If the area bounded by the curve $$2y^2 = 3x$$, lines $$x + y = 3$$, $$y = 0$$ and outside the circle $$(x-3)^2 + y^2 = 2$$ is A, then $$4(\pi + 4A)$$ is equal to _____.

$$y^2 = \frac{3x}{2}, \ x + y = 3, \ y = 0$$

$$2y^2 = 3(3 - y)$$

$$2y^2 + 3y - 9 = 0$$

$$2y^2 - 3y + 6y - 9 = 0$$

$$(2y - 3)(y + 2) = 0; \ y = 3/2$$

$$\text{Area } \left( \int_0^{3/2} (x_R - x_2) dy \right) - A_1$$

$$= \int_0^{3/2} \left( (3 - y) - \frac{2y^2}{3} \right) dy - \frac{\pi}{8}(2)$$

$$A = \left( 3y - \frac{y^2}{2} - \frac{2y^3}{9} \right)_0^{3/2} - \frac{\pi}{4}$$

$$4A + \pi = 4 \left[ \frac{9}{2} - \frac{9}{8} - \frac{3}{4} \right] = \frac{21}{2} = 10.50$$

$$\therefore 4(4A + \pi) = 42$$