Very strange shape

This is a more exact answer.

We want to integrate $y={ x }^{ 2 }$ and the limits would be the intersection to the $2$ graphs. Then we would minus the resultant value from the area of the semicircle of the equation ${ x }^{ 2 }+{ y }^{ 2 }=1$ to get the answer.

Intersection of the two graphs: $\sqrt { 1-{ x }^{ 2 } } ={ x }^{ 2 }$

$x=\pm\sqrt { \frac { 1 }{ 2 } \left[ \sqrt { 5 } -1 \right] }$

So integrating: $\int _{ -\sqrt { \frac { 1 }{ 2 } \left[ \sqrt { 5 } -1 \right] } }^{ \sqrt { \frac { 1 }{ 2 } \left[ \sqrt { 5 } -1 \right] } }{ { x }^{ 2 } } dx\approx 0.32391$

And therefore, the answer is: $\frac { 1 }{ 2 } \pi -0.32391=1.25=\alpha \quad (3s.f)$ $\left\lceil \alpha \right\rceil =\boxed{2}$

First of all, a mistake in wording.I hope you mean area of the intersection, not parabola.

More of Logic than of maths, if time is a concern

The area of circle is $\pi$ sq. units img

The shaded area is $\alpha$ . It's less than half area of circle and bigger than quarter area (note that $\angle BEC > 90^\circ$ )

Then we have $\displaystyle \dfrac{\pi}{4} < \alpha < \dfrac{\pi}{2}$

Thus $\lceil \alpha \rceil = 1$ or $\lceil \alpha \rceil = 2$

Answer is eventually $\boxed{2}$ again, in at the most $2$ attempts.