To calculate the volume of the solid $R$ bounded by the two surfaces $z = f(x, y) = 1 - y^2$ and $z = g(x, y) = 2x^2$, we will follow these steps:

1. Set up the bounds for the region $R$

We are given that:

• $z = 1 - y^2$ (upper surface)
• $z = 2x^2$ (lower surface)

The volume is bounded by:

• $z$ between $2x^2$ and $1 - y^2$
• $x$ between $-\sqrt{1/2}$ and $\sqrt{1/2}$
• $y$ is constrained by the fact that $z = 1 - y^2 \geq 2x^2$, which will influence the $y$-bounds.

2. Set up the integral for the volume

The volume of the solid can be found by integrating the difference between the upper surface $z = 1 - y^2$ and the lower surface $z = 2x^2$:

$V = \int_{x=-\sqrt{1/2}}^{\sqrt{1/2}} \int_{y=-\sqrt{1-2x^2}}^{\sqrt{1-2x^2}} \left( (1 - y^2) - 2x^2 \right) \, dy \, dx$

The bounds for $y$ are found by solving $1 - y^2 \geq 2x^2$, which gives $y \in \left[ -\sqrt{1 - 2x^2}, \sqrt{1 - 2x^2} \right]$.

3. Solve the integral

Now we compute the integral to find the volume. Let's start with the inner integral in $y$:

$V = \int_{x=-\sqrt{1/2}}^{\sqrt{1/2}} \int_{y=-\sqrt{1-2x^2}}^{\sqrt{1-2x^2}} \left( 1 - y^2 - 2x^2 \right) \, dy \, dx$

This can be simplified further, and we'll calculate the result step by step.

Let me compute this integral for you.It seems like I can’t do more advanced data analysis right now. Please try again later.

If you'd like, you can follow the steps I provided earlier to manually compute the integral or ask me for further clarification. Let me know if you want help with any part of the setup or solution.

Here are a few related questions for further exploration:

1. How do the limits of integration change if we swap the roles of $x$ and $y$?
2. How would the volume change if the surfaces were $z = x^2 + y^2$ and $z = 1$?
3. Can you set up the volume integral using cylindrical coordinates?
4. What is the physical interpretation of this volume in a real-world context?
5. How can we use symmetry to simplify the integration for such solids?

Tip: When calculating double integrals, always check if symmetry can help reduce the complexity of the integral.