Let the point (p, p+1) lie inside the region | vedclass

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(D) The region $E$ is bounded by $y \geq 3-x$ and $y \leq \sqrt{9-x^2}$ for $0 \leq x \leq 3$.The point $(p, p+1)$ lies on the line $y = x+1$.To find

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$3$

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(D) The region $E$ is bounded by $y \geq 3-x$ and $y \leq \sqrt{9-x^2}$ for $0 \leq x \leq 3$.The point $(p, p+1)$ lies on the line $y = x+1$.To find the range of $p$,we find the intersection of $y = x+1$ with the boundaries of $E$.$1$. Intersection with $y = 3-x$:$x+1 = 3-x \implies 2x = 2 \implies x = 1$.Thus,$p = 1$ (this is $a$).$2$. Intersection with $y = \sqrt{9-x^2}$:$x+1 = \sqrt{9-x^2} \implies (x+1)^2 = 9-x^2 \implies x^2+2x+1 = 9-x^2 \implies 2x^2+2x-8 = 0 \implies x^2+x-4 = 0$.Using the quadratic formula,$x = \frac{-1 \pm \sqrt{1+16}}{2} = \frac{-1 \pm \sqrt{17}}{2}$.Since $x \geq 0$,we take $x = \frac{-1+\sqrt{17}}{2}$.Thus,$p = \frac{-1+\sqrt{17}}{2}$ (this is $b$).So,$p \in \left(1, \frac{-1+\sqrt{17}}{2}\right)$,where $a = 1$ and $b = \frac{-1+\sqrt{17}}{2}$.We need to calculate $b^2+b-a^2$:Since $b^2+b-4 = 0$,we have $b^2+b = 4$.Therefore,$b^2+b-a^2 = 4 - (1)^2 = 4-1 = 3$.

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

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