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.