We are asked to evaluate the definite integral $\int_{-4}^3 [3-x-(x^2-a)] dx - \frac{\pi 3^2}{2}$.

1. Problem Description

We are asked to evaluate the definite integral $\int_{-4}^3 [3-x-(x^2-a)] dx - \frac{\pi 3^2}{2}$.

2. Solution Steps

First, we simplify the integrand:

$3 - x - (x^2 - a) = 3 - x - x^2 + a = -x^2 - x + (3+a)$.

Then we calculate the definite integral:

$\int_{-4}^3 (-x^2 - x + (3+a)) dx = [-\frac{x^3}{3} - \frac{x^2}{2} + (3+a)x]_{-4}^3$

$= (-\frac{3^3}{3} - \frac{3^2}{2} + (3+a)3) - (-\frac{(-4)^3}{3} - \frac{(-4)^2}{2} + (3+a)(-4))$

$= (-\frac{27}{3} - \frac{9}{2} + 9 + 3a) - (\frac{64}{3} - \frac{16}{2} - 12 - 4a)$

$= (-9 - \frac{9}{2} + 9 + 3a) - (\frac{64}{3} - 8 - 12 - 4a)$

$= (-\frac{9}{2} + 3a) - (\frac{64}{3} - 20 - 4a)$

$= -\frac{9}{2} + 3a - \frac{64}{3} + 20 + 4a$

$= 7a + 20 - \frac{9}{2} - \frac{64}{3} = 7a + \frac{120 - 27 - 128}{6} = 7a + \frac{-35}{6}$.

Now, we subtract $\frac{\pi 3^2}{2} = \frac{9\pi}{2}$ from the result:

$7a - \frac{35}{6} - \frac{9\pi}{2} = 7a - \frac{35}{6} - \frac{27\pi}{6} = 7a - \frac{35 + 27\pi}{6}$.

3. Final Answer

$7a - \frac{35 + 27\pi}{6}$