_2^5 {\left[ {\int_1^3 {\sqrt {2x + y} dx} } \right]dy} = \int_2^5 {\left( {\frac{2}{3}{{\left( {6 + y} \right)}^{3/2}} - \frac{2}{3}{{\left( {2 + y} \right)}^{3/2}}} \right)dy} \cr & = \frac{2}{3}\int_2^5 {\left( {{{\left( {6 + y} \right)}^{3/2}} - {{\left( {2 + y} \right)}^{3/2}}} \right)dy} \cr & {\text{integrating}} \cr & = \frac{2}{3}\left( {\frac{{{{\left( {6 + y} \right)}^{5/2}}}}{{5/2}} - \frac{{{{\left( {2 + y} \right)}^{5/2}}}}{{5/2}}} \right)_2^5 \cr & = \frac{4}{{15}}\left( {{{\left( {6 + y} \right)}^{5/2}} - {{\left( {2 + y} \right)}^{5/2}}} \right)_2^5 \cr & {\text{evaluate}} \cr & = \frac{4}{{15}}\left( {{{\left( {6 + 5} \right)}^{5/2}} - {{\left( {2 + 5} \right)}^{5/2}}} \right) - \frac{4}{{15}}\left( {{{\left( {6 + 2} \right)}^{5/2}} - {{\left( {2 + 2} \right)}^{5/2}}} \right) \cr & {\text{simplifying}} \cr & = \frac{4}{{15}}\left( {{{11}^{5/2}} - {7^{5/2}}} \right) - \frac{4}{{15}}\left( {{8^{5/2}} - {4^{5/2}}} \right) \cr & = \frac{4}{{15}}\left( {{{11}^{5/2}} - {7^{5/2}} - {8^{5/2}} + 32} \right) \cr} $$