Chapter 13 - Multiple Integration - 13.1 Double Integrals over Rectangular Regions - 13.1 Exercises - Page 971: 21

$$\frac{4}{3}$$

$$\eqalign{ & \iint\limits_R {\sqrt {\frac{x}{y}} }dA;\,\,\,\,\,\,\,R = \left\{ {\left( {x,y} \right):0 \leqslant x \leqslant 1,\,\,\,\,1 \leqslant y \leqslant 4} \right\} \cr & {\text{Convert to an iterated integral substituting the region }}R \cr & = \int_1^4 {\int_0^1 {\sqrt {\frac{x}{y}} } } dxdy \cr & = \int_1^4 {\int_0^1 {\frac{{\sqrt x }}{{\sqrt y }}} } dxdy \cr & = \int_1^4 {\left[ {\int_0^1 {\frac{{\sqrt x }}{{\sqrt y }}} dx} \right]} dy \cr & {\text{solve the inner integral}}{\text{, treat }}y{\text{ as a constant}} \cr & = \frac{1}{{\sqrt y }}\int_0^1 {\sqrt x } dx \cr & = \frac{1}{{\sqrt y }}\left( {\frac{2}{3}{x^{3/2}}} \right)_0^1 \cr & {\text{evaluating the limits for the variable }}x \cr & = \frac{2}{{3\sqrt y }}\left( {{{\left( 1 \right)}^{3/2}} - {{\left( 0 \right)}^{3/2}}} \right) \cr & {\text{simplifying}} \cr & = \frac{2}{{3\sqrt y }} \cr & \cr & = \int_1^4 {\left[ {\int_0^1 {\frac{{\sqrt x }}{{\sqrt y }}} dx} \right]} dy = \int_1^4 {\frac{2}{{3\sqrt y }}} dy \cr & = \frac{2}{3}\int_1^4 {\frac{1}{{\sqrt y }}} dy \cr & = \frac{2}{3}\int_1^4 {{y^{ - 1/2}}} dy \cr & {\text{integrating}} \cr & = \frac{2}{3}\left( {\frac{{{y^{1/2}}}}{{1/2}}} \right)_1^4 \cr & = \frac{4}{3}\left( {\sqrt y } \right)_1^4 \cr & {\text{evaluate}} \cr & = \frac{4}{3}\left( {\sqrt 4 - \sqrt 1 } \right) \cr & = \frac{4}{3}\left( {2 - 1} \right) \cr & = \frac{4}{3} \cr} $$