Animated Solution for Mathematics - Definite Integration: 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 ________.

Welcome, future engineer. Today, we are not just solving an equation; we are exploring a geometric landscape. Imagine you are standing in the first quadrant of the Cartesian plane.

You have a region $E$ defined by two powerful boundaries. The first is a straight line, $x+y=3$, which acts as a floor. The second is a circular arc, $x^2+y^2=9$, which acts as a ceiling.

Our region $E$ is the space trapped between this floor and this ceiling. It is a beautiful, curved sliver of space where $0\le x\le 3$. Before we touch any algebra, visualize this: the line cuts through the circle, and we are looking at the area above the line and below the circle.

The Path of $(p,p+1)$

Now, consider a point $(p,p+1)$ moving through this arena. This is not a random point; it is a traveler constrained to a specific path.

If we set $x=p$ and $y=p+1$, we can eliminate the parameter $p$ to reveal the path: $y=x+1$. This is a line with a slope of $1$ and a $y$-intercept of $1$.

As $p$ varies, the point slides along this line. However, it must stay within our region $E$. This means the point enters the region at some value $a$ and exits at some value $b$. Our goal is to find these boundaries, $a$ and $b$, and then compute the value of $b^2+b-a^2$.

Finding the Boundaries

To find the entry point $a$, we look for where our path $y=x+1$ hits the lower boundary, $x+y=3$. Substituting $y=x+1$ into the line equation, we get:

This simplifies to $2x=2$, or $x=1$. Thus, our lower bound is $a=1$. This is the moment our traveler enters the arena.

Now, for the exit point $b$. The traveler hits the upper boundary, the circle $x^2+y^2=9$. Substituting $y=x+1$ into the circle equation, we get:

Expanding this, we find $x^2+x^2+2x+1=9$, which simplifies to $2x^2+2x-8=0$. Dividing by $2$, we arrive at the elegant quadratic equation:

The value $b$ is the positive root of this equation.

Algebraic Elegance

Here is where you separate the calculator-dependent student from the true mathematician. You might be tempted to use the quadratic formula to find $b=\frac{-1+\sqrt{1+16}}{2}$. Please, resist!

Look at the expression we need to evaluate: $b^2+b-a^2$. Since $b$ is a root of $p^2+p-4=0$, we know by definition that:

This implies that $b^2+b=4$. We have found the value of the first part of our expression without ever calculating $b$!

Finally, we know $a=1$, so $a^2=1$. Substituting these into our target expression, we get $4-1=3$.

The beauty of this problem lies not in the arithmetic, but in the recognition of the algebraic structure. You have navigated the constraints, identified the path, and used the properties of the roots to find the final answer of 3.