Answer
$$A = 54$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 9 - {x^2}{\text{; }}\left[ {0,6} \right] \cr
& {\text{First check if the graph crosses the }}x - {\text{axis in the given interval}} \cr
& 9 - {x^2} = 0 \cr
& {\text{solve for }}x \cr
& - {x^2} = - 9 \cr
& x = \pm 3 \cr
& {\text{the graph crosses the }}x{\text{ - axis at }}x = \pm 3{\text{ and }}x = 3{\text{ is in the given interval }}\left[ {0,6} \right],{\text{ }} \cr
& f\left( 2 \right) = 9 - {\left( 2 \right)^2} = 5,{\text{ }}f\left( x \right){\text{ is positive for }}\left( {0,3} \right) \cr
& {\text{then}} \cr
& {\text{the definite integral of the total area will be}}: \cr
& A = \int_0^3 {\left( {9 - {x^2}} \right)} dx + \left| {\int_3^6 {\left( {9 - {x^2}} \right)} dx} \right| \cr
& A = \left( {9x - \frac{{{x^3}}}{3}} \right)_0^3 + \left| {\left( {9x - \frac{{{x^3}}}{3}} \right)_3^6} \right| \cr
& A = \left( {9\left( 3 \right) - \frac{{{{\left( 3 \right)}^3}}}{3}} \right) - \left( {9\left( 0 \right) - \frac{{{{\left( 0 \right)}^3}}}{3}} \right) + \left| {\left( {9\left( 6 \right) - \frac{{{{\left( 6 \right)}^3}}}{3}} \right) - \left( {9\left( 3 \right) - \frac{{{{\left( 3 \right)}^3}}}{3}} \right)} \right| \cr
& {\text{simplifying}} \cr
& A = 18 - 0 + \left| { - 18 - 18} \right| \cr
& A = 18 + 36 \cr
& A = 54 \cr} $$
