Interpret the curves

We are given: $|y|=1-x^2$ and $x^2+y^2=1.$

Since $|y|=1-x^2$, this represents two parabolas: $y=1-x^2 \quad (y\ge 0),$ $y=x^2-1 \quad (y\le 0).$

The circle is the unit circle centered at the origin.

Find points of intersection

Because of symmetry about both axes, it is enough to work in the upper half-plane and then double.

In the upper half-plane, intersect $y=1-x^2$ with $x^2+y^2=1.$

Substitute $y=1-x^2$ into the circle: $x^2+(1-x^2)^2=1.$

Expanding, $x^2+1-2x^2+x^4=1$ $x^4-x^2=0$ $x^2(x^2-1)=0.$

Hence, $x=0,\ \pm 1.$

Corresponding upper-half points are:

• $x=0 \Rightarrow y=1$
• $x=\pm 1 \Rightarrow y=0$

Determine which curve is above the other

For $0\le x\le 1$:

• parabola: $y=1-x^2$
• circle: $y=\sqrt{1-x^2}$

Check at $x=\tfrac12$:

Evaluate the integrals

So, $\alpha=4\left(\int_0^1\sqrt{1-x^2}\,dx-\int_0^1(1-x^2)\,dx\right).$

Now, $\int_0^1\sqrt{1-x^2}\,dx=\frac{\pi}{4}$ because it is the area of a quarter of the unit circle.

Also, $\int_0^1(1-x^2)\,dx=\left[x-\frac{x^3}{3}\right]_0^1=1-\frac13=\frac23.$

Hence, $\alpha=4\left(\frac{\pi}{4}-\frac23\right)=\pi-\frac{8}{3}.$

Compute $9\alpha$

$9\alpha=9\left(\pi-\frac{8}{3}\right)=9\pi-24.$

Comparing with $9\alpha=\beta\pi+\gamma,$ we get $\beta=9,\qquad \gamma=-24.$

Find $|\beta-\gamma|$

$|\beta-\gamma|=|9-(-24)|=33.$

Option check

The correct option is: $\boxed{\text{C: }33}$