Answer
$$A = \frac{{3\pi }}{4}$$
Work Step by Step
$$\eqalign{
& y = \frac{{\sqrt {9 - {x^2}} }}{3} \cr
& {\text{The area is enclosed by }}y{\text{ and the first quadrant}}{\text{; thus the area is }} \cr
& \,\,\,\,\,A = \int_0^3 {\frac{{\sqrt {9 - {x^2}} }}{3}} dx \cr
& {\text{Use the Trigonometric Substitutions:}} \cr
& {\text{We set}}{\text{, }}x = 3\sin \theta ,{\text{ }}dx = 3\cos \theta d\theta \cr
& {\text{ }}x = 3\sin \theta ,\,\,\,\theta = {\sin ^{ - 1}}\left( {\frac{x}{3}} \right) \cr
& x = 0 \to \theta = {\sin ^{ - 1}}\left( {\frac{0}{3}} \right) = 0 \cr
& x = 3 \to \theta = {\sin ^{ - 1}}\left( {\frac{3}{3}} \right) = \frac{\pi }{2} \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \,\,\,\,\,A = \int_0^{\pi /2} {\frac{{\sqrt {9 - {{\left( {3\sin \theta } \right)}^2}} }}{3}} \left( {3\cos \theta } \right)d\theta \cr
& \,\,\,\,\,A = \int_0^{\pi /2} {\frac{{\sqrt {9 - 9{{\sin }^2}\theta } }}{3}} \left( {3\cos \theta } \right)d\theta \cr
& \,\,\,\,\,A = \int_0^{\pi /2} {3\sqrt {1 - {{\sin }^2}\theta } } \left( {\cos \theta } \right)d\theta \cr
& \,\,\,\,\,A = 3\int_0^{\pi /2} {\sqrt {{{\cos }^2}\theta } } \left( {\cos \theta } \right)d\theta \cr
& \,\,\,\,\,A = 3\int_0^{\pi /2} {{{\cos }^2}\theta } d\theta \cr
& \,\,\,\,\,A = 3\int_0^{\pi /2} {\left( {\frac{{1 + \cos 2\theta }}{2}} \right)} d\theta \cr
& \,\,\,\,\,A = \frac{3}{2}\int_0^{\pi /2} {\left( {1 + \cos 2\theta } \right)} d\theta \cr
& {\text{Integrating, we get}} \cr
& \,\,\,\,\,A = \frac{3}{2}\left( {\theta + \frac{1}{2}\sin 2\theta } \right)_0^{\pi /2} \cr
& \,\,\,\,\,A = \frac{3}{2}\left( {\frac{\pi }{2} + \frac{1}{2}\sin 2\left( {\frac{\pi }{2}} \right)} \right) - \frac{3}{2}\left( {0 + \frac{1}{2}\sin 2\left( 0 \right)} \right) \cr
& \,\,\,\,\,A = \frac{3}{2}\left( {\frac{\pi }{2} + 0} \right) - \frac{3}{2}\left( 0 \right) \cr
& \,\,\,\,\,A = \frac{{3\pi }}{4} \cr} $$
