Question

The volume of the region 𝑅={(𝑥, 𝑦, 𝑧) ∈ ℝ × ℝ × ℝ ∶ 𝑥 ² + 𝑦 ² ≤ 4, 0 ≤ 𝑧 ≤ 4 − 𝑦} is

• A. 16𝜋-16
• B. 16𝜋
• C. 8𝜋
• D. 16𝜋 + 4

Answer 1

The Correct Option is B

To find the volume of the region \( R = \{(x, y, z) \in \mathbb{R} \times \mathbb{R} \times \mathbb{R} \mid x^2 + y^2 \leq 4, 0 \leq z \leq 4 - y\} \), we first need to understand the geometry of the region.

The condition \( x^2 + y^2 \leq 4 \) describes a cylinder with radius 2 centered along the z-axis. This is because the equation \((x^2 + y^2 = 4)\) outlines a circle of radius 2 in the xy-plane, and \( x^2 + y^2 \leq 4 \) represents the interior of this circle along the entire z-axis.

The region also has the constraint \( 0 \leq z \leq 4 - y \). This indicates that for each fixed value of \( y \), \( z \) ranges from 0 to \( 4 - y \). This constraint suggests a slant cutoff in the volume. At \( y = 0 \), the maximum \( z \) is 4, and at \( y = 4 \), \( z \) becomes zero.

Now, we'll calculate the volume using integration: 

1. Identify the bounds for integration:
   • For \( x \) and \( y \), we have \( x^2 + y^2 \leq 4 \), which is a circle of radius 2. We can switch to polar coordinates for simplicity: \( x = r \cos \theta \) and \( y = r \sin \theta \).
   • The bounds for \( r \) are 0 to 2, and for \( \theta \), they are \( 0 \) to \( 2\pi \).
2. Integrate with respect to \( z \):
   • \( z \) ranges from 0 to \( 4-y = 4 - r\sin\theta \).
   • The volume integral can be set up in cylindrical coordinates as: \(V = \int_{\theta = 0}^{2\pi} \int_{r = 0}^{2} \int_{z = 0}^{4 - r\sin\theta} r \, dz \, dr \, d\theta\)
3. Calculate the integral:
   • First integrate with respect to \( z \): \(\int_{z = 0}^{4 - r\sin\theta} 1 \, dz = (4 - r\sin\theta)\)
   • Next, integrate with respect to \( r \): \(\int_{r = 0}^{2} r(4 - r\sin\theta) \, dr = \int_{r = 0}^{2} (4r - r^2 \sin\theta) \, dr\)
   • Evaluate the integral: \(\left[ 2r^2 - \frac{r^3}{3}\sin\theta \right]_{0}^{2} = 8 - \frac{8}{3}\sin\theta\)
   • Finally, integrate with respect to \( \theta \): \(\int_{\theta = 0}^{2\pi} \left( 8 - \frac{8}{3}\sin\theta \right) \, d\theta = \left[ 8\theta + \frac{8}{3}\cos\theta \right]_{0}^{2\pi}\)
   • Evaluate this integral: \(= 8(2\pi) + \frac{8}{3}(\cos(2\pi) - \cos(0)) = 16\pi\)

Therefore, the volume of the given region \( R \) is 16𝜋, which matches the correct option.