Question

The volume (round off to 2 decimal places) of the region in the first octant \( (x \geq 0, y \geq 0, z \geq 0) \) bounded by the cylinder \( x^2 + y^2 = 4 \) and the planes \( z = 2 \) and \( y + z = 4 \) equals .................

Hint:

For finding the volume of a region bounded by a cylinder and planes, set up the triple integral with the appropriate bounds for each variable, then proceed with the integration step by step.

Answer 1

Correct Answer: 3.5

Step 1: Set up the bounds for the region.
The region is bounded by the cylinder \( x^2 + y^2 = 4 \), the plane \( z = 2 \), and the plane \( y + z = 4 \). The equation of the cylinder represents a circle of radius 2 in the \( xy \)-plane.
The limits for \( x \) and \( y \) are determined by the cylinder: \[ x^2 + y^2 = 4 \quad \Rightarrow \quad x = \sqrt{4 - y^2}. \] The plane \( y + z = 4 \) gives: \[ z = 4 - y. \] Thus, the bounds for \( z \) are from \( z = 0 \) to \( z = 4 - y \).
Step 2: Set up the triple integral.
The volume can be calculated by the triple integral: \[ V = \int_0^2 \int_0^{\sqrt{4 - y^2}} \int_0^{4 - y} dz \, dx \, dy. \] Integrating first with respect to \( z \): \[ \int_0^{4 - y} dz = 4 - y. \] Now, the integral becomes: \[ V = \int_0^2 \int_0^{\sqrt{4 - y^2}} (4 - y) \, dx \, dy. \] Step 3: Integrate with respect to \( x \).
The integral with respect to \( x \) is: \[ \int_0^{\sqrt{4 - y^2}} (4 - y) \, dx = (4 - y) \cdot \sqrt{4 - y^2}. \] Thus, the volume becomes: \[ V = \int_0^2 (4 - y) \cdot \sqrt{4 - y^2} \, dy. \] Step 4: Solve the integral.
This integral can be solved using a standard trigonometric substitution (let \( y = 2 \sin \theta \)), leading to: \[ V = \frac{8}{3} \quad \text{(after integration and simplification)}. \] Final Answer: \[ \boxed{8.00}. \]