Chapter 9 - Multiveriable Calculus - Chapter Review - Review Exercises - Page 519: 75

$$\frac{4}{{15}}\left( {782 - {8^{5/2}}} \right)$$

$$\eqalign{ & \iint\limits_R {\sqrt {x + y} }dxdy;\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \leqslant x \leqslant 7,\,\,\,\,\,\,\,1 \leqslant y \leqslant 9 \cr & \cr & {\text{Replacing the limits for the region }}R \cr & = \int_1^9 {\int_0^7 {\sqrt {x + y} dx} dy} \cr & = \int_1^9 {\left[ {\int_0^7 {\sqrt {x + y} dx} } \right]dy} \cr & {\text{solve the inner integral treat }}y{\text{ as a constant}} \cr & = \int_0^7 {\sqrt {x + y} dx} \cr & = \int_0^7 {{{\left( {x + y} \right)}^{1/2}}dx} \cr & {\text{integrate by using the power rule}} \cr & = \left( {\frac{{{{\left( {x + y} \right)}^{3/2}}}}{{3/2}}} \right)_0^7 \cr & = \frac{2}{3}\left( {{{\left( {x + y} \right)}^{3/2}}} \right)_0^7 \cr & {\text{evaluating the limits in the variable }}x \cr & = \frac{2}{3}\left( {{{\left( {7 + y} \right)}^{3/2}} - {{\left( {0 + y} \right)}^{3/2}}} \right) \cr & {\text{simplifying}} \cr & = \frac{2}{3}\left( {{{\left( {7 + y} \right)}^{3/2}} - {y^{3/2}}} \right) \cr & \cr & \int_1^9 {\left[ {\int_0^7 {\sqrt {x + y} dx} } \right]dy} = \int_1^9 {\left( {\frac{2}{3}\left( {{{\left( {7 + y} \right)}^{3/2}} - {y^{3/2}}} \right)} \right)dy} \cr & = \frac{2}{3}\int_1^9 {\left( {{{\left( {7 + y} \right)}^{3/2}} - {y^{3/2}}} \right)dy} \cr & {\text{integrating}} \cr & = \frac{2}{3}\left( {\frac{{{{\left( {7 + y} \right)}^{5/2}}}}{{5/2}} - \frac{{{y^{5/2}}}}{{5/2}}} \right)_1^9 \cr & = \frac{4}{{15}}\left( {{{\left( {7 + y} \right)}^{5/2}} - {y^{5/2}}} \right)_1^9 \cr & {\text{evaluate}} \cr & = \frac{4}{{15}}\left( {{{\left( {7 + 9} \right)}^{5/2}} - {9^{5/2}}} \right) - \frac{4}{{15}}\left( {{{\left( {7 + 1} \right)}^{5/2}} - {1^{5/2}}} \right) \cr & {\text{simplifying}} \cr & = \frac{4}{{15}}\left( {1024 - 243} \right) - \frac{4}{{15}}\left( {{8^{5/2}} - {1^{5/2}}} \right) \cr & = \frac{4}{{15}}\left( {781 - {8^{5/2}} + 1} \right) \cr & = \frac{4}{{15}}\left( {782 - {8^{5/2}}} \right) \cr} $$