Answer
$${\text{area}} = 4\sqrt 2 - \frac{2}{3}\ln \left( {6 + \sqrt {32} } \right) + \ln \root 3 \of 6 $$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = \sqrt {9{x^2} - 4} ,\,\,\,\,y = 0,\,\,\,x = 2 \cr
& {\text{For }}y = 0\,\,\sqrt {9{x^2} - 4} = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9{x^2} - 4 = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9{x^2} = 4 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \pm \frac{2}{3} \cr
& {\text{Taking }}x = \frac{2}{3} \cr
& {\text{We can represents the enclosed by the curves as}} \cr
& {\text{area}} = \int_{2/3}^2 {\sqrt {9{x^2} - 4} } dx \cr
& {\text{integrate by tables using the formula }} \cr
& \int {\sqrt {{u^2} - {a^2}} } du = \frac{u}{2}\sqrt {{u^2} - {a^2}} - \frac{{{a^2}}}{2}\ln \left| {u + \sqrt {{u^2} - {a^2}} } \right| + C \cr
& \cr
& {\text{write }}\int_{2/3}^2 {\sqrt {9{x^2} - 4} } dx{\text{ in terms of }}u \cr
& \,\,\,\,\,\,u = 3x,\,\,\,du = 3dx,\,\, \cr
& \,\,\,\,\,\,x = 2,\,\,u = 6{\text{ and }}x = 2/3,\,\,\,u = 2 \cr
& \int_{2/3}^2 {\sqrt {9{x^2} - 4} } dx = \int_2^6 {\sqrt {{u^2} - {2^2}} \left( {\frac{1}{3}} \right)} du \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{3}\int_2^6 {\sqrt {{u^2} - {2^2}} } du \cr
& {\text{area}} = \frac{1}{3}\int_2^6 {\sqrt {{u^2} - {2^2}} } du \cr
& {\text{area}} = \frac{1}{3}\left[ {\frac{u}{2}\sqrt {{u^2} - 4} - \frac{4}{2}\ln \left| {u + \sqrt {{u^2} - 4} } \right|} \right]_2^6 \cr
& {\text{evaluate the limits and simplify}} \cr
& {\text{area}} = \frac{1}{3}\left[ {\frac{6}{2}\sqrt {{6^2} - 4} - 2\ln \left| {6 + \sqrt {{6^2} - 4} } \right|} \right] - \frac{1}{3}\left[ {\frac{6}{2}\sqrt {{2^2} - 4} - 2\ln \left| {6 + \sqrt {{2^2} - 4} } \right|} \right] \cr
& {\text{area}} = \frac{1}{3}\left[ {3\sqrt {32} - 2\ln \left| {6 + \sqrt {32} } \right|} \right] - \frac{1}{3}\left[ {\frac{6}{2}\sqrt 0 - 2\ln \left| {6 + \sqrt 0 } \right|} \right] \cr
& {\text{area}} = \sqrt {32} - \frac{2}{3}\ln \left( {6 + \sqrt {32} } \right) + \frac{2}{3}\ln \sqrt 6 \cr
& {\text{area}} = 4\sqrt 2 - \frac{2}{3}\ln \left( {6 + \sqrt {32} } \right) + \ln \root 3 \of 6 \cr} $$
