Question

The volume of the solid bounded by the surfaces \[ x = 1 - y^2, \quad x = y^2 - 1, \quad \text{and the planes} \quad z = 0 \quad \text{and} \quad z = 2 \] (round off to 2 decimal places) is ........

Hint:

When calculating the volume of a solid, use double integrals to compute the area of the base and multiply by the height.

Answer 1

Correct Answer: 5.3

Step 1: Find the bounds for the region in the \( xy \)-plane.
The equations \( x = 1 - y^2 \) and \( x = y^2 - 1 \) represent two parabolas. The volume is enclosed by the curves, and the bounds for \( y \) are determined by finding the intersection points of the curves: \[ 1 - y^2 = y^2 - 1 \quad \Rightarrow \quad 2y^2 = 2 \quad \Rightarrow \quad y = \pm 1. \] Thus, the region in the \( xy \)-plane is bounded by \( -1 \leq y \leq 1 \).
Step 2: Set up the double integral.
The volume is given by the double integral: \[ V = \int_{-1}^{1} \int_{y^2 - 1}^{1 - y^2} 2 \, dx \, dy. \]
Step 3: Compute the integral.
We compute the integral to get the volume: \[ V = 2 \times \text{(area of the region)}. \] After evaluating the integral, we find: \[ \boxed{8.00}. \]