Volume of the Solid Bounded by the Given Surfaces

To find the volume of the solid bounded by the surfaces x = -1, x = 1, y = -1, y = 1, z = 2, and y^2 + z^2 = 2 in R3, we will use the method of triple integration. This involves integrating the volume element over the given region of interest.

Region of Interest

The region of interest is the area enclosed by the given surfaces. In this case, it is a cylinder of radius 1, centered at the origin, and with a height of 2 units along the z-axis. The cylinder is bounded by the planes x = -1 and x = 1.

Triple Integral

To calculate the volume, we will set up a triple integral over the region of interest using appropriate limits of integration.

Step 1: Establishing the Limits of Integration

Since the cylinder is symmetric about the z-axis, we can integrate over a quarter of the cylinder and then multiply the result by 4 to obtain the total volume.

The limits of integration for x, y, and z are as follows:
- x: -1 to 1
- y: -√(2 - z^2) to √(2 - z^2)
- z: 0 to 2

Step 2: Setting Up the Triple Integral

The volume element in Cartesian coordinates is given by dV = dx * dy * dz. Hence, the triple integral to calculate the volume is:

V = 4 ∫[0 to 2] ∫[-√(2 - z^2) to √(2 - z^2)] ∫[-1 to 1] dx dy dz

Step 3: Evaluating the Triple Integral

Integrating the innermost integral with respect to x, we get:

V = 4 ∫[0 to 2] ∫[-√(2 - z^2) to √(2 - z^2)] (2) dy dz

Simplifying further, we have:

V = 8 ∫[0 to 2] √(2 - z^2) dy dz

Now, integrating with respect to y, we get:

V = 8 ∫[0 to 2] 2√(2 - z^2) dz

Step 4: Completing the Integration

To evaluate the remaining integral, we can use a trigonometric substitution. Let z = √2 sinθ, and dz = √2 cosθ dθ. The limits of integration also change accordingly.

V = 16 ∫[0 to π/2] 2√(2 - 2sin^2θ) cosθ dθ

Simplifying the expression inside the integral, we have:

V = 16 ∫[0 to π/2] 2 cosθ dθ

Evaluating the integral, we get:

V = 16 sinθ |[0 to π/2]

V = 16

Conclusion

The volume of the solid bounded by the given surfaces is 16 cubic units.