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

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

Now by observation $(p,p+1)$ lies on $x^2+y^2=9$

So, solving $y-x=1$ and $x^2+y^2=9,$ we get $x^2+x-4=0$

$x=\frac{-1\pm \sqrt{1+4.4}}{2}=\frac{\pm \sqrt{17}-1}{2}$

Hence, $P\in (1,\frac{\sqrt{17}-1}{2})$

Now on comparing with $p\in (a,b)$ we get $\Rightarrow a=1,b^2+b=4$

Hence the values of $b^2+b+a^2=4+1=5$