Let the point $$p, p+1$$ lie inside the region $$E = \{x, y: 3-x \le y \le \sqrt{9-x^2}, 0 \le x \le 3\}$$. If the set of all values of $$p$$ is the interval $$(a, b)$$, then $$b^2 + b - a^2$$ is equal to ______.

The region $$E = \{(x,y): 3-x \le y \le \sqrt{9-x^2},\; 0 \le x \le 3\}$$ is bounded below by the line $$y = 3-x$$ and above by the circle $$x^2+y^2 = 9$$ (upper semicircle) in the first quadrant.

The point $$(p, p+1)$$ lies inside E, so:

The first condition requires $$3-p \le p+1 \implies 2 \le 2p \implies p \ge 1$$

Additionally, $$p+1 \le \sqrt{9-p^2} \implies (p+1)^2 \le 9-p^2$$

$$ p^2+2p+1 \le 9-p^2 \implies 2p^2+2p-8 \le 0 \implies p^2+p-4 \le 0 $$

$$ p \le \frac{-1+\sqrt{17}}{2} $$

Also, $$0 \le p \le 3$$ (automatically satisfied)

So the interval is $$\left(1,\; \frac{-1+\sqrt{17}}{2}\right)$$ where $$a = 1$$ and $$b = \frac{-1+\sqrt{17}}{2}$$.

Computing $$b^2 + b - a^2$$:

$$ b^2 + b = b(b+1) = \frac{-1+\sqrt{17}}{2} \cdot \frac{1+\sqrt{17}}{2} = \frac{(-1+\sqrt{17})(1+\sqrt{17})}{4} = \frac{17-1}{4} = 4 $$

$$ b^2 + b - a^2 = 4 - 1 = 3 $$

The answer is 3.