Find Area Under x=2Sin^2(y) & y=x^2 Graphs

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The discussion focuses on finding the area between the curves defined by the equations y = x^2 and x = 2 Sin^2(y). The user correctly identifies the need to set the equations equal to find points of intersection but struggles with simplifying the equation for y. The hint provided suggests using the identity cos(2y) = 1 - 2sin^2(y) to rearrange and express y as a function of x, which is crucial for determining the area between the curves.

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Find the area of the regions shown in the figures.

These are the graphs used :

y = x^2
x = 2 Sin ^2 (y)

I know that I need to set the two equations equal to each other in order to find the points of intersection, but I run into some trouble when trying to simplify it for y.

This is what I tried

Since y=x^2
Sqrt.(y)= x

So now I have
x= Sqrt.(y)
x= 2 Sin ^2 (y)

Sqrt.(y) - 2 Sin ^2 (y) = 0

So based on the graph that I am given in the book I can't really tell which is the top or right curve to subtract the bottom or left curve.

After this step I would square both sides and be left with (Sqrt.(y) -2 Sin(^2y)^2) = 0, but I am not sure if that is allowed. Not sure on how to proceed from here. I would appreciate some help. Thanks!
 
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Hint: [itex]cos(2y)=1-2sin^{2}(y)[/itex]. Rearrange and find y as a function of x.
 

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