Answer
$$A = \frac{{81}}{2}$$
Work Step by Step
$$\eqalign{
& y = 8 - x \Rightarrow x = 8 - y \cr
& {\text{The intersection point between }}x = 8 - y{\text{ and }}x = \frac{{{{\left( {y - 2} \right)}^2}}}{3},{\text{ is}} \cr
& 8 - y = \frac{{{{\left( {y - 2} \right)}^2}}}{3} \cr
& 24 - 3y = {y^2} - 4y + 4 \cr
& {y^2} - 4y + 3y + 4 - 24 = 0 \cr
& {y^2} - y - 20 = 0 \cr
& \left( {y - 5} \right)\left( {y + 4} \right) = 0 \cr
& {y_1} = - 4,{\text{ }}{y_2} = 5 \cr
& {\text{The area of the region is given by:}} \cr
& A = \int_{ - 4}^5 {\left[ {8 - y- \frac{{{{\left( {y - 2} \right)}^2}}}{3}} \right]dy} \cr
& {\text{Integrating}} \cr
& A = \left[ {8y- \frac{1}{2}{y^2} - \frac{{{{\left( {y - 2} \right)}^3}}}{9}} \right]_{ - 4}^5 \cr
& A = \left[ {8\left( 5 \right) - \frac{1}{2}{{\left( 5 \right)}^2} - \frac{{{{\left( {5 - 2} \right)}^3}}}{9}} \right] - \left[ {8\left( { - 4} \right) - \frac{1}{2}{{\left( { - 4} \right)}^2} - \frac{{{{\left( { - 4 - 2} \right)}^3}}}{9}} \right] \cr
& {\text{Simplifying}} \cr
& A = \frac{{49}}{2} + 16 \cr
& A = \frac{{81}}{2} \cr} $$
