The given problem involves solving an integral with respect to the equations:

1. \( x = y^2 \)
2. \( x = 1 - y^2 \)

We are likely looking for the area between these two curves. To find this area, the steps are as follows:

### 1. Set the two equations equal to find the points of intersection.
\[
y^2 = 1 - y^2
\]
\[
2y^2 = 1
\]
\[
y^2 = \frac{1}{2}
\]
\[
y = \pm \frac{1}{\sqrt{2}}
\]

So, the curves intersect at \( y = -\frac{1}{\sqrt{2}} \) and \( y = \frac{1}{\sqrt{2}} \).

### 2. Set up the integral.
The area between these curves is the difference between the upper function \( x = 1 - y^2 \) and the lower function \( x = y^2 \) over the range of integration \( y = -\frac{1}{\sqrt{2}} \) to \( y = \frac{1}{\sqrt{2}} \).

The area \( A \) is given by:
\[
A = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \left( (1 - y^2) - y^2 \right) \, dy
\]
\[
A = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (1 - 2y^2) \, dy
\]

### 3. Solve the integral.
Break it down into two parts:
\[
A = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} 1 \, dy - 2 \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} y^2 \, dy
\]

For the first integral:
\[
\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} 1 \, dy = \left[ y \right]_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2}} - \left( -\frac{1}{\sqrt{2}} \right) = \frac{2}{\sqrt{2}} = \sqrt{2}
\]

For the second integral:
\[
\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} = \frac{\left( \frac{1}{\sqrt{2}} \right)^3}{3} - \left( \frac{-\frac{1}{\sqrt{2}}}{3} \right) = \frac{2}{3} \cdot \frac{1}{2\sqrt{2}} = \frac{1}{3\sqrt{2}}
\]

Thus, the total area is:
\[
A = \sqrt{2} - 2 \times \frac{1}{3\sqrt{2}} = \sqrt{2} - \frac{2}{3\sqrt{2}} = \frac{3}{3\sqrt{2}} - \frac{2}{3\sqrt{2}} = \frac{1}{3\sqrt{2}}
\]

So, the area between the curves is:
\[
A = \frac{1}{\sqrt{2}} \approx 0.707
\]

Would you like further details, or do you have any questions?

### Related questions:
1. How do we compute areas between more complex curves?
2. What if the bounds of integration are not symmetric?
3. Can we use a different method, such as polar coordinates, for similar problems?
4. How do we find the intersection points for curves of higher degree?
5. How does this method apply to three-dimensional surfaces?

### Tip:
Always check the limits of integration when solving area problems, as curves may have different behaviors across intervals.

Learn how to calculate the area between the curves x = y² and x = 1 - y² using definite integrals. This problem covers key calculus concepts such as finding points of intersection, setting up integrals, and solving for the area between two curves. Suitable for Grades 11-12 and early college.

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