Let R denote the region between the curves y = x2 and y = 2 − x2. Set up an integral used to find the volume of the solid whose base is the region R and whosecross-sections, perpendicular to the x-axis, are isosceles right triangles (i.e. equal base andheight) with the base lying on the region. You do not have to evaluate the integral.

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.

Let R denote the region between the curves y = x^2 and y = 2 − x^2. Set up an integral used to find the volume of the solid whose base is the region R and whose cross-sections, perpendicular to the x-axis, are isosceles right triangles (i.e. equal base and height) with the base lying on the region. You do not have to evaluate the integral.

Set up the integral to find the volume of a solid with a base between the curves y = x^2 and y = 2 - x^2, where cross-sections perpendicular to the x-axis are isosceles right triangles. This problem involves integral calculus, parabola equations, and calculating cross-sectional areas.

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