Question

The volume of the region \[ R = \left\{ (x, y, z) \in \mathbb{R}^3 : x + y + z \leq 3, y^2 \leq 4x, 0 \leq x \leq 1, y \geq 0, z \geq 0 \right\} \] is

Hint:

When finding the volume of a region defined by inequalities, set up the bounds for each variable and use a triple integral to compute the volume.

Answer 1

Correct Answer: 2.1 - 2.3

Step 1: Understanding the inequalities.
We are given the region \( R \) described by three inequalities involving \( x \), \( y \), and \( z \). The volume of the region can be found by setting up an appropriate triple integral.
Step 2: Setting up the integral.
To calculate the volume, we first solve for the bounds of the integration. From the inequality \( y^2 \leq 4x \), we get the bounds for \( y \) in terms of \( x \). The limits for \( x \) are from 0 to 1, as given. The upper bound for \( z \) is determined from \( x + y + z \leq 3 \).
Step 3: Computing the volume.
The volume integral can be expressed as: \[ V = \int_0^1 \int_0^{2\sqrt{x}} \int_0^{3 - x - y} dz \, dy \, dx. \] Evaluating this integral gives the volume as approximately between 2.1 and 2.3.

Step 4: Conclusion.
Thus, the volume of the region is approximately \( 2.1 \) to \( 2.3 \).