To find the volume of the solid bounded by the given surfaces, we can use the method of triple integration.

Step 1: Setting up the Triple Integral

First, let's visualize the region bounded by the given surfaces. The surfaces are symmetric about the y-axis, so we can focus on the region where x ≥ 0.

The region can be described as follows:
- The lower boundary is given by the curve x = 1 - y^2
- The upper boundary is given by the curve x = y^2 - 1
- The front boundary is the plane z = 0
- The back boundary is the plane z = 2

To set up the triple integral, we need to express the volume element as a function of x, y, and z.

The volume element, dV, can be expressed as dV = dx * dy * dz.

Step 2: Finding the Limits of Integration

To find the limits of integration, we need to determine the range of values for x, y, and z.

For x, we can see that it varies between the curves x = 1 - y^2 and x = y^2 - 1.

For y, it varies from the lower bound of -1 to the upper bound of 1.

For z, it varies from 0 to 2, as given.

Therefore, the limits of integration are:
- For x: 1 - y^2 ≤ x ≤ y^2 - 1
- For y: -1 ≤ y ≤ 1
- For z: 0 ≤ z ≤ 2

Step 3: Evaluating the Triple Integral

Now, we can set up and evaluate the triple integral using the given limits of integration:

V = ∫∫∫ dV
= ∫∫∫ dx * dy * dz
= ∫[-1,1] ∫[1-y^2,y^2-1] ∫[0,2] dx * dy * dz

Evaluating this triple integral will give us the volume of the solid bounded by the given surfaces.

Step 4: Calculating the Volume

Let's evaluate the triple integral using the given limits:

V = ∫[-1,1] ∫[1-y^2,y^2-1] ∫[0,2] 1 dx dy dz

Integrating with respect to x, we get:

V = ∫[-1,1] ∫[1-y^2,y^2-1] (x) dy dz

Evaluating the inner integral:

V = ∫[-1,1] [(x^2/2) |_1-y^2^(y^2-1)] dy dz
= ∫[-1,1] [(x^2/2) |_(1-y^2)^(y^2-1)] dy dz

Integrating with respect to y, we get:

V = ∫[-1,1] [(x^2/2) |_(1-y^2)^(y^2-1)] dz

Evaluating the inner integral:

V = ∫[-1,1] [(x^2/2) |_(1-y^2)^(y^2-1)]