We are tasked with setting up an integral to find the volume of a solid with a base in the region between the curves $y = x^2$ and $y = 2 - x^2$. The cross-sections perpendicular to the x-axis are isosceles right triangles, where the base lies along the region between the curves.

Step 1: Determine the bounds of integration

The curves are given by:

1. $y = x^2$ (parabola opening upwards)
2. $y = 2 - x^2$ (parabola opening downwards)

To find the limits of integration, we set the two equations equal to find where the curves intersect: $x^2 = 2 - x^2$ $2x^2 = 2 \quad \Rightarrow \quad x^2 = 1 \quad \Rightarrow \quad x = \pm 1$ So, the curves intersect at $x = -1$ and $x = 1$. Therefore, the limits of integration will be from $x = -1$ to $x = 1$.

Step 2: Determine the length of the base of the triangles

At any point $x$, the base of the triangle is the vertical distance between the two curves, which is: $\text{Base of the triangle} = \left( 2 - x^2 \right) - \left( x^2 \right) = 2 - 2x^2$

Step 3: Area of the cross-sectional isosceles right triangle

Since the cross-sections are isosceles right triangles, the height of each triangle is equal to its base. For an isosceles right triangle, the area is given by: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ Since the height is equal to the base, the area simplifies to: $\text{Area} = \frac{1}{2} \times \text{base}^2 = \frac{1}{2} \times \left( 2 - 2x^2 \right)^2$

Step 4: Set up the volume integral

The volume is found by integrating the area of the cross-sections over the interval $x = -1$ to $x = 1$. Therefore, the volume $V$ is: $V = \int_{-1}^{1} \frac{1}{2} \left( 2 - 2x^2 \right)^2 \, dx$

This is the integral that represents the volume of the solid. You do not need to evaluate it.

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Would you like more details or have any questions?

Here are five related questions to further explore the topic:

1. How do you find the volume of a solid with a different type of cross-section?
2. What changes in the integral if the cross-section was an equilateral triangle instead of an isosceles right triangle?
3. How do you find the centroid of the region bounded by the curves $y = x^2$ and $y = 2 - x^2$?
4. What if the cross-sections were squares instead of triangles? How would the integral change?
5. How would the problem change if the cross-sections were perpendicular to the y-axis instead of the x-axis?

Tip: When setting up volume integrals with cross-sections, always pay close attention to how the cross-sectional area is expressed, as this directly influences the integral structure.