Calculate the volume of the solid R
 bounded by the two surfaces z=f(x,y)=1−y2
 and z=g(x,y)=2x2
.

generated image
The solid can be described as:

R={(x,y,z)|−12−−√≤x≤12−−√,
≤y≤
,
≤z≤
}
,

and its volume is .

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.

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.

This page provides a detailed solution to calculate the volume of a solid bounded by the surfaces z = 1 − y^2 and z = 2x^2 using double integration. Follow step-by-step instructions, including setting up bounds and solving the integral. Suitable for college-level calculus students.

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