Answer
$A = 9$
Work Step by Step
$$\eqalign{
& {\text{Let the functions }}y = \sqrt x ,{\text{ }}y = \frac{1}{3}x,{\text{ on the interval }}0 \leqslant x \leqslant 16 \cr
& {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr
& \cr
& {\text{Find the intersection points}} \cr
& y = y \cr
& \sqrt x = \frac{1}{3}x \cr
& {\text{Solve for }}x \cr
& {\left( {\sqrt x } \right)^2} = \frac{1}{3}x \cr
& x = \frac{1}{9}{x^2} \cr
& 9x = {x^2} \cr
& {x^2} - 9x = 0 \cr
& x\left( {x - 9} \right) = 0 \cr
& x = 0,{\text{ }}x = 9 \cr
& {\text{We have the intervals }}\left[ {0,9} \right]{\text{ and }}\left[ {9,16} \right] \cr
& \cr
& {\text{We can find the area integrating with respect to }}x \cr
& A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx{\text{ }}\left( {\bf{1}} \right){\text{ }}\left( {{\text{see page 439}}} \right) \cr
& {\text{From the graph }} \cr
& \sqrt x \geqslant \frac{1}{3}x{\text{ on }}\left[ {0,9} \right]{\text{ and }}\frac{1}{3}x \geqslant \sqrt x {\text{ on }}\left[ {9,16} \right] \cr
& {\text{Therefore}} \cr
& A = \int_0^9 {\left( {\sqrt x - \frac{1}{3}x} \right)} dx + \int_9^{16} {\left( {\frac{1}{3}x - \sqrt x } \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {\frac{{{x^{3/2}}}}{{3/2}} - \frac{1}{3}\left( {\frac{{{x^2}}}{2}} \right)} \right]_0^9 + \left[ {\frac{1}{3}\left( {\frac{{{x^2}}}{2}} \right) - \frac{{{x^{3/2}}}}{{3/2}}} \right]_9^{16} \cr
& A = \left[ {\frac{{2{x^{3/2}}}}{3} - \frac{{{x^2}}}{6}} \right]_0^9 + \left[ {\frac{{{x^2}}}{6} - \frac{{2{x^{3/2}}}}{3}} \right]_9^{16} \cr
& {\text{Evaluate the limits}} \cr
& A = \left[ {\frac{{2{{\left( 9 \right)}^{3/2}}}}{3} - \frac{{{{\left( 9 \right)}^2}}}{6}} \right] + \left[ {\frac{{{{\left( {16} \right)}^2}}}{6} - \frac{{2{{\left( {16} \right)}^{3/2}}}}{3}} \right] - \left[ {\frac{{{{\left( 9 \right)}^2}}}{6} - \frac{{2{{\left( 9 \right)}^{3/2}}}}{3}} \right] \cr
& A = \frac{9}{2} + 0 - \left( { - \frac{9}{2}} \right) \cr
& A = 9 \cr} $$
