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

Given region, <br/><br/>$E=\left\{(x, y): 3-x \leq y \leq \sqrt{9-x^2}, 0 \leq x \leq 3\right\}$ <br/><br/>Since, point $(p, p+1)$ lie on line $y=x+1$ <br/><br/>$\therefore$ Point of intersection of $y=x+1$ and $y=3-x$ <br/><br/>i.e., $x+1=3-x$ <br/><br/>$\Rightarrow$ $2 x=2 \Rightarrow x=1$ <br/><br/>and $y=2$ <br/><br/>and point of intersection of $y=x+1$ and <br/><br/>$y=\sqrt{9-x^2}$ <br/><br/>i.e., $(x+1)^2=9-x^2$ <br/><br/>$\begin{array}{lc} &\Rightarrow x^2+1+2 x=9-x^2 \\\\ &\Rightarrow 2 x^2+2 x-8=0 \\\\ &\Rightarrow x^2+x-4=0 \\\\ &\Rightarrow x=\frac{-1 \pm \sqrt{1+4(1)(4)}}{2} \\\\ &\Rightarrow x=\frac{-1 \pm \sqrt{17}}{2}\end{array}$ <br/><br/>$\begin{array}{lll}\Rightarrow & x=\frac{-1+\sqrt{17}}{2}, & \text { (Since, } x \in[0,3]) \\\\ \therefore & p \in \left(1, \frac{-1+\sqrt{17}}{2}\right)\end{array}$ <br/><br/>$\Rightarrow a=1, b=\frac{-1+\sqrt{17}}{2}$ <br/><br/>$\begin{aligned} & \therefore b^2+b-a \\\\ &= \frac{1+17-2 \sqrt{17}}{4}+\frac{(-1+\sqrt{17})}{2}-1 \\\\ &= \frac{18-2 \sqrt{17}-2+2 \sqrt{17}-4}{4} \\\\ &= \frac{12}{4}=3\end{aligned}$