Solution Please??

I do not think this could be a legit question for the GRE because should require calculus notions which are beyond the GRE would ask to you

Also, it is not specified IF the region of the circle is not common to the region bounded by the parabola or is common I.E if the parabola touch the circle in one or two points on the circumference

From the equation of the circle we do know that \(x^2+y^2=2\) and 2 is the radius

\(r^2=2\)

\(r=\sqrt{ 2}\)

And the answer, theoretically, would be C

The correct answer is actually A : The area is 1.903333 approximately (rounding pi to 3.14).
This is out of the scope for the GRE because you need to know integrals, however, the rest of the math is pretty much GRE math.
Let's solve this beautiful problem. I recommend opening Desmos for this one.

We have a parabola X^2 and a circle with center (0,0) and radius (sqrt of 2). (sorry I don't know how to enter that character in my computer)

First let's limit our finding to just half of the circle because that makes our life easier, the total area for half of the circle is pi ((sqrt of two)^2*pi)/2))

Now let's use calculus to find the area under the curve from the parabola and the x-axis from -1 to 1, (this because the area is enclosed there and it will help us later) so the definite integral of x^2dx is equal to x^3/3, using the first and second fundamental theorem of calculus (F(x)=intf(t)dt with boundaries a and b is equal to F'(x)=F(a)-F(b) where a is the upper boundary and b the lower boundary) the area under the curve is equal to 2/3.

Now there sill some area we have to account for, the side areas which are equal let's label them x and the area we want let's label it y
We know that 2/3+2x+y=pi
We need to solve for x in order to solve the problem.
A very cunning way to solve it is to inscribe a square in the circle, notice that the are not covered by the square is equal to 2x, the area we are looking for!! (cool right)
So if we inscribe a square in the circle, it's diagonal is equal to the diameter of the circle which is 2*(sqrt of two), using the properties of 45 45 90, we find the length of the side by dividing by sqrt 2, giving us 2. The area of the square is 4.
If we susbstract the area of the circle and the area of the square we now that the area not covered by the square is= 2.28 which we can divide in 4 equal spaces= 0.57 which is equal to the 2x in our previous formula.

Now we can solve the problem
2/3+0.57+y=3.14 (pi)
y=1.9033333
The area under the parabola and the square.
Correct answer is A

Great EXPALANTION. Thank yo sir

However, too much for a GRE question as I said.

Many Thanks