$${x^2} + {y^2} = 36$$ and $${y^2} = 9x$$

$$ \therefore $$ $${x^2} + 9x - 36 = 0$$

$$ \Rightarrow x = 3, - 12$$

Required Area,

$$A = \pi {{{(6)}^2}} - 2\left[ {{A_1} + {A_2}} \right]$$

$$A = \pi {{{(6)}^2} - 2\left[ {\int_0^3 {\sqrt {9x} dx + \int_3^6 {\sqrt {36 - {x^2}} } dx} } \right]} $$

$$ = 36\pi - $$$$2\left[ {\left[ {3 \times {2 \over 3}{x^{{3 \over 2}}}} \right]_0^3 + \left[ {{x \over 2}\sqrt {36 - {x^2}} + {{36} \over 2}{{\sin }^{ - 1}}{x \over 6}} \right]_3^6} \right]$$

$$ = 36\pi - $$$$2\left[ {\left[ {2{x^{{3 \over 2}}}} \right]_0^3 + \left[ {{x \over 2}\sqrt {36 - {x^2}} + 18{{\sin }^{ - 1}}{x \over 6}} \right]_3^6} \right]$$

$$ = 36\pi - $$$$2\left[ {\left[ {6\sqrt 3 - 0} \right] + \left[ {\left( {0 + 18{{\sin }^{ - 1}}{6 \over 6}} \right) - \left( {{3 \over 2} \times 3\sqrt 3 + 18{{\sin }^{ - 1}}{3 \over 6}} \right)} \right]} \right]$$

$$ = 36\pi - $$$$2\left[ {\left[ {6\sqrt 3 } \right] + \left[ {\left( {{{18\pi } \over 2}} \right) - \left( {{3 \over 2} \times 3\sqrt 3 + {{18\pi } \over 6}} \right)} \right]} \right]$$

$$ = 36\pi - $$ $$\left( {12\sqrt 3 + 18\pi - 9\sqrt 3 - 6\pi } \right)$$

= $$24\pi - 3\sqrt 3 $$

Note :
(i) $$\int {\sqrt {{x^2} + {a^2}} dx = {1 \over 2}\left[ {x\sqrt {{x^2} + {a^2}} + {a^2}\log |x + \sqrt {{x^2} + {a^2}} |} \right]} + C$$

(ii) $$\int {\sqrt {{a^2} - {x^2}} dx = {1 \over 2}\left[ {x\sqrt {{a^2} - {x^2}} + {a^2}{{\sin }^{ - 1}}\left( {{x \over a}} \right)} \right]} + C$$

(iii) $$\int {\sqrt {{x^2} - {a^2}} dx = {1 \over 2}\left[ {x\sqrt {{x^2} - {a^2}} - {a^2}\log |x + \sqrt {{x^2} - {a^2}} |} \right]} + C$$