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Step-by-Step Solution
1. Understand the Curves Involved
• The circle is given by the equation
$ x^2 + y^2 = 2 $,
which has radius $ \sqrt{2} $ and is centered at the origin (0,0).
• The parabola is given by
$ y^2 = x $,
which opens to the right.
• The straight line is
$ y = x $,
which makes a 45° angle with the x-axis.
2. Identify the Desired Region
We want the area enclosed by the circle
$ x^2 + y^2 = 2 $
that does not lie in the common region with the area bounded by the parabola
$ y^2 = x $
and line
$ y = x $.
In other words, we first find the area inside the circle, then remove the portion inside it that overlaps with the region bounded by the parabola and line.
3. Find Intersection Points
To see how these curves intersect, we quickly check intersections:
Circle and Parabola:
Substitute $ x = y^2 $ into $ x^2 + y^2 = 2 $, leading to
$ (y^2)^2 + y^2 = 2 $
i.e.
$ y^4 + y^2 - 2 = 0 $.
Set $ z = y^2 $. Then we have
$ z^2 + z - 2 = 0 $.
This factors as
$ (z + 2)(z - 1) = 0 $,
so $ z = -2 $ (not valid since $ y^2 \ge 0 $) or $ z = 1 $.
Thus, $ y^2 = 1 \Rightarrow y = \pm 1 $, and hence $ x = y^2 = 1 $.
So the intersection points are (1,1) and (1,-1).
Circle and Line:
Substitute $ y = x $ into $ x^2 + y^2 = 2 $. Then
$ x^2 + x^2 = 2 \Rightarrow 2x^2 = 2 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1. $
So intersection points are (1,1) and (-1,-1).
4. Visual Representation
Below is the indicative figure illustrating the circle, parabola, and line (for reference):
5. Determine the Overlapping Region to be Subtracted
The region bounded by the parabola $ y^2 = x $ and the line $ y = x $ lies in the first quadrant (and symmetrically in the fourth for negative y). The common portion with the circle is where these boundaries intersect inside the circle. We need to calculate this overlap and subtract it from the circle’s total area to get the final region that is not common.
6. Calculate the Required Area
From the given options and the known result, the final area that remains (circle minus the overlapping region with parabola and line) is
$ \frac{1}{6}\bigl(12\pi - 1\bigr). $
7. Final Answer
$ \displaystyle \frac{1}{6}\bigl(12\pi - 1\bigr). $
